Machines

Suppose a machine overcomes a load L by the application of effort E.

Let,

the displacement of effort = d_E and

the displacement of the load = d_L in time t.

$\text{Work Input} = effort \times \text{displacement of effort} \\[0.5em] = E \times d_E \\[0.5em] \text{Work Output} = load \times \text{displacement of load} \\[0.5em] = L \times d_L \\[0.5em] \text{Efficiency η} = \dfrac{\text{work output}}{\text{work input}} \\[0.5em] η = \dfrac{L \times d_L}{E \times d_E} \\[0.5em] η = \dfrac{L}{E} \times \dfrac{d_L}{d_E} \\[0.5em] = \dfrac{L}{E} \times \dfrac{1}{\dfrac{d_E}{d_L}} \\[0.5em]$

But we know that,

$\dfrac{L}{E} = M.A. \\[0.5em]$

and

$\dfrac{d_E}{d_L} = V.R. \\[0.5em]$

Substituting the above two values in the formula for efficiency we get

$η = \dfrac{M.A.}{V.R.} \\[0.5em] \Rightarrow M.A. = V.R. \times η \\[0.5em]$

Hence, the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

A pair of scissors which is used to cut a piece of cloth has blades longer than the handles i.e. it has effort arm shorter than load arm so that the blades move longer on the cloth when the handles are moved a little.

Shears used for cutting metals have short blades and long handles i.e., it's effort arm is longer than the load arm and acts as a force multiplier which helps us to overcome large resistive force by a small effort.

(a) The weight W of the scale is greater than E.

The effort arm is 30 cm and the load arm is 10 cm. In order to balance the scale — weight (W) of scale has to be more than effort (E).

(b) Given,

Load arm = 10 cm
Effort arm = 30 cm

As we know that,

$M.A. = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Substituting the values in the formula we get,

$M.A. = \dfrac{\text{30}}{\text{10}} \\[0.5em] \Rightarrow M.A. = 3$

The Class II levers have mechanical advantage always greater than 1 and the reason for this is that, the load and effort are on same side of the fulcrum but in opposite directions and the effort arm is always greater than the load arm.

Therefore, mechanical advantage is always greater than 1.

This type of levers act as a force multiplier so less effort is needed to overcome a large load.

Example — a nut cracker

In order to increase the mechanical advantage, we can increase the length of effort arm.

A Class II lever acts as a force multiplier as the mechanical advantage is always greater than 1 and the effort arm is longer than the load arm.

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.

So, the load and the effort are on the same side of the fulcrum but in opposite directions and the effort arm is always longer than the load arm.

Therefore, the mechanical advantage is always greater than 1.

Draw a labelled diagram of a Class II lever. Give one example of such a lever.

The figure below represents a Class II lever.

A nut cracker is an example of a class II lever.

(a) When the mechanical advantage is more the 1, the lever serves as force multiplier i.e. it enables us to overcome a large resistive force (load) by a small effort.

(b) When the mechanical advantage is equal to 1, the lever has effort arm and load arm of equal lengths like in a physical balance with both arms equal in length.

(c) When the mechanical advantage is less than 1, the levers are used to obtain the gain in speed. This means that the displacement of load is more as compared to the displacement of effort.

Sugar tongs is an example of class III lever.

(a) A seesaw

(b) A common balance

(c) A nutcracker

(d) Forceps.

A single fixed Pulley used to lift a bucket of water from a well.

Reason — When a single fixed pulley is used to lift a bucket of water from a well, it only changes the direction of the force required to lift the bucket.

Output energy = Input energy

Reason — In an ideal machine, the output energy is equal to the input energy. Ideal machines do not account for any energy losses due to friction, inefficiencies, or other factors.

Due to friction, mechanical advantage of a machine increases.

Reason — Friction reduces the mechanical advantage of a machine because frictional forces oppose motion, which means they can decrease the efficiency of a machine by dissipating energy as heat. This results in a decrease in the mechanical advantage of the machine.

No unit

Reason — As velocity ratio is a ratio between similar quantities (distances) so it has no unit.

M.A. = η x V.R.

Reason — The mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

Efficiency = Work input / Work output

Reason — Efficiency is the ratio of useful work output to the total work input.

Greater than 1

Reason — Mechanical advantage (M.A.) is defined as the ratio of the load force to the effort force. When the effort force required is less than the load force, the mechanical advantage is greater than one, indicating that the machine amplifies the input force to overcome the load.

A machine can have mechanical advantage greater than the velocity ratio.

Reason — In actual practice, mechanical advantage for all practical machines is always less than its velocity ratio (i.e., MA < VR) or the output work is always less than the 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.

Effort between fulcrum and load.

Reason — In class III lever, i.e., when effort is between fulcrum and load the mechanical advantage is less than 1.

M.A. > 1

Reason — Mechanical advantage (M.A.) is defined as the ratio of effort arm to load arm. In class II levers, as effort arm is always longer than the load arm hence M.A. > 1.

Class I lever

Reason — As fulcrum is between load and effort hence the figure shows class I lever.

Class II or class III Lever

Reason — In class II or class III levers, effort and load are on the same side of fulcrum .

