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Machines: ICSE Class 10 Physics solutions, notes
Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 3 Machines: ICSE Class 10 Physics (Concise/Selina) which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed.
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Summary
Machines are tools that help us do work more easily. They can make a heavy object feel lighter to lift, which means they act as a force multiplier. This is like having your strength increased by the machine. Machines can also change where you apply your force to a more convenient spot. Sometimes, they change the direction of the force, like pulling down on a rope to lift something up. Machines can also help achieve a gain in speed, where the object being moved (the load) travels a greater distance or faster than the part where you apply your force (the effort).
When we talk about machines, we use special terms. The object we want to move or the resistance we want to overcome is called the load. The force we apply to the machine is called the effort. Mechanical advantage, or MA, tells us how much the machine multiplies our effort. It is found by dividing the load by the effort. If MA is greater than one, the machine is a force multiplier. If MA is less than one, it gives a gain in speed. Velocity ratio, or VR, compares the distance the effort moves to the distance the load moves. Work input is the work done on the machine by the effort, and work output is the work done by the machine on the load. Efficiency measures how much of the work input is turned into useful work output. For a perfect, or ideal, machine, efficiency is 100%, and the MA equals the VR. However, real machines always lose some energy due to friction or the weight of their parts, so their efficiency is always less than 100%, and their MA is less than their VR. The relationship is MA equals efficiency multiplied by VR.
A lever is a simple machine made of a rigid bar that pivots around a fixed point called a fulcrum. Levers work on the principle of moments: the load multiplied by its distance from the fulcrum (load arm) equals the effort multiplied by its distance from the fulcrum (effort arm). The MA of a lever is the effort arm divided by the load arm. There are three classes of levers. Class I levers have the fulcrum between the effort and the load, like a seesaw. Their MA can be greater than, less than, or equal to one. Class II levers have the load between the fulcrum and the effort, like a wheelbarrow. Their MA is always greater than one, making them force multipliers. Class III levers have the effort between the fulcrum and the load, like a fishing rod. Their MA is always less than one, so they provide a gain in speed. Our bodies also use levers, such as when we nod our head (Class I) or lift our body on our toes (Class II).
Pulleys are wheels with a grooved rim for a rope or string. A single fixed pulley has its axle fixed in place. Its MA is ideally one, and its VR is one. It doesn’t multiply force but is useful for changing the direction of effort. A single movable pulley has an axle that moves with the load. Ideally, it has an MA of two and a VR of two, acting as a force multiplier. Often, pulleys are used in combinations. In a block and tackle system, which uses multiple pulleys in upper (fixed) and lower (movable) blocks, the MA and VR are ideally equal to the number of rope segments supporting the load. This system can lift very heavy loads with less effort. The weight of the movable pulleys and friction reduce the actual MA and efficiency of pulley systems.
Workbook solutions (Concise/Selina)
Exercise A
MCQs
1. Which of the following is not an example of a machine used as a force multiplier ?
(a) A jack used to lift a car.
(b) A single fixed pulley used to lift a bucket of water from a well.
(c) A bar used to lift a heavy stone.
(d) A wheel barrow used to carry a load.
Answer: (b) A single fixed pulley used to lift a bucket of water from a well.
2. For an ideal machine :
(a) Output energy = Input energy
(b) Output energy > Input energy
(c) Output energy < Input energy
(d) Work output > Work input
Answer: (a) Output energy = Input energy
3. Which of the following statements is incorrect?
(a) For a machine, velocity ratio does not change.
(b) Due to friction, mechanical advantage of a machine increases.
(c) For an ideal machine, work output is equal to the work input.
(d) For an ideal machine, mechanical advantage is numerically equal to the velocity ratio.
Answer: (b) Due to friction, mechanical advantage of a machine increases.
4. The unit of velocity ratio is :
(a) Nm
(b) Joule
(c) Newton
(d) No unit
Answer: (d) No unit
5. The correct relationship between mechanical advantage (M.A.), velocity ratio (V.R.) and efficiency (η) is :
(a) Μ.Α. = η × V.R.
(b) V.R. = η × Μ.Α.
(c) η = Μ.Α. × V.R.
(d) None of these
Answer: (a) Μ.Α. = η × V.R.
6. Which of the following relations is wrong?
(a) Work input = Effort x Displacement of effort
(b) Work output = Load × Displacement of load
(c) Efficiency = Work input / Work output
(d) Mechanical advantage = Velocity ratio x Efficiency
Answer: (c) Efficiency = Work input / Work output
7. If the effort needed is less than the load, then mechanical advantage of the machine is :
(a) Greater than 1
(b) Less than 1
(c) Equal to 1
(d) None of these
Answer: (a) Greater than 1
8. Select the incorrect statement :
(a) A machine always has efficiency less than 100%
(b) The mechanical advantage of a machine can be less than 1.
(c) A machine can have mechanical advantage greater than its velocity ratio.
(d) A machine can be used as a speed multiplier.
Answer: (c) A machine can have mechanical advantage greater than its velocity ratio.
9. The lever for which mechanical advantage is less than 1 has:
(a) Fulcrum at the mid point between load and effort.
(b) Load between effort and fulcrum.
(c) Effort between fulcrum and load.
(d) Load and effort acting at the same point.
Answer: (c) Effort between fulcrum and load.
10. Class II levers are designed to have :
(a) M.A. = V.R.
(b) M.A. > V.R.
(c) M.A. > 1
(d) M.A. < 1
Answer: (c) M.A. > 1
11. The lever shown in the figure is :
(a) Class I lever
(b) Class II lever
(c) Class III lever
(d) None of these
Answer: (a) Class I lever
12. Which class of levers have effort and load on the same side of fulcrum.
(a) Class I lever
(b) Class II lever
(c) Class III lever
(d) Class II or Class III lever
Answer: (d) Class II or Class III lever
Very Short Questions
1. (a) What do you understand by a simple machine?
Answer: (a) A machine is a device by which we can either obtain a gain in speed or overcome a large resistive force (or load) at some point by applying a small force (or effort) at a convenient point and in a desired direction.
(b) State the principle of an ideal machine.
Answer: (b) For an ideal machine, Output energy = Input energy. This means the useful work done by an ideal machine (i.e., output energy) can never be greater than the work done on the machine (i.e., input energy), and its efficiency is 100%.
2. Name a machine for each of the following use :
(a) to multiply the force,
Answer: (a) A jack is used to lift a car (acts as a force multiplier).
(b) to change the point of application of force,
Answer: (b) The rear wheel of a cycle is rotated by applying effort on the pedal attached to the toothed wheel which is joined to the rear wheel with the help of a chain. Thus, the point of application of effort is changed from the rear wheel to the pedal.
(c) to change the direction of force,
Answer: (c) A single fixed pulley is used to lift a bucket of water from a well by applying effort in the downward direction instead of upwards.
(d) to obtain the gain in speed.
Answer: (d) When a pair of scissors is used to cut a cloth, its blades move longer on cloth, while its handles move a little (acts as a speed multiplier).
3. How is mechanical advantage related to velocity ratio for (i) an ideal machine, (ii) a practical machine?
Answer: (i) For an ideal machine (free from friction, etc.), work output is equal to work input, so the efficiency is equal to 1 (or 100%) and the mechanical advantage is numerically equal to the velocity ratio (M.A. = V.R.).
(ii) In actual practice, mechanical advantage for all practical machines is always less than its velocity ratio (i.e., M.A. < V.R.) due to some loss of input energy against the force of friction etc. The relationship is M.A. = η × V.R., where η is the efficiency and is less than 1.
4. A machine works as a (i) force multiplier, (ii) speed multiplier. In each case state whether the velocity ratio is more than or less than 1.
Answer: (i) When a machine acts as a force multiplier, the effort needed is less than the load (M.A. > 1). If the velocity ratio of a machine is more than 1, i.e., the displacement of load is less than the displacement of effort, the machine acts as a force multiplier.
(ii) When a machine acts as a speed multiplier, it gives a gain in speed (M.A. < 1). A machine in which the displacement of load is more than the displacement of effort, will have velocity ratio less than 1 and such a machine gives gain in speed.
5. How is mechanical advantage related with velocity ratio for an actual machine? State whether the efficiency of such a machine is equal to 1, less than 1 or more than 1.
