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ICSE Class 10 Physics Question 12 of 19

Machines — Question 7

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Question 7

Draw a diagram of 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) velocity ratio and

(iii) the efficiency of the combination in the ideal situation.

Answer

The diagram is shown below:

Draw a diagram of 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. Machines, Concise Physics Class 10 Solutions.

(i) In equilibrium,

Effort E = T3       (1)

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

T1 = L2\bold{\dfrac{L}{2}}       (2)

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

2T2 = T1

T2= T12\dfrac{T_1}{2}

Substituting value of T1 from equation 2,

T2 = L22\bold{\dfrac{L}{2^2}}       (3)

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

2T3 = T2

T3 = T22\dfrac{T_2}{2}

Substituting value of T2 from equation 3,

T3 = L23\bold{\dfrac{L}{2^3}}       (4)

In equilibrium, T3 = E

From equation 4,

Load L = 23 x T3       (5)

Mechanical advantage M.A.=LEM.A.=23×T3T3⇒M.A.=23\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 dL = 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, dE = 23d.

As we know,

Velocity Ratio V.R.=Distance moved by effort Distance moved by load=dEdL\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,

=23×dd=23Efficiency(η)=M.A.V.R.=2323=1= \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]