Given,

Total length of a crowbar = 120 cm

Effort arm = 120 – 20 = 100 cm

Load arm = 20 cm

We know that,

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em] M.A. = \dfrac{100}{20} \\[0.5em] \Rightarrow M.A. = 5 \\[0.5em]$

Given,
Load arm = 15 cm

Effort arm = 7.5 cm

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em] M.A. = \dfrac{7.5}{15} \\[0.5em] \Rightarrow M.A. = 0.5 \\[0.5em]$

Given,

Load arm = 5 cm

Effort arm = 10 cm

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em] M.A. = \dfrac{10}{5} \\[0.5em] \Rightarrow M.A. = 2 \\[0.5em]$

We also know that,

$\text{Effort} = \dfrac{Load}{M.A.} \\[0.5em]$

Substituting the values in the formula we get,

$\text{Effort} = \dfrac{5}{2} \\[0.5em] \Rightarrow \text{Effort} = \text {2.5 kgf}$

(a) The diagram represents a Class I lever.

(b) Given,

AB = 1m

AF = Load arm =0.4 m

BF = Effort arm = 0.6 m

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em] M.A. = \dfrac{0.6}{0.4} \\[0.5em] \Rightarrow M.A. = 1.5 \\[0.5em]$

(c) We know that,

$\text{Effort} = \dfrac{Load}{M.A} \\[0.5em]$

Given,

Load = 15 kgf

Substituting the values in the formula we get,

$\text{Effort} = \dfrac{15}{1.5} \\[0.5em] \Rightarrow \text{Effort} = \text{10 kgf} \\[0.5em]$

(a) Below is the diagram of the crowbar showing the fulcrum (F), load (L) and effort (E) with their directions:

(b) Crowbar belongs to class I lever.

(c)

Given,

(i) Total length = 1.5 m

Effort arm = 1 m

$\text{Load arm} = \text{Total length} - \text{Effort arm} \\[0.5em] \text{Load arm} = 1.5m - 1m \\[0.5em] \Rightarrow \text{Load arm} = 0.5 m \\[0.5em]$

(ii) Effort arm = 1 m

(iii)

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em] M.A. = \dfrac{1}{0.5} \\[0.5em] \Rightarrow M.A. = 2 \\[0.5em]$

(iv) Given,

Load = 75 kgf

We know that,

$\text{Effort} =\dfrac{\text{Load}}{\text{M.A.}} \\[0.5em]$

Substituting the values in formula we get,

$\text{Effort} =\dfrac{75}{2} \\[0.5em] \Rightarrow \text{Effort} = 37.5 \text{ kgf} \\[0.5em]$

(a)

(i) We know that,

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Given,

Load arm = 8.0 cm

Effort arm = 2 cm

Substituting the values in the formula we get,

$M.A. = \dfrac{2}{8} \\[0.5em] \Rightarrow M.A. = 0.25 \\[0.5em]$

(ii) We know,

$Load = M.A. \times Effort \\[0.5em]$

Given,

Effort = 10 kgf

Substituting the values in the formula we get,

$Load = 0.25 \times 10 \\[0.5em] \Rightarrow Load = 2.5 kgf \\[0.5em]$

(b) As the mechanical advantage is less than 1 i.e. M.A. < 1. Hence, the pair of scissors acts as a speed multiplier as there is larger displacement of load than displacement of effort.

(a) The principle of moments states that —

Moment of the load about the fulcrum = Moment of the effort about the fulcrum

Given,

Load = 18kgf and 18 kgf load is placed at 60 cm from the support.

Effort = W and W kgf weight is placed at 250 cm from the support.

Substituting the values in the formula we get,

$18 \times 60 = W \times 250 \\[0.5em] \Rightarrow W = \dfrac{18 \times 60}{250} \\[0.5em] \Rightarrow W = 4.32 \\[0.5em]$

∴ W = 4.32 kgf

(b) The principle of moments states that —

Moment of the load about the fulcrum = Moment of the effort about the fulcrum

Given,

Load = 18kgf and 18 kgf load is placed at 60 cm from the support.

W = 5 kgf and 5 kgf weight is placed at d cm from the support.

Substituting the values in the formula we get,

$18 \times 60 = 5\times d \\[0.5em] \Rightarrow d = \dfrac{18 \times 60}{5} \\[0.5em] \Rightarrow d = 216 cm \\[0.5em]$

∴ d = 216 cm

(c) It belongs to class I lever as fulcrum is in between the effort and the load.

(a) It belongs to Class II lever as length of the lever is equal to the effort arm and the effort arm is also more than the load arm.

(b) Below is the diagram of the lever with position of fulcrum F and directions of both the load L and effort E labelled:

(c) We know that,

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Given,

Load arm = 5 cm

Effort arm = 9 cm

Substituting the values in the formula we get,

$M.A. = \dfrac{9}{5} \\[0.5em] \Rightarrow M.A. = 1.8 \\[0.5em]$

Relation between mechanical advantage, efficiency and velocity ratio is

$M.A. = \eta \times V.R. \\[0.5em]$

Given, η = 100% = 1

So,

$M.A. = 1 \times V.R. \\[0.5em]$

Therefore, Mechanical advantage = Velocity ratio = 1.8

(d) When efficiency reduces to 50%, its mechanical advantage reduces however, its velocity ratio remains the same. So,

M.A. = 0.5 x 1.8 = 0.9
V.R. = 1.8

(a) The diagram shown belongs to class II lever as Load is in between Fulcrum and Effort.