Answer: For an actual machine, the mechanical advantage (M.A.) is related to the velocity ratio (V.R.) by the formula M.A. = η × V.R., where η is the efficiency. In actual practice, mechanical advantage for all practical machines is always less than its velocity ratio (i.e., M.A. < V.R.) or output work is always less than input work, so the efficiency is less than 1 (i.e., η < 1) due to some loss of input energy against the force of friction etc. The efficiency of such a machine is less than 1.
6. Write down a relation expressing the mechanical advantage of a lever.
Answer: The mechanical advantage of a lever is equal to the ratio of the length of its effort arm to the length of its load arm. M.A. = (Effort arm FA) / (Load arm FB).
7. Give one example each of a class I lever where mechanical advantage is (a) more than 1, and (b) less than 1.
Answer: (a) For a class I lever where mechanical advantage is more than 1, an example is shears (used for cutting thin metal sheets) which have much longer handles (effort arm) as compared to its blades (load arm).
(b) For a class I lever where mechanical advantage is less than 1, an example is a pair of scissors whose blades are longer than its handles, used to cut a piece of cloth so that the blades move longer on the cloth when the handles are moved a little.
8. Classify the following into levers as class I, class II or class III:
(a) a door
Answer: (a) a door is an example of a class II lever.
(b) a catapult
Answer: (b) a catapult is an example of a class I lever.
(c) claw hammer
Answer: (c) a claw hammer is an example of a class I lever.
(d) a wheel barrow
Answer: (d) a wheel barrow is an example of a class II lever.
(e) a fishing rod.
Answer: (e) a fishing rod is an example of a class III lever.
(f) sugar tongs
Answer: (f) sugar tongs are an example of a class III lever.
9. What type of lever is formed by a human body while (a) raising a load on the palm, and (b) raising the weight of body on toes?
Answer: (a) Raising a load by forearm (load on the palm) is an example of a class III lever, where the elbow joint acts as fulcrum F at one end, biceps exerts the effort E in the middle and load L on the palm is at the other end.
(b) Raising the weight of the body on toes is an example of a class II lever, where the fulcrum F is at the toes at one end, the load L (i.e., weight of the body) is in the middle and effort E by muscles is at the other end.
10. Give an example of each class of lever in a human body.
Answer: (1) Class I lever in the human body: The action of nodding of head. In this action, the spine acts as the fulcrum F, load L is at its front part, while effort E is at its rear part.
(2) Class II lever in the human body: Raising the weight of the body on toes. The fulcrum F is at the toes at one end, the load L (i.e., weight of the body) is in the middle and effort E by muscles is at the other end.
(3) Class III lever in the human body: Raising a load by forearm. The elbow joint acts as fulcrum F at one end, biceps exerts the effort E in the middle and load L on the palm is at the other end.
11. Complete the following sentences :
(a) Mechanical advantage = ………. x velocity ratio
Answer: (a) Mechanical advantage = efficiency × velocity ratio
(b) In class II lever, effort arm is ………. than the load arm.
Answer: (b) In class II lever, effort arm is always longer than the load arm.
(c) A pair of scissors is a ………. multiplier.
Answer: (c) A pair of scissors, whose blades are longer than its handles, is used to obtain gain in speed, so it acts as a speed multiplier. (If handles are longer than blades, it can act as a force multiplier, e.g. shears). The question is ambiguous, but based on the example on page 6 for speed gain, it is a speed multiplier. The answer key on page 13 states “speed”.
Short Answer Type Questions
1. State four ways in which machines are useful to us.
Answer: The various functions of machines are useful to us in the following four ways:
(1) In lifting a heavy load by applying less effort, i.e., as a force multiplier.
(2) In changing the point of application of effort to a convenient point.
(3) In changing the direction of effort to a convenient direction.
(4) For obtaining a gain in speed (i.e., a greater movement of load by a smaller movement of effort).
2. What is the purpose of a jack in lifting a car by it?
Answer: A jack is used to lift a car by applying less effort; it acts as a force multiplier. The effort needed is much less than the load (the car’s weight), so the machine acts as a force multiplier.
3. What do you understand by an ideal machine? How does it differ from a practical machine?
Answer: An ideal machine is that in which there is no loss of energy in any manner. Here the work output is equal to the work input, i.e., the efficiency of an ideal machine is 100%.
It differs from a practical machine in that in an actual (practical) machine, the output energy is always less than the input energy i.e., there is always some loss of energy during its operation. The loss of energy in a machine is due to reasons such as: (1) the moving parts in it are neither weightless nor smooth (or frictionless), (2) the string in it (if any) is not perfectly elastic, and (3) its different parts are not perfectly rigid. Thus, the efficiency of a practical machine is always less than 100%.
4. Explain the term mechanical advantage. State its unit.
Answer: The ratio of load to effort is called the mechanical advantage of the machine, i.e., Mechanical advantage (M.A.) = Load (L) / Effort (E).
Since mechanical advantage is the ratio of two similar quantities (forces), so it has no unit.
5. Define the term velocity ratio. State its unit.
Answer: The ratio of the velocity of effort to the velocity of load is called the velocity ratio of machine, i.e., Velocity ratio (V.R.) = Velocity of effort / Velocity of load . It is also defined as the ratio of the displacement of effort to the displacement of load , i.e., V.R. = E / L.
Since velocity ratio is also the ratio of two similar quantities (distances), so it has no unit just like M.A.
6. Define the term efficiency of a machine. Give two reasons for a machine not to be 100% efficient?
Answer: Efficiency of a machine is the ratio of the work done on the load by the machine to the work done on the machine by the effort. In other words, efficiency is the ratio of work output to work input. It is denoted by the symbol η (eta).
Thus, Efficiency η = Work output/Work input
Two reasons for a machine not to be 100% efficient are:
(1) The moving parts in it are neither weightless nor smooth (or frictionless), leading to energy loss in overcoming friction.
(2) The string in it (if any) is not perfectly elastic, or its different parts are not perfectly rigid, leading to energy loss.
The most prominent loss in energy is in overcoming the force of friction between the moving parts of a machine.
7. When does a machine act as (a) a force multiplier, (b) a speed multiplier. Can a machine act as a force multiplier and a speed multiplier simultaneously?
Answer: (a) A machine acts as a force multiplier when the effort needed is less than the load, i.e., its mechanical advantage is greater than 1.
(b) A machine acts as a speed multiplier when it provides a gain in speed, i.e., a greater movement of load by a smaller movement of effort. In this case, its mechanical advantage is less than 1.
A machine cannot be used as a force multiplier as well as a speed multiplier simultaneously. For a machine used as a force multiplier, effort < load, while for a machine used to obtain gain in speed, effort > load.
8. (a) State the relationship between mechanical advantage, velocity ratio and efficiency.
Answer: (a) The relationship between mechanical advantage (M.A.), velocity ratio (V.R.) and efficiency (η) is M.A. = V.R. × η or Efficiency η = M.A. / V.R.
(b) Name the term that will not change for a machine of a given design.
Answer: (b) For a given design of a machine, its velocity ratio does not change.
9. State one reason why mechanical advantage is less than the velocity ratio for an actual machine.
Answer: In an actual machine, mechanical advantage is less than the velocity ratio because there is always some loss of input energy in overcoming the force of friction between the moving parts of the machine. This means the work output is always less than the work input, so efficiency is less than 1, and since M.A. = η × V.R., M.A. < V.R.
10. What is a lever? State its principle.
Answer: A lever is a rigid, straight (or bent) bar which is capable of turning about a fixed axis.
A lever works on the principle of moments according to which at the equilibrium position of the lever, moment of load about the fulcrum must be equal to the moment of effort about the fulcrum and the two moments must always be in opposite directions. That is, Load × load arm = Effort × effort arm.
11. Name the three classes of levers and state how are they distinguished. Give two examples of each class.
Answer: The three classes of levers are: Class I levers, Class II levers, and Class III levers. They are distinguished by the relative positions of effort, load and fulcrum.
(1) Class I levers: In this type of levers, the fulcrum F is in between the effort E and the load L.
Examples: A seesaw, a pair of scissors.
(2) Class II levers: In this type of levers, the fulcrum F and the effort E are at the two ends of the lever and the load L is somewhere in between the effort E and the fulcrum F. The load and effort are on the same side of the fulcrum but acting in opposite directions.
