Question

To lift a load of 30 kgf, Suhas uses a single fixed pulley, while Radha uses a single movable pulley. The displacement of efforts in both the cases are equal. In an ideal situation calculate the ratio of:
(a) the efforts in the two cases.
(b) the potential energy gained by the loads in the two cases.
(c) the efficiencies in the two cases.

Hint:

For pulleys, remember: a single fixed pulley only changes the direction of effort (MA=1), while a single movable pulley halves the effort (MA=2).

Answer 1

Step 1: Analyze the setup for Suhas and Radha:
- Load (\(L\)) = 30 kgf
- Suhas: Single fixed pulley. Ideal Mechanical Advantage (\(MA_S\)) = 1. Velocity Ratio (\(VR_S\)) = 1.
- Radha: Single movable pulley. Ideal Mechanical Advantage (\(MA_R\)) = 2. Velocity Ratio (\(VR_R\)) = 2.
- Let the displacement of effort be \(d_E\) for both.
(a) Ratio of efforts:
Effort (\(E\)) = Load (\(L\)) / MA.
- Effort by Suhas: \(E_S = L / MA_S = 30 \text{ kgf} / 1 = 30 \text{ kgf}\)
- Effort by Radha: \(E_R = L / MA_R = 30 \text{ kgf} / 2 = 15 \text{ kgf}\)
- The ratio \(E_S : E_R = 30 : 15 = 2 : 1\).
(b) Ratio of potential energy gained:
Potential Energy Gained (\(PE\)) = Work done on Load = \(L \times d_L\), where \(d_L\) is the displacement of the load.
From the Velocity Ratio formula, \(d_L = d_E / VR\).
- Displacement of Suhas's load: \(d_{LS} = d_E / VR_S = d_E / 1 = d_E\).
- Displacement of Radha's load: \(d_{LR} = d_E / VR_R = d_E / 2\).
Now, calculate the potential energy gained:
- \(PE_S = L \times d_{LS} = 30 \times d_E\).
- \(PE_R = L \times d_{LR} = 30 \times (d_E / 2)\).
- The ratio \(PE_S : PE_R = (30 \times d_E) : (30 \times d_E / 2) = 1 : 1/2 = 2 : 1\).
(c) Ratio of efficiencies:
The problem states it is an "ideal situation". In an ideal situation, there is no energy loss due to friction or other factors. The efficiency of any ideal machine is 100\%, or 1.
- Efficiency for Suhas (\(\eta_S\)) = 1.
- Efficiency for Radha (\(\eta_R\)) = 1.
- The ratio \(\eta_S : \eta_R = 1 : 1\).