(b) The mechanical advantage of the lever increases when load is shifted towards the fulcrum, as less effort is required to lift the load.

(a) Load arm AF = 20 cm

(b) Effort arm CF = 60 cm

(c) We know that,

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Substituting the values in the formula we get,

$M.A. = \dfrac{60}{20} \\[0.5em] \Rightarrow M.A. = 3 \\[0.5em]$

(d) We know that,

$\text{Effort} = \dfrac{\text{Load}}{\text{M.A.}} \\[0.5em]$

Given,

Load = 30 kgf + 15 kgf = 45 kgf

Substituting the values in the formula we get,

$\text{Effort} = \dfrac{45}{3} \\[0.5em] \Rightarrow \text{Effort} = 15\text{ kgf} \\[0.5em]$

A fishing rod is an example of a Class III lever.

Here,

• Fulcrum : at the handle (where the rod pivots)

• Effort : applied by the fisher's hand along the rod

• Load : the fish at the end of the rod

In Class III levers : Fulcrum — Effort — Load

The principle of moments states that —

Moment of the load about the fulcrum = Moment of the effort about the fulcrum

∴ Effort × Effort Arm = Load × Load Arm .......... (i)

Given :

Load (L) = 10 N

Load arm (distance between fulcrum and fish) = 1.5 m

Effort arm (distance between fulcrum and applied effort) = 0.5 m

From (i)

Effort x 0.5 = 10 x 1.5

⇒ Effort = $\dfrac{15}{0.5}$ = 30 N

(a) The principle of moments states that —

Moment of the load about the fulcrum = Moment of the effort about the fulcrum

(b) The above diagram belongs to Class III lever. The example of Class III lever is a sugar tongs.

(c)

(i) We know that,

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Given,

Load arm = 490 cm + 10 cm = 500 cm

Effort arm = 10 cm

Substituting the values in the formula we get,

$M.A. = \dfrac{10}{500} \\[0.5em] \Rightarrow M.A. = \dfrac{1}{50} \\[0.5em]$

(ii) We know that,

$\text{Effort} = \dfrac{\text{Load}}{\text{M.A.}} \\[0.5em]$

Given,

Load = 50 kgf

Substituting the values in the formula we get,

$\text{Effort} = \dfrac{50}{\dfrac{1}{50}} \\[0.5em] \Rightarrow \text{Effort} = 2500 N \\[0.5em]$

(i) We know that,

$\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

Given,

Load arm = 20 cm

Effort arm = 15 cm

Substituting the values in the formula we get,

$M.A. = \dfrac{15}{20} \\[0.5em] \Rightarrow M.A. = 0.75 \\[0.5em]$

(ii) We know that,

$\text{Effort} = \dfrac{\text{Load}}{\text{M.A.}} \\[0.5em]$

Given,

Load = 1.5 kgf

Substituting the values in the formula we get,

$\text{Effort} = \dfrac{1.5}{0.75} \\[0.5em] \Rightarrow \text{Effort} = 2 \text{ kgf} \\[0.5em]$

Machines are useful to us in —

(a) Changing the point of application of effort to a convenient point.

(b) Lifting a heavy load by applying a less effort.

(c) Changing the direction of effort to a convenient direction.

(d) For obtaining a gain in speed.

A jack is a mechanical lifting device used to lift heavy loads.

A jack works as a force multiplier, as with a little effort on the jack we are able to lift heavy load like a car. It acts as a force multiplier.

An ideal machine is that where there is no loss of energy i.e. the work output is equal to the work input. An ideal machine is 100% efficient.

In a practical machine output energy is always less than the input energy i.e. there is some loss of energy during the operation which is not the case in an ideal machine.

The ratio of the load to the effort is called mechanical advantage of the machine.

$\text {Mechanical Advantage} = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

As mechanical advantage is a ratio of similar quantities so it has no unit.

Velocity ratio is defined as the ratio of the velocity of effort to the velocity of the load.

$\text {Velocity Ratio} = \dfrac{\text {Velocity of effort}}{\text {Velocity of load}} \\[0.5em]$

As velocity ratio is a ratio of two similar quantities, so it has no unit.

The efficiency of a machine is defined as the ratio of the work done on load by the machine to the work done on the machine by the effort.

$\eta = \dfrac{\text {Work output}}{\text {Work Input}} \\[0.5em]$

It has no unit as it is the ratio of two similar quantities.

A machine is not always 100% efficient due to —

(a) friction

(b) weight of moving parts of machine of a given design.

Due to the above reasons, the velocity ratio of the machine does not change but it's mechanical advantage decreases so its efficiency decreases.

(a) A machine acts as a force multiplier when the effort arm is longer than the load arm. The mechanical advantage in such cases is greater than 1.

(b) A machine acts as a speed multiplier when the effort arm is shorter than the load arm. The mechanical advantage in such cases is less than 1.