Examples: A nut cracker, a bottle opener.
(3) Class III levers: In this type of levers, the fulcrum F and the load L are at the two ends of the lever and the effort E is somewhere in between the fulcrum F and the load L. Effort and load are on the same side of the fulcrum but acting in opposite directions.
Examples: Sugar tongs, a knife.
12. Both a pair of scissors and a pair of pliers belong to the same class of levers. Name the class of lever. Which one has mechanical advantage less than 1?
Answer: Both a pair of scissors and a pair of pliers belong to Class I levers (fulcrum is between effort and load). A pair of scissors, whose blades are longer than its handles, used to cut a piece of cloth (or paper) so that the blades move longer on the cloth (or paper) when the handles are moved a little, has mechanical advantage less than 1 because its effort arm is shorter than its load arm, and it is used to obtain gain in speed.
13. Fig. 3.12 shows a lemon crusher.
(a) In the diagram, mark the position of the fulcrum F and the line of action of load L and effort E.
Answer: (a) In the diagram of a lemon crusher, the fulcrum F is at the hinge, the load L is the lemon being crushed between the arms, and the effort E is applied at the ends of the handles.
(b) Name the class of lever.
Answer: (b) A lemon crusher is an example of a class II lever.
14. The diagram below shows a rod lifting a stone.
(a) Mark position of fulcrum F and draw arrows to show the directions of load L and effort E.
Answer: (a) For a rod lifting a stone, if it’s used as a crowbar, the fulcrum F would be a point on the ground or a support near the stone. The load L is the stone, acting downwards. The effort E is applied at the other end of the rod, usually downwards.
(b) What class of lever is the rod?
Answer: (b) The rod used in this manner is a class I lever, as the fulcrum is between the load and the effort.
(c) Give one more example of the same class of lever stated in part (b).
Answer: (c) Another example of a class I lever is a pair of pliers or a seesaw.
15. State the kind of lever which always has mechanical advantage less than 1. Draw a labelled diagram of such a lever.
Answer: Class III levers always have mechanical advantage less than 1. This is because in class III levers, the effort arm is always smaller than the load arm.
16. Explain why mechanical advantage of class III lever is always less than 1.
Answer: In class III levers, the effort E is somewhere in between the fulcrum F and the load L. This means the effort arm (distance from fulcrum to effort) is always smaller than the load arm (distance from fulcrum to load). Since mechanical advantage of a lever is the ratio of effort arm to load arm (M.A. = Effort arm / Load arm), and effort arm < load arm, the M.A. is always < 1.
17. Class III levers have mechanical advantage less than 1. Why are they then used?
Answer: Class III levers have mechanical advantage less than 1, which means effort is more than the load. They are used because with levers of class III, we do not get gain in force, but we get gain in speed, i.e., a larger displacement of load is obtained by a smaller displacement of effort. For example, the blade of a knife moves longer by a small displacement of its handle.
18. State the class of levers and the relative positions of load (L), effort (E) and fulcrum (F) in (a) a bottle opener, and (b) sugar tongs.
Answer: (a) A bottle opener is a class II lever. The fulcrum F is at one end (the tip resting on the cap), the load L is the bottle cap being lifted (in the middle), and the effort E is applied at the other end of the handle.
(b) Sugar tongs are a class III lever. The fulcrum F is at the joined end, the effort E is applied in the middle of the arms, and the load L (sugar cube) is held at the tips (other end).
19. Indicate the positions of load L, effort E and fulcrum F in the forearm shown alongside in the figure given. Name the class of lever.
Answer: In the forearm raising a load kept on a palm:
The fulcrum F is the elbow joint.
The effort E is exerted by the biceps muscle, acting upwards, attached to the forearm near the elbow.
The load L is the weight held in the palm, acting downwards.
The effort E is between the fulcrum F and the load L.
This is an example of a class III lever.
Long Answer Questions
1. Derive a relationship between mechanical advantage, velocity ratio and efficiency of a machine.
Answer: Suppose a machine overcomes a load L by the application of an effort E, in time t. Let the displacement of effort be dE and the displacement of load be dL.
Work input = effort × displacement of effort = E × dE
Work output = load × displacement of load = L × dL
By definition, Efficiency η = (work output) / (work input).
From the above equations,
η = (L × dL) / (E × dE)
η = (L / E) × (dL / dE)
η = (L / E) × (1 / (dE / dL))
But L / E = Mechanical Advantage (M.A.) and dE / dL = Velocity Ratio (V.R.).
Therefore, Efficiency η = M.A. / V.R.
or M.A. = V.R. × η.
Thus, the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.
2. Explain why scissors for cutting cloth may have blades longer than the handles, but shears for cutting metals have short blades and long handles.
Answer: Scissors for cutting cloth may have blades longer than their handles (that is, the load arm is greater than the effort arm). Such levers are used to obtain a gain in speed because when the velocity ratio is less than 1, the displacement of the load is greater than the displacement of the effort. This means the blades move a longer distance on the cloth when the handles are moved only a little. In this case, the mechanical advantage (M.A.) is less than 1.
Shears used for cutting thin metal sheets have much longer handles compared to their blades (that is, the effort arm is greater than the load arm). Such a lever acts as a force multiplier, enabling us to overcome a large resistive force (the load) with only a small effort. Here, the mechanical advantage (M.A.) is greater than 1.
3. Fig. 3.14, shows a uniform metre rule of weight W supported on a fulcrum at the 60 cm mark by applying the effort E at the 90 cm mark.
(a) State with reason whether the weight W of the rule is greater than, less than or equal to the effort E.
Answer: (a) The weight W of the uniform metre rule acts at its centre, i.e., at the 50 cm mark. The fulcrum F is at the 60 cm mark. The effort E is applied at the 90 cm mark.
Load arm (for W) = distance from fulcrum to 50 cm mark = 60 cm – 50 cm = 10 cm.
Effort arm (for E) = distance from fulcrum to 90 cm mark = 90 cm – 60 cm = 30 cm.
By the principle of moments, W × Load arm = E × Effort arm
W × 10 = E × 30
W/E = 30/10 = 3
So, W = 3E.
Thus, the weight W of the rule is greater than the effort E.
Reason: The weight W of a uniform metre rule acts at the 50 cm mark. Since the distance of the weight of the rule from the fulcrum F (10 cm) is less than that of the effort E from the fulcrum F (30 cm), for the rule to be balanced, the weight W of the rule must be greater than the effort E.
(b) Find the mechanical advantage in an ideal case.
Answer: (b) In this arrangement, the effort E is overcoming the weight W of the rule. So, W is the load.
Mechanical Advantage (M.A.) = Load / Effort = W / E.
From part (a), W = 3E.
So, M.A. = 3E / E = 3.
Alternatively, M.A. = Effort arm / Load arm = 30 cm / 10 cm = 3.
The mechanical advantage is 3.
4. Which type of lever has mechanical advantage always more than 1? Give reason with one example. What change can be made in this lever to increase its mechanical advantage?
Answer: Class II levers have mechanical advantage always more than 1.
Reason: In this type of levers, the load L is somewhere in between the effort E and the fulcrum F. This means the effort arm is always longer than the load arm. Since M.A. = Effort arm / Load arm, and Effort arm > Load arm, M.A. is always greater than 1.
Example: A nut cracker. The fulcrum is at the hinge, the nut (load) is placed between the arms, and effort is applied at the ends of the handles. The effort arm is longer than the load arm.
To increase the mechanical advantage of a class II lever, one can either increase its effort arm (by applying effort further from the fulcrum) or decrease its load arm (by placing the load closer to the fulcrum).
5. Draw a diagram of a lever which is always used as a force multiplier. How is the effort arm related to the load arm in such a lever?
Answer: A lever which is always used as a force multiplier is a class II lever, as its M.A. is always > 1. In a class II lever, the effort arm is always longer than the load arm.
6. Explain why mechanical advantage of a class II lever is always more than 1.
Answer: In class II levers, the fulcrum F and the effort E are at the two ends of the lever and the load L is somewhere in between the effort E and the fulcrum F. The load and effort are on the same side of the fulcrum but acting in opposite directions. Here the effort arm (distance from fulcrum to effort) is always longer than the load arm (distance from fulcrum to load). Since mechanical advantage (M.A.) of a lever is given by M.A. = Effort arm / Load arm, and the effort arm is always longer than the load arm, the M.A. is always greater than 1.