No, it is not possible for a machine to act as a force multiplier and speed multiplier simultaneously.

For a machine used as a force multiplier, effort < load while for a machine used to obtain a gain in speed, effort > load. Hence, it is not possible for a single machine to act as both, simultaneously.

(a) The mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

M.A. = V.R. x η

(b) The term that does not change for a machine of a given design is its velocity ratio.

The mechanical advantage of an actual machine is always less than the velocity ratio as there is a loss of energy due to friction.

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 in the equilibrium position of levers, moment of load about the fulcrum must be equal to the moment of effort about the fulcrum and the two moments must be in opposite directions.

The three classes of levers are —

(i) Class I levers — In this type of the levers, the fulcrum F is in between the effort E and the load L.

Example — a seesaw, a pair of scissors

(ii) Class II levers — In these 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.

Example — a nutcracker, a bottle opener.

(iii) Class III levers — In these 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.

Example — sugar tongs, foot treadle.

A pair of scissors and a pair of pliers both belong to class I lever.

As the blades of the scissors are longer than its handles, so the effort arm is shorter than the load arm. Hence, the mechanical advantage of a pair of scissors is less than 1.

(a) Labelled diagram of lemon crusher is shown below:

(b) It is a class II lever.

(a) Below diagram shows the position of fulcrum F and the directions of load L and effort E:

(b) The rod is a Class II lever as the load is between fulcrum and effort.

(c) A nut cracker is also an example of a Class II lever.

The Class III levers have mechanical advantage always less than 1.

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.

The effort and the load are on the same side of fulcrum but in opposite directions and the effort arm is always smaller than the load arm. Therefore, the mechanical advantage of class III lever is always less than 1.

Class III levers have mechanical advantage less than 1 so as effort arm is always less than the load arm, so 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.

(a) A bottle opener is an example of Class II lever, as the fulcrum F and the effort E are at the two ends of the lever and the load is in between the fulcrum F and the effort E.

(b) Sugar tongs is an example of Class III lever, as the fulcrum F and the load L are at the two ends of the lever and the effort E is in between the fulcrum F and the load L.

The positions of load L, effort E and fulcrum F in the forearm are shown below:

It is a Class III lever.

(a) A machine is a device by which we can either 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 or by which we can obtain a gain in speed.

(b) 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. its efficiency is 100%.

(a) In order to multiply the force — a bar is used to lift a heavy stone.

(b) In order to change the point of application of force — The rear wheel of a cycle is rotated by applying the effort on the pedal attached to the toothed wheel which is joined to the rear wheel with the help of a chain.

(c) In order to change the direction of force — to lift a bucket full of water from the well, a single fixed pulley is used by applying the effort in the downward direction instead of applying it upwards when the bucket is lifted up without the use of pulley.

(d) In order to obtain gain in speed — A knife. The blade of a knife moves longer by a small displacement of its handle.

(i) For an ideal machine (free from friction etc.), work output is equal to the work input, so the efficiency is equal to 1 ( or 100% ) and the mechanical advantage is numerically equal to the velocity ratio. So,

For an ideal machine, M.A. = V.R.

(ii) In the case of a practical machine, the mechanical advantage is always less than its velocity ratio or the output work is always less than the input work as some of the input energy is lost due to the force of friction etc. So,

For a practical machine, M.A. < V.R.

(i) When a machine works as a force multiplier then the displacement of the load is less than the displacement of effort. Hence, the velocity ratio is more than 1.

(ii) When a machine works as a speed multiplier then the displacement of load is more than displacement of effort. Hence, the velocity ratio is less than 1.

For an actual machine, the mechanical advantage is always less than its velocity ratio or the output work is always less than the input work.

The efficiency of such a machine is always less than 1 as some part of input energy is lost against the force of friction.

The expression of the mechanical advantage of a lever is:

$M.A. = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]$

(a) For a class I lever, the mechanical advantage is more than 1 in the case of shears used for cutting the thin metal sheets.

(b) For a class I lever, the mechanical advantage is less than 1 in case of a pair of scissors whose blades are longer than its handles.

(a) A door — Class II

(b) A catapult — Class I

(c) Claw hammer — Class I

(d) A wheel barrow — Class II

(e) A fishing rod — Class III

(f) sugar tongs — Class III

(a) When a human body raises a load on the palm then it is an example of a Class III lever. The elbow joint acts as fulcrum F at one end, biceps exerts the effort in the middle and load L on the palm is at the other end.

(b) When raising the weight of body on toes then it is a Class II lever. The fulcrum F is at toes at one end, the load L (i.e. weight of the body) is in the middle and the effort E by muscles is at the other end.

The examples of each class of lever in a human body are:

(i) The action of nodding of the head is a Class I lever. In this case the spine acts as the fulcrum F, load L is at its front part while effort E is at its rear part.

(ii) The action of raising the weight of the body on toes is an example of Class II lever. In this case, the fulcrum F is at toes at one end, the load L (i.e. the weight of the body) is in the middle and effort E by muscles is at the other end.

(iii) The action of raising a load by forearm is an example of Class III lever. In this case the elbow joint acts as fulcrum F at one end, biceps exerts the effort E in the middle and a load L on the palm is at the other end.

(a) Mechanical advantage = efficiency × velocity ratio

(b) In Class II lever, effort arm is longer than the load arm.

(c) A pair of scissors is speed multiplier

The single movable pulley acts as a force multiplier and has a mechanical advantage equal to 2. Below is the labelled diagram of a single movable pulley:

In the case of a single pulley with a mechanical advantage greater than 1, the force needs to be applied in an upward direction which is very inconvenient.

In order to change the direction of force applied without altering its mechanical advantage, we have to use a single movable pulley along with a fixed pulley.

The load is attached to the axle of movable pulley A and the effort is applied in the downward direction at the free end of the string passing over the fixed pulley B. One can also use his own weight as effort which will be quite convenient.

Below is the diagram of an arrangement of two pulleys, one fixed and other moveable:

The ideal mechanical advantage of a single movable pulley with a fixed pulley system is 2. The ideal mechanical advantage of 2 can be achieved by assuming there is no friction in the pulley bearings, between the string and surface of the rim of the pulley and that string and the pulley are massless.

(a) A → Movable pulley, B → Fixed pulley

(b) Tension on each strand of string is as shown in the diagram below:

(c) The purpose of fixed pulley B is to change the direction of effort to be applied from upward to downward.

(d)(i) When tension in the string is T then the effort E balances the tension T. So, E = T.

(d)(ii) When the load is balanced by the tension T in two segments of the string and the effort E balances the tension T at the free end, so

L = T + T = 2 T and

E = T,

Hence, 2E = L.

(e) When distance moved by effort is d then distance moved by load is d/2 so the velocity ratio becomes 2.

(f) If the efficiency of the system is 100%, then effort applied is equal to half the load. Hence, it acts as a force multiplier and the mechanical advantage is 2.

(a) Pulleys A and B are movable pulleys while pulley C is a fixed pulley.

(b) The direction of load (L), effort (E) and tension T₁ and T₂ in the two strings are marked in the below diagram:

(c) The relation between magnitudes of L and E to the tension T₁ is as follows:

The magnitude of effort E = T₁
The magnitude of load L = 2² T₁ = 4T₁.

(d) The mechanical advantage is 2² = 4 as there are two movable pulleys and we know,

M.A. = 2ⁿ where, n is the number of movable pulleys.

The velocity ratio = 2² = 4 as there are two movable pulleys and we know,

V.R. = 2ⁿ where, n is the number of movable pulleys.

(e) The assumptions made are:

1. The pulleys A and B are weightless
2. There is no friction between the bearings of the pulleys or the axle.

The diagram is shown below:

(i) In equilibrium,

Effort E = T₃       (1)

Tension T₁ in the string passing over the pulley A is given as 2T₁ = L

T₁ = $\bold{\dfrac{L}{2}}$       (2)

Tension T₂ in the string passing over the pulley B is given as

2T₂ = T₁

T₂= $\dfrac{T_1}{2}$

Substituting value of T₁ from equation 2,

T₂ = $\bold{\dfrac{L}{2^2}}$       (3)

Tension T₃ in the string passing over the pulley C is given as

2T₃ = T₂

T₃ = $\dfrac{T_2}{2}$

Substituting value of T₂ from equation 3,

T₃ = $\bold{\dfrac{L}{2^3}}$       (4)

In equilibrium, T₃ = E

From equation 4,

Load L = 2³ x T₃       (5)

$\text{Mechanical advantage M.A.} = \dfrac{L}{E} \\[0.5em] M.A. = \dfrac{2^3 \times T_3}{T^3} \\[0.5em] \Rightarrow M.A. = 2^3$

(ii) As we know, one end of each string passing over a movable pulley is fixed, so the other end of string moves up twice the distance moved by the axle of the movable pulley.

If the load L attached to the pulley A moves a distance d,

then d_L = d

Now, the string connected to the axle of pulley B, moves up by a distance,

2 times d = 2d.

Then the string connected to the axle of the pulley C, moves up by a distance,

2 times 2d = 2²d

Then the end of the string passing over the fixed pulley D, moves up by a distance,

2 times 2²d = 2³d.

Hence, d_E = 2³d.

As we know,

$\text{Velocity Ratio V.R.} = \dfrac{\text{Distance moved by effort }}{\text{Distance moved by load}} \\[0.5em] = \dfrac{d_E}{d_L} \\[0.5em]$

(iii) Substituting the values in the formula we get,

$= \dfrac {2^3 \times d}{d} \\[0.5em] = 2^3 \\[1.5em] \text{Efficiency} (\eta) = \dfrac{M.A.}{V.R.} \\[0.5em] = \dfrac{2^3}{2^3} = 1 \\[0.5em]$

The diagram of a block and tackle system of pulleys having a velocity ratio of 5 with directions of the load L, effort E and tension T in each strand is shown below:

as a force multiplier

Reason — Pulleys are commonly used as force multipliers in machines. They allow a person to lift or move heavy objects with less effort by distributing the force required over multiple ropes and pulleys.

helps to apply the effort in a convenient direction.