7. Draw a labelled diagram of a class II lever. Give one example of such a lever.
Answer: A labelled diagram of a class II lever shows the fulcrum F at one end, the load L in between, and the effort E at the other end, with the effort arm being longer than the load arm. Example: A wheel barrow.
8. What is the use of a lever if its mechanical advantage is (a) more than 1, (b) equal to 1, and (c) less than 1.
Answer: (a) If the mechanical advantage of a lever is more than 1, it acts as a force multiplier. This means less effort is needed to overcome a large load. Example: A crowbar used to lift a heavy stone.
(b) If the mechanical advantage of a lever is equal to 1, the effort needed is equal to the load. Such a lever is generally used to change the direction of effort as there is no gain in force or speed. Example: A physical balance with both arms equal in length.
(c) If the mechanical advantage of a lever is less than 1, it is used to obtain a gain in speed. This means a larger displacement of load is obtained by a smaller displacement of effort, but the effort needed is more than the load. Example: A pair of scissors with blades longer than handles, used to cut cloth.
9. Draw a labelled sketch of a class III lever. Give one example of this kind of lever.
Answer: A labelled sketch of a class III lever shows the fulcrum F at one end, the effort E in between, and the load L at the other end, with the effort arm being shorter than the load arm. Example: A fishing rod.
10. Draw diagrams to illustrate the positions of fulcrum, load and effort, in each of the following:
(a) A seesaw
Answer: (a) A seesaw is a class I lever. The fulcrum (F) is in the middle, with load (L) (person on one side) and effort (E) (person on the other side) on either side of the fulcrum.
(b) A common balance
Answer: (b) A common balance (physical balance) is a class I lever. The fulcrum (F) is at the center support, the load (L) is on one pan, and the effort (E) (weights) is on the other pan.
(c) A nut cracker
Answer: (c) A nut cracker is a class II lever. The fulcrum (F) is at the hinge, the load (L) (nut) is placed between the arms, and the effort (E) is applied at the ends of the handles.
(d) Forceps.
Answer: (d) Forceps (like sugar tongs) are class III levers. The fulcrum (F) is at the joined end, the effort (E) is applied in the middle by squeezing, and the load (L) is held at the tips.
Numerical Questions
1. A crowbar of length 120 cm has its fulcrum situated at a distance of 20 cm from the load. Calculate the mechanical advantage of the crowbar.
Answer: Given:
Total length of crowbar = 120 cm
Distance of fulcrum from load (Load arm) = 20 cm
First, we find the length of the effort arm.
Effort arm = Total length of crowbar – Load arm
Effort arm = 120 cm – 20 cm = 100 cm
Now, we calculate the mechanical advantage (M.A.).
M.A. = Effort arm / Load arm
M.A. = 100 cm / 20 cm
=> M.A. = 5
The mechanical advantage of the crowbar is 5.
2. A pair of scissors has its blades 15 cm long, while its handles are 7.5 cm long. What is its mechanical advantage?
Answer: In a pair of scissors, the pivot is the fulcrum. The length of the blades is the load arm, and the length of the handles is the effort arm.
Given:
Length of blades (Load arm) = 15 cm
Length of handles (Effort arm) = 7.5 cm
Mechanical advantage (M.A.) = Effort arm / Load arm
M.A. = 7.5 cm / 15 cm
=> M.A. = 0.5
The mechanical advantage of the pair of scissors is 0.5.
3. A force of 5 kgf is required to cut a metal sheet. A pair of shears used for cutting the metal sheet has its blades 5 cm long, while its handles are 10 cm long. What effort is needed to cut the sheet?
Answer: Given:
Force required to cut the sheet (Load) = 5 kgf
Length of blades (Load arm) = 5 cm
Length of handles (Effort arm) = 10 cm
Using the principle of moments for a lever:
Load × Load arm = Effort × Effort arm
5 kgf × 5 cm = Effort × 10 cm
=> 25 kgf cm = Effort × 10 cm
=> Effort = 25 kgf cm / 10 cm
=> Effort = 2.5 kgf
The effort needed to cut the sheet is 2.5 kgf.
4. Fig. 3.15 below shows a lever in use.
(a) To which class of lever does it belong?
(b) If AB = 1 m, AF = 0.4 m, find its mechanical advantage.
(c) Calculate the value of E.
Answer: Given:
Load (L) = 15 kgf
Total length (AB) = 1 m = 100 cm
Distance AF = 0.4 m = 40 cm
(a) The fulcrum F is located between the load L (at A) and the effort E (at B). Therefore, it is a Class I lever.
(b) To find the mechanical advantage, we first determine the load arm and effort arm.
Load arm = AF = 40 cm
Effort arm = FB = AB – AF = 100 cm – 40 cm = 60 cm
Mechanical Advantage (M.A.) = Effort arm / Load arm
M.A. = 60 cm / 40 cm
=> M.A. = 1.5
(c) To calculate the effort E, we use the formula for mechanical advantage:
M.A. = Load / Effort
1.5 = 15 kgf / E
=> E = 15 kgf / 1.5
=> E = 10 kgf
The value of effort E is 10 kgf.
5. A man uses a crowbar of length 1.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance of 1 m from his hand.
(a) Draw a diagram of the arrangement showing the fulcrum (F), load (L) and effort (E) with their directions.
(b) State the kind of lever.
(c) Calculate: (i) load arm, (ii) effort arm, (iii) mechanical advantage, and (iv) the effort needed.
Answer:
(a)
(b) The fulcrum (F) is between the effort (E) and the load (L). This is a Class I lever.
(c) Given:
Total length = 1.5 m
Load (L) = 75 kgf
Distance from hand to edge (Effort arm) = 1 m
(i) Calculate the load arm:
Load arm = Total length – Effort arm
Load arm = 1.5 m – 1 m = 0.5 m
(ii) The effort arm is given as 1 m.
(iii) Calculate the mechanical advantage (M.A.):
M.A. = Effort arm / Load arm
M.A. = 1 m / 0.5 m
=> M.A. = 2
(iv) Calculate the effort needed (E):
M.A. = Load / Effort
2 = 75 kgf / E
=> E = 75 kgf / 2
=> E = 37.5 kgf
6. A pair of scissors is used to cut a piece of cloth by keeping it at a distance of 8.0 cm from its rivet and applying an effort of 10 kgf by fingers at a distance of 2.0 cm from the rivet.
(a) Find: (i) the mechanical advantage of the scissors, and (ii) the load offered by the cloth.
(b) How does the pair of scissors act: as a force multiplier or as a speed multiplier?
Answer: Given:
Distance of cloth from rivet (Load arm) = 8.0 cm
Distance of fingers from rivet (Effort arm) = 2.0 cm
Effort (E) = 10 kgf
(a)
(i) Mechanical Advantage (M.A.):
M.A. = Effort arm / Load arm
M.A. = 2.0 cm / 8.0 cm
=> M.A. = 0.25
(ii) Load offered by the cloth (L):
M.A. = Load / Effort
0.25 = L / 10 kgf
=> L = 0.25 × 10 kgf
=> L = 2.5 kgf
(b) Since the mechanical advantage (0.25) is less than 1, the pair of scissors acts as a speed multiplier.
7. A 4 m long rod of negligible weight is supported at a point 125 cm from its one end and a load of 18 kgf is suspended at a point 60 cm from the support on the shorter arm.
(a) If a weight W is placed at a distance of 250 cm from the support on the longer arm to balance the rod, find W.
(b) If a weight 5 kgf is kept to balance the rod, find its position.
(c) To which class of lever does it belong?
Answer: Given:
Rod length = 4 m = 400 cm
Support (fulcrum) is at 125 cm from one end.
Shorter arm length = 125 cm
Longer arm length = 400 cm – 125 cm = 275 cm
Load = 18 kgf at 60 cm from the support on the shorter arm.
This setup works on the principle of moments.
Anticlockwise moment (due to the load) = Load × distance
Anticlockwise moment = 18 kgf × 60 cm = 1080 kgf cm
(a) A weight W is placed at 250 cm from the support on the longer arm.