Reason — A single fixed pulley changes the direction of the force needed to lift an object. However, it doesn't provide a mechanical advantage greater than 1 or a velocity ratio less than 1, nor does it guarantee 100% efficiency.

1

Reason — The velocity ratio of a single fixed pulley is 1. If the point of application of effort E moves a distance d, then the load L also moves the same distance d upwards.

i.e. if d_E = d_L = d then V.R. = 1

$V.R. = \dfrac{d_E}{d_L} = \dfrac{d}{d} = 1 \\[0.5em]$

2

Reason — The mechanical advantage in the ideal case is 2 as the load can be lifted by applying the effort equal to half the load i.e., the single movable pulley acts as a force multiplier.

All the given options are correct.

Reason — The velocity ratio of a single fixed pulley is always 1 and that of a single movable pulley is always 2.

The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2ⁿ whereas the velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting load.

2ⁿ

Reason — In general, if there are n movable pulleys with one fixed pulley, the mechanical advantage is then 2ⁿ.

(3) and (4)

Reason — The correct differences are:

M.A.< V.R.

Reason — In actual practice there is some friction between the parts of the machine or deformation of parts, which causes the mechanical advantage to be less than the velocity ratio.

2:1

Reason — We know that in case of a single fixed pulley, the effort applied is equal to the load.

Hence,

$E_f = L \\[0.5em]$

In the case of a single movable pulley the effort needed to lift a load is equal to half the load.

$E_m = \dfrac{L}{2} \\[1em] E_m = \dfrac{50 \text{ kgf}}{2} \\[1em] \Rightarrow E_m = 25\text{ kgf} \\[1em]$

Hence, the ratio of efforts applied by the respective pulley is

$\dfrac{E_f}{E_m} = \dfrac{2}{1} \\[0.5em]$

Therefore, E_f : E_m = 2 : 1

5

Reason — In a block and tackle system of 5 pulleys, 5 strands will be supporting the load.

4

Reason — The number of strands supporting the load in case of block and tackle system of 4 pulleys will be 4.

$\eta = 1 - \dfrac{w}{nE}$

Reason — For a system of n pulleys, let w be the total weight of the lower block along with the pulleys in it. In the balanced position,

E = T and L + w = nT

or L = nT - w = nE - w

M.A. = $\dfrac{L}{E} = \dfrac{nE-w}{E} = n - \dfrac{w}{E}$

Thus, the mechanical advantage is less than the ideal value n.

The velocity ratio does not change, it remains n, i.e., V.R. = n

Hence, efficiency $\eta = \dfrac{M.A.}{V.R.} = \dfrac{n - \dfrac{w}{E} }{n} = 1 - \dfrac{w}{nE}$.

assertion is true but reason is false.

Explanation

Assertion (A) is true. Class III levers always have a mechanical advantage less than 1, indicating that the load force is greater than the effort force.

Reason (R) is false. In class III levers, the load arm is always longer than the effort arm. Therefore, the statement that the effort arm is always greater than the load arm is incorrect.

Given,

Load = m x g = 6 x 10 = 60 N

Effort = 70 N

As we know that,

$M.A. = \dfrac{\text{Load}}{\text{Effort}} \\[0.5em]$

Substituting the values in the formula we get,

$M.A. = \dfrac{60}{70} \\[0.5em] M.A. = \dfrac{6}{7} \\[0.5em] \Rightarrow M.A. = 0.857 \\[0.5em]$

(a) As we know,

$\text {Power} = \text {Force} \times \text {velocity} \\[0.5em]$

and

$\text {Velocity} = \dfrac{\text{Change in distance}}{\text{Change in time}} \\[0.5em] \text {velocity} = \dfrac{8}{4} \\[0.5em] So, \text {velocity} = 2 m/sec \\[0.5em]$

Given,

$\text {Tension in pulley} = 100 \times 10 \\[0.5em] \text {Force} = \text {Tension} = 1000 N \\[0.5em]$

Substituting the values in the formula for power we get,

$\text {Power} = 1000 \times 2 \\[0.5em] \Rightarrow \text {Power} = 2000 W \\[0.5em]$

(b) We know,

$\eta = \dfrac{M.A.}{V.R.} \\[0.5em]$

Given,

$\text {Load} = 75 \times 10 \\[0.5em] \text {Load} = 750 N \\[0.5em]$

We know,

$M.A. = \dfrac{Load}{Effort}$ Substituting the values for M.A. we get,

$M.A. = \dfrac{750}{1000} \\[0.5em] \Rightarrow M.A. = 0.75 \\[0.5em]$

When the effort moves by a distance d downwards, the load moves by the same distance upwards. So we get,

V.R. = 1

Substituting the values of M.A. and V.R. in the formula for efficiency we get,

$\eta = \dfrac{0.75}{1} \\[0.5em] \Rightarrow \eta = 0.75 \\[0.5em]$

Hence, efficiency = 75 %

(c) As the load moves same distance upwards as much as effort moves downwards. So, the height achieved by the load is 8 m.

We know that in case of a single fixed pulley, the effort applied is equal to the load.

Hence,

$E_f = L \\[0.5em]$

In the case of a movable pulley the effort needed to lift a load is equal to half the load.

$E_m = \dfrac{L}{2} \\[0.5em] E_m = \dfrac{50 kgf}{2} \\[0.5em] \Rightarrow E_m = 25kgf \\[0.5em]$

Hence, the ratio of efforts applied by the respective pulley is

$\dfrac{E_f}{E_m} = \dfrac{2}{1} \\[0.5em]$

Therefore, E_f : E_m = 2 : 1

(i) Given,

Effort = 25 kgf

Load = 75 kgf

As we know that,

$M.A. = \dfrac{\text{Load}}{\text{Effort}} \\[0.5em]$

Substituting the values in the formula we get,

$M.A. = \dfrac{75}{25} \\[0.5em] M.A. = \dfrac{3}{1} \\[0.5em] \Rightarrow M.A. = 3 \\[0.5em]$

(ii) As we know,

V.R. = n = 3

(iii) Also, $\eta = \dfrac{M.A.}{V.R.} \\[0.5em] \eta = \dfrac{3}{3} \\[0.5em] \Rightarrow\eta = 1 \\[0.5em]$

Therefore, efficiency = 100%

(a) Given,

Effort = 1000 N

Load = 4500 N

Number of pulleys = n = 5

As we know that,

$M.A. = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

Substituting the values in the formula we get,

$M.A. = \dfrac{4500}{1000} \\[0.5em] M.A. = \dfrac{45}{10} \\[0.5em] \Rightarrow M.A. = 4.5 \\[0.5em]$

(b) As we know,

V.R. = n = 5

(c) And, $\eta = \dfrac{M.A.}{V.R.} \\[0.5em] \eta = \dfrac{4.5}{5} \\[0.5em] \Rightarrow\eta = 0.9 \\[0.5em]$

Therefore, efficiency = 90%

(a) The below diagram shows the tackle along with the direction of load L and effort E marked:

(b) As we know that in a system of 5 pulleys, if the load moves up through a distance 1 m, each segment of the string is loosened by a length 1m so the effort end moves through a distance 5 times 1m i.e. 5m

(c) Five strands of tackle are supporting the load.

(d) As we know,

$M.A. = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

Load = 5m
Effort = 1m

Substituting the values in the formula for M.A. we get,

$M.A. = \dfrac{5}{1} \\[0.5em] \Rightarrow M.A. = 5 \\[0.5em]$

Given,

Effort = 60 N

Load = 150 N

As we know,

$M.A. = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

Substituting the values in the formula for M.A. we get,

$M.A. = \dfrac{150}{60} \\[0.5em] \Rightarrow M.A. = 2.5 \\[0.5em]$

Therefore, Mechanical advantage = 2.5

For an ideal case the M.A. should be equal n, where n is the number of pulleys of the pulley system.

In this case as the mechanical advantage is less than 3 ( i.e. the number of pulleys ) so the pulley system is not ideal.

(a) The diagram below shows the point of application and direction of the effort E:

(b) As we know that in a system of 4 pulleys, if the load moves up through a distance d, each segment of the string is loosened by a length d so the effort end moves through a distance 4 times d.

$\text{Velocity ratio} = \dfrac{d_E}{d_L} \\[0.5em] \text{Velocity ratio} = \dfrac{4d}{d} \\[0.5em] \Rightarrow \text{Velocity ratio}= 4 \\[0.5em]$

(c) As we know,

$M.A. = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

As M.A. = n = 4

Substituting the values in the formula we get,

$\dfrac{Load}{Effort} = 4 \\[0.5em] \Rightarrow Load = 4 \times Effort$

(d) Assumptions made are —

(i) There is no friction in the pulley bearings.
(ii) Weight of lower pulleys is negligible.

(iii) The effort is applied downwards.

(a) There are 4 strands of tackle supporting the load.

(b) Tension T in each strand is as shown in the diagram below:

(c) As we know,

$M.A. = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

Load = 4T

Effort = T

Substituting the values in the formula for M.A. we get,

$M.A. = \dfrac{4T}{T} \\[0.5em]$

Hence, M.A. = 4

(d) In a system of 4 pulleys, if the load moves up through a distance 1m, each segment of the string is loosened by a length 1m so the effort end moves through a distance 4 times 1m. Hence, we get 4m as the distance moved by the effort arm.

(e) As we know,

$M.A. = \dfrac{\text {Load}}{\text {Effort}} \\[0.5em]$

M.A = 4

Load = 100 N

Substituting the values in the formula for M.A. we get,

$4 = \dfrac{100}{\text{Effort}} \\[0.5em] \text{Effort} = \dfrac{100}{4} \\[0.5em] \Rightarrow \text{Effort} = 25N \\[0.5em]$

(f) V.R = n (number of pulleys) = 4

Given,

V.R = n (number of pulleys) = 3

Hence, number of pulleys = 3

Below is a labelled diagram of the system indicating the points of application and the directions of load L and effort E:

(a) As we know,

$\eta = 1 - \dfrac{w}{nE}$

Given,

Effort = 200 kgf

Efficiency = 60% = 0.6

Substituting the values in the formula for efficiency we get,

$0.6 = 1 - \dfrac{w}{3 \times 200} \\[0.5em] \dfrac{w}{600} = 1 - 0.6 \\[0.5em] w = 600 \times 0.4 \\[0.5em] \Rightarrow w = 240 \text{ kgf}$