Clockwise moment = W × 250 cm
For balance, Clockwise moment = Anticlockwise moment
W × 250 cm = 1080 kgf cm
=> W = 1080 / 250
=> W = 4.32 kgf
(b) A weight of 5 kgf is used to balance the rod. Let its position be ‘x’ cm from the support.
Clockwise moment = 5 kgf × x cm
For balance, 5 × x = 1080
=> x = 1080 / 5
=> x = 216 cm (or 2.16 m) from the support on the longer arm.
(c) The fulcrum (support) is between the load and the effort (balancing weight). Therefore, it is a Class I lever.
8. A lever of length 9 cm has its load arm 5 cm long and the effort arm is 9 cm long.
(a) To which class does it belong?
(b) Draw a diagram of the lever showing the position of the fulcrum F and directions of both the load L and effort E.
(c) What is the mechanical advantage and velocity ratio if the efficiency is 100%?
(d) What will be the mechanical advantage and velocity ratio if the efficiency becomes 50%?
Answer: (a) Since the effort arm (9 cm) is the full length of the lever, the fulcrum must be at one end. The load arm is 5 cm, so the load is between the fulcrum and the effort. This is a Class II lever.
(b)
(c) When efficiency (η) = 100% (or 1):
Velocity Ratio (V.R.) = Effort arm / Load arm = 9 cm / 5 cm = 1.8
For an ideal machine (η = 100%), Mechanical Advantage (M.A.) = V.R.
So, M.A. = 1.8
(d) When efficiency (η) = 50% (or 0.5):
The Velocity Ratio (V.R.) of a lever is determined by its geometry and does not change with efficiency.
So, V.R. = 1.8
Mechanical Advantage (M.A.) = η × V.R.
M.A. = 0.5 × 1.8
=> M.A. = 0.9
9. Fig. 3.16 below shows a lever in use:
(a) To which class of levers does it belong?
(b) Without changing the dimensions of the lever, if the load is shifted towards the fulcrum what happens to the mechanical advantage of the lever?
Answer:
(a) The diagram shows the fulcrum (F) at one end, the load (L) in the middle, and the effort (E) at the other end. This arrangement is characteristic of a Class II lever.
(b) The formula for mechanical advantage is M.A. = Effort arm / Load arm.
If the load is shifted towards the fulcrum, the length of the ‘Load arm’ decreases. Since the ‘Effort arm’ remains unchanged, the value of the ratio (M.A.) will increase.
10. Fig. 3.17 below shows a wheel barrow of mass 15 kg carrying a load of 30 kgf with its centre of gravity at A. The points B and C are the centre of wheel and tip of the handle such that the horizontal distance AB = 20 cm and AC = 40 cm.
Find: (a) the load arm, (b) the effort arm, (c) the mechanical advantage, and (d) the minimum effort required to keep the leg just off the ground.
Answer: In a wheelbarrow, the wheel’s axle (B) is the fulcrum. The load acts at the center of gravity (A), and the effort is applied at the handles (C).
Total Load (L) = Weight of wheelbarrow + Weight of load
Total Load (L) = 15 kgf + 30 kgf = 45 kgf
(a) The load arm is the distance from the fulcrum (B) to the load (A).
Load arm = AB = 20 cm
(b) The effort arm is the distance from the fulcrum (B) to the effort (C). From the diagram, BC = AB + AC.
Effort arm = BC = 20 cm + 40 cm = 60 cm
(c) The mechanical advantage (M.A.) is the ratio of the effort arm to the load arm.
M.A. = Effort arm / Load arm
M.A. = 60 cm / 20 cm
=> M.A. = 3
(d) The minimum effort (E) required is calculated using the M.A. formula.
M.A. = Load / Effort
3 = 45 kgf / E
=> E = 45 kgf / 3
=> E = 15 kgf
11. Ramesh went for fishing with his fishing rod. Identify the class of lever for a fishing rod. If a fish exerts a load of 10 N and the distance between fulcrum and fish is 1.5 m, and the effort applied is 0.5 m from the fulcrum, what is the force required by the fisher to pull the fish?
Answer: A fishing rod is a Class III lever, where the fulcrum is at one end (held by one hand), the effort is applied in between, and the load (the fish) is at the other end.
Given:
Load (L) = 10 N
Distance from fulcrum to fish (Load arm) = 1.5 m
Distance from fulcrum to effort (Effort arm) = 0.5 m
Using the principle of moments:
Effort × Effort arm = Load × Load arm
E × 0.5 m = 10 N × 1.5 m
=> E × 0.5 = 15 N m
=> E = 15 / 0.5
=> E = 30 N
The force required by the fisher is 30 N.
12. Fig. 3.18 below shows the use of a lever.
(a) State the principle of moments as applied to the above lever.
(b) To which class of lever does it belong? Give an example of this class of lever.
(c) If FA = 10 cm, AB = 490 cm, calculate: (i) the mechanical advantage, and (ii) the minimum effort required to lift the load (= 50 N).
Answer: (a) The principle of moments states that for the lever to be in equilibrium, the moment of the effort about the fulcrum F (Effort × FA) must be equal to the moment of the load about the fulcrum F (Load × FB). Since they cause rotation in opposite directions, the anticlockwise moment equals the clockwise moment.
(b) The diagram shows the effort (E) is applied between the fulcrum (F) and the load (L). This is a Class III lever. An example is a pair of sugar tongs or a fishing rod.
(c) Given:
Load (L) = 50 N
Effort arm = FA = 10 cm
Load arm = FB = FA + AB = 10 cm + 490 cm = 500 cm
(i) Mechanical Advantage (M.A.):
M.A. = Effort arm / Load arm
M.A. = 10 cm / 500 cm
=> M.A. = 1/50 or 0.02
(ii) Minimum effort (E) required:
M.A. = Load / Effort
0.02 = 50 N / E
=> E = 50 N / 0.02
=> E = 2500 N
13. A fire tongs has its arms 20 cm long. It is used to lift a coal of weight 1.5 kgf by applying an effort at a distance of 15 cm from the fulcrum. Find: (i) the mechanical advantage of the fire tongs, and (ii) the effort needed.
Answer: A fire tongs is a Class III lever. The fulcrum is the hinge.
Given:
Load (L) = 1.5 kgf
Length of arms (Load arm) = 20 cm
Distance of effort from fulcrum (Effort arm) = 15 cm
(i) Mechanical Advantage (M.A.):
M.A. = Effort arm / Load arm
M.A. = 15 cm / 20 cm
=> M.A. = 0.75
(ii) Effort needed (E):
M.A. = Load / Effort
0.75 = 1.5 kgf / E
=> E = 1.5 kgf / 0.75
=> E = 2 kgf
Exercise B
MCQs
1. The pulley system is used in a machine :
(a) for gain in energy
(b) as a force multiplier
(c) as a force multiplier and for a gain in energy
(d) none of these
Answer: (b) as a force multiplier
2. A single fixed pulley is used because it :
(a) has mechanical advantage greater than 1
(b) has velocity ratio less than 1
(c) gives 100% efficiency
(d) helps to apply the effort in a convenient direction
Answer: (d) helps to apply the effort in a convenient direction
3. The figure given below shows a fixed pulley used by a boy to lift a load of 200 N through a vertical height of 5 m in 10 s. The velocity ratio of the pulley is :
(a) 0
(b) 1
(c) 2
(d) none
Answer: (b) 1
4. The mechanical advantage of an ideal single movable pulley is :
(a) 1
(b) 2
(c) less than 2
(d) less than 1
Answer: (b) 2
5. Which of the following statements is incorrect ?
(a) The velocity ratio of a single fixed pulley is always 1.
(b) The velocity ratio of a single movable pulley is always 2.
(c) The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2″.
(d) The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load.
Answer: (a) The velocity ratio of a single fixed pulley is always 1.