As we know,

$\text{Load} = (n \times E) - w \\[0.5em] \text{Load} = (3 \times 200) - w \\[0.5em] \text{Load} = 600 - 240 \\[0.5em] \Rightarrow \text{Load} = 360kgf$

(b) The velocity ratio of system is V.R = number of pulleys = 3

As we know,

$V.R. = \dfrac{d_E}{d_L} \\[0.5em]$

Substituting the values in the formula for V.R. we get,

$3 = \dfrac{d_E}{d_L} \\[0.5em] d_L= \dfrac{d_E}{3} \\[0.5em] d_L= \dfrac{60}{3} \\[0.5em] \Rightarrow d_L= 20 cm \\[0.5em]$

The four pulleys can be used as three movable pulleys with one fixed pulley as shown in the diagram below:

Assumptions are —

(i) The weight of the pulleys and the string is massless.

(ii) There is no friction between the bearings of the pulley.

A pulley, which has its axis of rotation stationary in position is called a fixed pulley.

A fixed pulley is used to change the direction of effort to be applied. It is used for lifting small load such as a water bucket.

The pulley that has no gain in mechanical advantage is a single fixed pulley.

A single fixed pulley is used only to change the direction of effort to be applied i.e. with its use the effort can be applied in a more convenient direction.

It is difficult to apply the effort upwards to lift a load up directly, but it becomes easier with the help of a fixed pulley, because the effort can be applied in the downward direction to raise the load up. Further to apply the effort downwards one can conveniently make use of his own weight also for the effort.

A pulley whose axis of rotation is movable is called a single movable pulley. The mechanical advantage in the ideal case is 2 as the load can be lifted by applying the effort equal to half the load. A single, movable pulley acts as a force multiplier.

The efficiency of a single movable pulley system is not 100% because of the following reasons:

1. There is friction in the pulley bearings or at the axle.
2. The weight of the pulley and string is not zero.

In the case of a single movable pulley the velocity ratio is always 2. The friction in the pulley bearing has no effect.

When the free end of string is pulled up by the effort through a distance x, the load is raised up through a distance x/2. The reason is that the segment of string on both sides of the pulley moves up by a distance x/2 i.e., if d_E = x, then d_L = x/2.

(a) For an actual single movable pulley, due to (i) friction in the pulley bearings or at the axle and (ii) the weight of the pulley and string, the effort needed to lift up a load L will be more than $\dfrac{\text{L}}{2}$, so the mechanical advantage will be less than 2. But the velocity ratio will remain as 2. Hence, for a single movable pulley, the velocity ratio is always more than the mechanical advantage.

(b) As some effort is wasted in overcoming the friction between the strings and the grooves of the pulley as a result the efficiency of a movable pulley is always less than 100%.

(c) In the case of a block and tackle system, the mechanical advantage increases with the increase in the number of pulleys.

$M.A. = \dfrac{Load}{Effort} \\[0.5em] M.A. = \dfrac{nT}{T} = n \\[0.5em]$

∴ M.A. = Total number of pulleys in both the blocks

(d) The lower block of a block and tackle pulley system must be of negligible weight as efficiency is reduced due to the weight of the lower block of the pulley.

$\eta = 1 - \dfrac{w}{nE} \\[0.5em]$

where, w is the total weight of the lower block along with the pulleys in it.

In case of an ideal single fixed pulley the effort needed is equal to the load itself and so the mechanical advantage is equal to 1.

As the amount of effort required is equal to the load itself so a single fixed pulley cannot be used as a force multiplier.

The velocity ratio of a single fixed pulley is 1. If the point of application of effort E moves a distance d, then the load L also moves the same distance d upwards.

i.e. if d_E = d_L = d then V.R. = 1

$V.R. = \dfrac{d_E}{d_L} = \dfrac{d}{d} = 1 \\[0.5em]$

In a single movable pulley, if the effort moves by a distance x upward, the load is raised to a height of $\dfrac{x}{2}$ as the segment of string on both sides of the pulley moves up by a distance $\dfrac{x}{2}$.

(a) Multiply force — a movable pulley is used

(b) Multiply speed — Class III lever

(c) Change the direction of force applied — single fixed pulley

(a) False

Reason — The velocity ratio of a single fixed pulley is always equal to 1, not more than 1.

(b) True

Reason — The velocity ratio of a single movable pulley is always 2.

(c) True

Reason — In a combination of n movable pulleys with a fixed pulley, each movable pulley contributes a velocity ratio of 2 because the rope supporting the load passes through two sections of the rope for each movable pulley. Therefore, if there are n movable pulleys, the total velocity ratio is 2ⁿ.

(d) True

Reason — In a block and tackle system, the velocity ratio is equal to the number of strands of the tackle supporting the load. Each strand of rope supports the load, and as the rope is pulled, each strand contributes to lifting the load. Therefore, the velocity ratio, which is the ratio of the distance moved by the effort to the distance moved by the load, is equal to the number of strands supporting the load.