6. If there are n movable pulleys with one fixed pulley, the mechanical advantage is given by :
(a) 2^n
(b) 2n + 1
(c) 2n – 1
(d) 2n + 2
Answer: (a) 2^n
7. Which of the following differences between a single fixed pulley and a single movable pulley is incorrect?
The differences are listed as pairs:
| Single Fixed Pulley | Single Movable Pulley |
|---|---|
| It is fixed to a rigid support. | It is not fixed to a rigid support. |
| The load moves in a direction opposite to the effort. | The load moves in the direction of the effort. |
| Its ideal mechanical advantage is 1. | Its ideal mechanical advantage is 2. |
| Its ideal velocity ratio is 1. | Its ideal velocity ratio is 2. |
| The axis of rotation does not move in space. | The axis of rotation also moves. |
(a) (5)
(b) (2) and (5)
(c) (3) and (5)
(d) (3) and (4)
Answer: (d) (3) and (4)
8. The relationship between mechanical advantage (M.A.) and velocity ratio (V.R.) for a practical machine is :
(a) M.A. = V.R.
(b) M.A. > V.R.
(c) M.A. < V.R.
(d) None of these
Answer: (c) M.A. < V.R.
9. A boy uses a single fixed pulley to lift a load of 50 kgf to some height. Another boy uses a single movable pulley to lift the same load to the same height. The ratio of the two efforts is :
(a) 1:2
(b) 2:1
(c) 1:1
(d) None of these
Answer: (b) 2:1
10. How many strands are supporting the load in case of block and tackle system of 5 pulleys ?
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (b) 5
11. How many strands are supporting the load in case of block and tackle system of 4 pulleys when effort is in the downward direction ?
(a) 5
(b) 4
(c) 3
(d) 6
Answer: (b) 4
12. For a system of n pulleys, the weight of the lower block alongwith pulleys is w, then its efficiency is given by :
(a) η = 1 – w/nE
(b) η = 1 + w/nE
(c) η = w/nE
(d) η = n – w/E
Answer: (a) η = 1 – w/nE
13. Assertion (A): The mechanical advantage of class III levers is always less than 1.
Reason (R): This is because the effort arm is always greater than the load arm.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
Very Short Questions
1. What is the ideal mechanical advantage of a single fixed pulley ? Can it be used as a force multiplier ?
Answer: In an ideal case for a single fixed pulley, the mechanical advantage is 1. In this arrangement, there is no gain in mechanical advantage, so it is not used as a force multiplier.
2. What is the velocity ratio of a single fixed pulley ?
Answer: The velocity ratio of a single fixed pulley is 1.
3. In a single fixed pulley, if the effort moves by a distance x downwards, by what height is the load raised upwards ?
Answer: If the point of application of effort E moves a distance x downwards, the load L also moves the same distance x upwards.
4. Name a machine which is used to :
Answer: (a) multiply force: A single movable pulley acts as a force multiplier. A lever of Class II also acts as a force multiplier.
(b) multiply speed: A lever of Class III is used to obtain gain in speed. A pair of scissors whose blades are longer than its handles is used to obtain gain in speed.
(c) change the direction of force applied: A single fixed pulley is used to change the direction of effort to a convenient direction.
5. State whether the following statements are true or false by writing T or F.
Answer: (a) The velocity ratio of a single fixed pulley is always more than 1. F
(b) The velocity ratio of a single movable pulley is always 2. T
(c) The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2ⁿ. T
(d) The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load. T
Short Questions
1. What is a fixed pulley ? State its one use.
Answer: A pulley which has its axis of rotation stationary in position, is called a fixed pulley. A fixed pulley is used only to change the direction of effort to a convenient direction, for example, to lift a small load such as a water bucket from a well.
2. Name the pulley which has no gain in mechanical advantage. Explain, why is such a pulley then used?
Answer: A single fixed pulley has no gain in mechanical advantage; in an ideal case, its mechanical advantage is 1. Such a pulley is used because it helps to change the direction of application of effort to a convenient direction. For instance, it is difficult to apply effort upwards to lift a load up directly, but it becomes easier with the help of a fixed pulley, because effort can now be applied in a downward direction to raise the load up. Further, to apply the effort downwards, one can conveniently make use of one’s own weight also for effort.
3. What is a single movable pulley ? What is its mechanical advantage in the ideal case ?
Answer: A pulley whose axis of rotation is movable (i.e., not fixed in position) is called a movable pulley. In the ideal case, neglecting friction and the weight of the pulley and string, the mechanical advantage of a single movable pulley is 2.
4. Give two reasons why the efficiency of a single movable pulley is not 100%.
Answer: The efficiency of an actual single movable pulley is not 100% (i.e., it is less than 2) due to:
(i) friction in the pulley bearings or at the axle, and
(ii) the weight of the pulley and string itself, which adds to the load to be lifted.
5. What is the velocity ratio of a single movable pulley ? How does the friction in the pulley bearing affect it?
Answer: The velocity ratio of a single movable pulley is 2. Friction in the pulley bearing reduces the mechanical advantage of the pulley, but for a given design, its velocity ratio does not change. Consequently, the efficiency (which is MA/VR) decreases due to friction.
6. In a single movable pulley, if the effort moves by a distance x upwards, by what height is the load raised?
Answer: In a single movable pulley, if the effort E moves a distance x upwards, the load L is raised up through a distance x/2.
7. Give reasons for the following :
(a) In a single movable pulley, the velocity ratio is always more than the mechanical advantage.
(b) The efficiency of a movable pulley is always less than 100%.
(c) In case of a block and tackle system, the mechanical advantage increases with the increase in the number of pulleys.
(d) The lower block of a block and tackle pulley system must be of negligible weight.
Answer: (a) This is because for an actual single movable pulley, due to friction in the pulley bearings or at the axle, and the weight of the pulley and string, the effort needed to lift a load L will be more than L/2, so the mechanical advantage will be less than 2. Since the velocity ratio remains 2, the velocity ratio (2) is always more than the actual mechanical advantage (which is < 2).
(b) This is because in an actual single movable pulley, there is energy loss due to (i) friction in the pulley bearings or at the axle, and (ii) the work done in lifting the weight of the pulley and string itself. Therefore, the work output is always less than the work input, making the efficiency less than 1 or 100%.
(c) This is because in a block and tackle system with ‘n’ total number of pulleys, the load is supported by ‘n’ segments of the string. In an ideal case, the mechanical advantage is equal to ‘n’. Thus, as ‘n’ increases, the mechanical advantage increases.
(d) This is because the weight of the lower block along with the pulleys in it (w) adds to the load to be overcome by the effort, reducing the mechanical advantage (M.A. = n – w/E). For greater efficiency, the pulleys in the lower block should be as light as possible.
Long Questions
1. Name the type of a single pulley that has an ideal mechanical advantage equal to 2. Draw a labelled diagram of the pulley mentioned by you.
Answer: The type of a single pulley that has an ideal mechanical advantage equal to 2 is a single movable pulley.
2. In which direction does the force need to be applied when a single pulley is used with a mechanical advantage greater than 1 ? How can you change the direction of force applied without altering its mechanical advantage ? Draw a labelled diagram of the system.
Answer: When a single movable pulley is used (which has a mechanical advantage greater than 1, ideally 2), the effort is to be applied in an upward direction.
The direction of force applied can be changed to a convenient downward direction by using the movable pulley along with a fixed pulley. In this arrangement, the mechanical advantage and velocity ratio remain the same as that of a single movable pulley (i.e., in an ideal case, M.A. = 2, V.R. = 2).
3. Draw a labelled diagram of an arrangement of two pulleys, one fixed and other movable. In the diagram, mark the directions of all forces acting on it. What is the ideal mechanical advantage of the system? How can it be achieved ?
Answer: An arrangement of two pulleys, one fixed and one movable, is used to lift a load where the effort is applied in a convenient downward direction.
In this system, the load L is attached to the axle of the movable pulley. The effort E is applied at the free end of the string passing over the fixed pulley. The tension T acts upwards on the load from the two segments of string around the movable pulley, and T also equals the effort E (if the fixed pulley is ideal). The ideal mechanical advantage of this system is 2.
This ideal mechanical advantage can be achieved if we neglect (1) the friction in the pulley bearings or at the axle, and (2) the weight of the movable pulley and string.
4. The diagram alongside shows a pulley arrangement.
(a) Name the pulleys A and B.
(b) In the diagram, mark the direction of tension on each strand of string.
(c) What is the purpose of the pulley B ?
(d) If the tension is T, deduce the relation between (i) T and E, and (ii) E and L.
(e) What is the velocity ratio of the arrangement ?
(f) Assuming that the efficiency of the system is 100%, what is the mechanical advantage ?
Answer: (a) Pulley A is a movable pulley. Pulley B is a fixed pulley.
(b) Tension on the strands supporting pulley A acts upwards. Tension on the strand going from pulley A over pulley B to the point of effort E is consistent throughout that part of the string.
(c) The purpose of the fixed pulley B is to change the direction of the applied effort E to a convenient downward direction.
(d) (i) Assuming T is the tension in the string segment where effort E is applied, and this string passes over the fixed pulley B, then E = T (neglecting friction in pulley B).
(ii) The movable pulley A is supported by two segments of the string, each with tension T (assuming the same string passes around A and then over B). So, the load L supported by pulley A is L = 2T. Since E = T, then L = 2E, or E = L/2.
(e) The velocity ratio of this arrangement (one movable and one fixed pulley) is 2.
(f) If the efficiency of the system is 100%, the mechanical advantage is equal to the velocity ratio. Therefore, the mechanical advantage is 2.
5. State four differences between a single fixed pulley and a single movable pulley.
Answer:
| Single Fixed Pulley | Single Movable Pulley |
| 1. It is fixed to a rigid support. | 1. It is not fixed to a rigid support. |
| 2. Its ideal mechanical advantage is 1. | 2. Its ideal mechanical advantage is 2. |
| 3. It is used to change the direction of effort. | 3. It is used as a force multiplier. |
| 4. The axis of rotation does not move in space. | 4. The axis of rotation also moves. |
6. The diagram alongside shows an arrangement of three pulleys A, B and C. The load is marked as L and the effort as E.
(a) Name the pulleys A, B and C.
(b) Mark in the diagram the directions of load (L), effort (E) and tension T₁ and T₂ in the two strings.
(c) How are the magnitudes of L and E related to the tension T₁?
(d) Calculate mechanical advantage and velocity ratio of the arrangement.
(e) What assumptions have you made in parts (c) and (d) ?
Answer: (a) Pulley A is a movable pulley. Pulley B is a movable pulley. Pulley C is a fixed pulley.
(b) Load L acts downwards. Effort E acts downwards (as implied by the fixed pulley C for convenience). Tension T₂ supports pulley A (and thus load L). Tension T₁ supports pulley B (which in turn supports pulley A).
(c) Assuming T₁ is the tension in the string where effort E is applied (so E = T₁ after passing over fixed pulley C).
Pulley B is supported by two segments of string with tension T₁. So, the force exerted downwards by pulley B (which is the tension T₂ in the string supporting pulley A) is T₂ = 2T₁.
Pulley A is supported by two segments of string with tension T₂. So, the load L = 2T₂.
Substituting T₂ = 2T₁, we get L = 2(2T₁) = 4T₁.
Since E = T₁, then L = 4E.
(d) Mechanical Advantage (M.A.) = L/E = 4E/E = 4.
This system has two movable pulleys (A and B) and one fixed pulley (C). The velocity ratio for ‘n’ movable pulleys used with one fixed pulley is 2ⁿ. Here n=2 (movable pulleys A and B).
So, Velocity Ratio (V.R.) = 2² = 4.
(e) The assumptions made are that the system is ideal, which means:
- The pulleys (A, B, and C) are weightless.
- There is no friction in the bearings of the pulleys or at their axles.
- The strings are inextensible and of negligible mass.
7. Draw a diagram of a combination of three movable pulleys and one fixed pulley to lift up a load. In the diagram, show the directions of load, effort and tension in each strand. Find:
(i) the mechanical advantage,
(ii) the velocity ratio, and
(iii) the efficiency of the combination, in the ideal situation.
Answer:
In this diagram, load L acts downwards. Effort E acts downwards. Tensions T₁, T₂, T₃ are shown in the respective strings.
(i) In the ideal situation, for a system of ‘n’ movable pulleys with one fixed pulley, the mechanical advantage M.A. = 2ⁿ. With three movable pulleys (n=3), M.A. = 2³ = 8.
(ii) The velocity ratio V.R. = 2ⁿ. With three movable pulleys (n=3), V.R. = 2³ = 8.
(iii) In the ideal situation, the efficiency η = M.A. / V.R. = 8 / 8 = 1, or 100%.
8. Draw a diagram of a block and tackle system of pulleys having a velocity ratio of 5. In your diagram indicate clearly the points of application and the directions of the load L and effort E. Also mark the tension T in each strand.
Answer: A block and tackle system with a velocity ratio of 5 will have a total of 5 pulleys (n=5). One common arrangement is to have 3 pulleys in the upper fixed block and 2 pulleys in the lower movable block, with the string attached to the hook of the lower block for downward effort application. Or, 3 pulleys in upper block and 2 in lower block, string attached to upper block. For VR=5, if effort is downwards, the string is usually attached to the lower block if upper block has more pulleys, or upper block if number of pulleys is same or lower block has one less. Fig 3.24 shows 5 pulleys where string is attached to lower block.
In the diagram:
- Load L is shown acting downwards, attached to the lower movable block.
- Effort E is shown acting downwards on the free end of the string.
- Tension T is marked on each of the 5 segments of the string supporting the load L.
Numericals
1. A woman draws water from a well using a fixed pulley. The mass of bucket and water together is 6 kg. The force applied by the woman is 70 N. Calculate the mechanical advantage. (Take g = 10 m s⁻²)
Answer: Given:
- Mass to be lifted = 6 kg
- Effort applied (E) = 70 N
- Acceleration due to gravity (g) = 10 m s⁻²
To find:
- Mechanical advantage (M.A.)
Process:
First, calculate the load (L) in newtons.
Load (L) = mass × g
=> L = 6 kg × 10 m s⁻²
=> L = 60 N
Now, calculate the mechanical advantage using the formula:
M.A. = Load (L) / Effort (E)
=> M.A. = 60 N / 70 N
=> M.A. = 0.857
The mechanical advantage is approximately 0.857.
2. A fixed pulley is driven by a 100 kg mass falling at a rate of 8.0 m in 4.0 s. It lifts a load of 75.0 kgf. Calculate :
(a) the power input to the pulley taking the force of gravity on 1 kg as 10 N.
(b) the efficiency of the pulley, and
(c) the height to which the load is raised in 4.0 s.
Answer: Given:
- Driving mass = 100 kg
- Distance moved by effort (dₑ) = 8.0 m
- Time (t) = 4.0 s
- Load (L) = 75.0 kgf
- Force of gravity = 10 N per kg
(a) Power input to the pulley:
First, calculate the effort (E). The effort is the force exerted by the falling mass.
Effort (E) = Driving mass × Force of gravity per kg
=> E = 100 kg × 10 N/kg
=> E = 1000 N
Next, calculate the work input.
Work input = Effort (E) × distance moved by effort (dₑ)
=> Work input = 1000 N × 8.0 m
=> Work input = 8000 J
Now, calculate the power input.
Power input = Work input / Time (t)
=> Power input = 8000 J / 4.0 s
=> Power input = 2000 W
(c) Height to which the load is raised:
For a single fixed pulley, the distance the load is raised is equal to the distance the effort moves.
Height of load (dₗ) = Distance of effort (dₑ)
=> dₗ = 8.0 m
(b) Efficiency of the pulley:
First, convert the load from kgf to N.
Load (L) = 75.0 kgf × 10 N/kgf
=> L = 750 N
Next, calculate the work output.
Work output = Load (L) × distance moved by load (dₗ)
=> Work output = 750 N × 8.0 m
=> Work output = 6000 J
Now, calculate the efficiency (η).
Efficiency (η) = (Work output / Work input) × 100%
=> η = (6000 J / 8000 J) × 100%
=> η = 0.75 × 100%
=> η = 75%
3. A single fixed pulley and a movable pulley both are separately used to lift a load of 50 kgf to the same height. Compare the efforts applied in an ideal situation.
Answer: Given:
- Load (L) = 50 kgf
- The situation is ideal (efficiency is 100%).
Case 1: Single Fixed Pulley
In an ideal single fixed pulley, the Mechanical Advantage (M.A.) is 1.
M.A. = Load (L) / Effort (E₁)
=> 1 = 50 kgf / E₁
=> E₁ = 50 kgf
Case 2: Single Movable Pulley
In an ideal single movable pulley, the Mechanical Advantage (M.A.) is 2.
M.A. = Load (L) / Effort (E₂)
=> 2 = 50 kgf / E₂
=> E₂ = 50 kgf / 2
=> E₂ = 25 kgf
Comparison: The effort required using the single fixed pulley (E₁) is 50 kgf, while the effort required using the single movable pulley (E₂) is 25 kgf. Therefore, the effort required for the movable pulley is half the effort required for the fixed pulley. The ratio of efforts (Fixed : Movable) is 50:25 or 2:1.
4. In a block and tackle system consisting of 3 pulleys, a load of 75 kgf is raised with an effort of 25 kgf. Find : (i) the mechanical advantage, (ii) the velocity ratio, and (iii) the efficiency.
Answer: Given:
- Number of pulleys (n) = 3
- Load (L) = 75 kgf
- Effort (E) = 25 kgf
(i) Mechanical advantage (M.A.):
M.A. = Load (L) / Effort (E)
=> M.A. = 75 kgf / 25 kgf
=> M.A. = 3
(ii) Velocity ratio (V.R.):
For a block and tackle system, the velocity ratio is equal to the total number of pulleys.
V.R. = n
=> V.R. = 3
(iii) Efficiency (η):
Efficiency (η) = M.A. / V.R.
=> η = 3 / 3
=> η = 1
In percentage, η = 1 × 100% = 100%.
5. A block and tackle system has 5 pulleys. If an effort of 1000 N is needed in downward direction to raise a load of 4500 N, calculate :
(a) the mechanical advantage,
(b) the velocity ratio, and
(c) the efficiency of the system.
Answer: Given:
- Number of pulleys (n) = 5
- Effort (E) = 1000 N
- Load (L) = 4500 N
(a) The mechanical advantage (M.A.):
M.A. = Load (L) / Effort (E)
=> M.A. = 4500 N / 1000 N
=> M.A. = 4.5
(b) The velocity ratio (V.R.):
V.R. = Number of pulleys (n)
=> V.R. = 5
(c) The efficiency of the system (η):
Efficiency (η) = (M.A. / V.R.) × 100%
=> η = (4.5 / 5) × 100%
=> η = 0.9 × 100%
=> η = 90%
6. In Fig. 3.31, draw a tackle to lift the load by applying the force in downward direction.
(a) Mark in the diagram the direction of load L and effort E.
(b) If the load is raised by 1 m, through what distance will the effort move ?
(c) State the number of strands of tackle supporting the load.
(d) What is the mechanical advantage of the system?
Answer:
(a) Direction of Load L and Effort E:
- Load L acts downwards from the center of the movable (lower) block.
- Effort E is applied downwards on the free end of the rope.
(b) Distance moved by effort:
The velocity ratio (V.R.) is equal to the number of strands supporting the load. In this setup, V.R. = 5.
V.R. = distance moved by effort (dₑ) / distance moved by load (dₗ)
=> 5 = dₑ / 1 m
=> dₑ = 5 × 1 m
=> dₑ = 5 m
The effort will move through a distance of 5 m.
(c) Number of strands supporting the load:
There are 5 strands supporting the movable block.
(d) Mechanical Advantage of the system:
Assuming the system is ideal (100% efficient), the mechanical advantage is equal to the velocity ratio.
Ideal M.A. = V.R. = 5.
7. A pulley system has a velocity ratio 3. Draw a diagram showing the point of application and direction of load (L), effort (E) and tension (T). It lifts a load of 150 N by an effort of 60 N. Calculate its mechanical advantage. Is the pulley system ideal? Give reason.
Answer:
Given:
- Velocity ratio (V.R.) = 3
- Load (L) = 150 N
- Effort (E) = 60 N
Mechanical Advantage (M.A.):
M.A. = Load (L) / Effort (E)
=> M.A. = 150 N / 60 N
=> M.A. = 2.5
Is the system ideal?
No, the pulley system is not ideal.
Reason: For an ideal system, the mechanical advantage must be equal to the velocity ratio (M.A. = V.R.). In this case, M.A. (2.5) is less than V.R. (3). This indicates that the efficiency is less than 100%, likely due to energy loss from friction in the pulleys and the weight of the movable block.
8. Fig. 3.32 shows a system of four pulleys. The upper two pulleys are fixed and the lower two are movable.
(a) Draw a string around the pulleys. Also show the point of application and direction in which the effort E is applied.
(b) What is the velocity ratio of the system ?
(c) How are load and effort of the pulley system related?
(d) What assumption do you make in arriving at your answer in part (c)?
Answer:
(a) Drawing of the string and effort.
(b) Velocity ratio of the system:
The number of strands supporting the movable block is 4.
Therefore, the Velocity Ratio (V.R.) = 4.
(c) Relation between load and effort:
For an ideal system, M.A. = V.R.
Since M.A. = L / E and V.R. = 4,
L / E = 4
=> L = 4E
The load is four times the effort.
(d) Assumption:
The assumption made is that the system is ideal, which means:
- There is no friction in the pulley bearings.
- The weight of the lower movable block and the string is negligible.
- The efficiency is 100%.
9. Fig. 3.33 shows a block and tackle system of pulleys used to lift a load.
(a) How many strands of tackle are supporting the load ?
(b) Draw arrows to represent tension T in each strand.
(c) What is the mechanical advantage of the system?
(d) When load is pulled up by a distance 1 m, how far does the effort end move?
(e) How much effort is needed to lift a load of 100 N ?
(f) What will be its V.R. if the weight of the movable block is doubled ?
Answer:
(a) Number of strands:
Counting the strands supporting the lower movable block in the diagram, there are 4 strands.
(b) Tension arrows:
(c) Mechanical advantage:
Assuming the system is ideal, the mechanical advantage (M.A.) equals the velocity ratio (V.R.), which is the number of strands.
M.A. = 4.
(d) Distance moved by effort:
V.R. = dₑ / dₗ
=> 4 = dₑ / 1 m
=> dₑ = 4 m
The effort end will move 4 m.
(e) Effort needed:
Assuming an ideal system:
E = Load (L) / M.A.
=> E = 100 N / 4
=> E = 25 N
(f) V.R. if weight of movable block is doubled:
The velocity ratio (V.R.) depends only on the physical arrangement (geometry) of the pulleys and strings, not on the weight of the blocks or friction. Therefore, the V.R. will remain unchanged at 4.
10. A block and tackle system has velocity ratio 3. Draw a labelled diagram of the system indicating the points of application and the directions of load L and effort E. A man can exert a pull of 200 kgf. (a) What is the maximum load he can raise with this pulley system if its efficiency is 60% ? (b) If the effort end moves a distance 60 cm, what distance does the load move?
Answer:
Given:
- Velocity ratio (V.R.) = 3
- Maximum effort (E) = 200 kgf
- Efficiency (η) = 60% = 0.6
- Distance moved by effort (dₑ) = 60 cm
(a) Maximum load (L):
First, find the mechanical advantage (M.A.).
η = M.A. / V.R.
=> M.A. = η × V.R.
=> M.A. = 0.6 × 3
=> M.A. = 1.8
Now, calculate the maximum load.
M.A. = Load (L) / Effort (E)
=> L = M.A. × E
=> L = 1.8 × 200 kgf
=> L = 360 kgf
The maximum load the man can raise is 360 kgf.
(b) Distance the load moves (dₗ):
V.R. = dₑ / dₗ
=> dₗ = dₑ / V.R.
=> dₗ = 60 cm / 3
=> dₗ = 20 cm
The load moves a distance of 20 cm.
11. You are given four pulleys and three strings. Draw a neat and labelled diagram to use them so as to obtain a maximum mechanical advantage equal to 8. In your diagram mark the directions of load L, effort E and tension in each strand. What assumptions have you made to obtain the required mechanical advantage ?
Answer:
The required diagram shows a system of one fixed and three movable pulleys, each with a separate string, to achieve M.A. = 2ⁿ = 2³ = 8.
Assumptions:
To obtain a mechanical advantage of 8, the following assumptions must be made:
- The system is ideal, meaning its efficiency is 100%.
- The pulleys are weightless.
- The strings are weightless and inextensible.
- There is no friction in the bearings or at the axles of the pulleys.
