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: Understanding the Concept:
In an ideal situation, the Mechanical Advantage (M.A.) of a single fixed pulley is 1, and for a single movable pulley, it is 2.
Step 2: Key Formula or Approach:
\(M.A. = \frac{\text{Load (L)}}{\text{Effort (E)}}\)
Step 3: Detailed Explanation:
Given Load, \(L = 30 \text{ kgf}\).
1. For Suhas (Fixed Pulley):
\(M.A. = 1 \implies E_{\text{suhas}} = L = 30 \text{ kgf}\).
2. For Radha (Movable Pulley):
\(M.A. = 2 \implies E_{\text{radha}} = \frac{L}{2} = \frac{30}{2} = 15 \text{ kgf}\).
Ratio = \(\frac{E_{\text{suhas}}}{E_{\text{radha}}} = \frac{30}{15} = \frac{2}{1}\).
Step 4: Final Answer:
The ratio of the efforts is 2 : 1.
(b)
Step 1: Understanding the Concept:
Potential Energy gained = \(mg \Delta h = L \times \text{displacement of load}\).
Step 2: Key Formula or Approach:
Velocity Ratio (\(V.R.\)) = \(\frac{\text{Displacement of Effort }(d_{E})}{\text{Displacement of Load }(d_{L})}\).
Step 3: Detailed Explanation:
Let the displacement of effort be \(d\) in both cases.
1. Suhas (Fixed Pulley): \(V.R. = 1 \implies d_{L} = d_{E} = d\).
Energy Gained (\(P.E._{1}\)) = \(L \times d\).
2. Radha (Movable Pulley): \(V.R. = 2 \implies d_{L} = \frac{d_{E}}{2} = \frac{d}{2}\).
Energy Gained (\(P.E._{2}\)) = \(L \times \frac{d}{2}\).
Ratio = \(\frac{P.E._{1}}{P.E._{2}} = \frac{L \times d}{L \times (d/2)} = \frac{2}{1}\).
Step 4: Final Answer:
The ratio of potential energy gained by the loads is 2 : 1.
(c)
Step 1: Understanding the Concept:
Efficiency (\(\eta\)) is defined as the ratio of Work Output to Work Input.
Step 2: Detailed Explanation:
The question specifies "in an ideal situation". In an ideal situation, there is no friction, and the string/pulley are weightless.
For any ideal machine, the efficiency is always 100% or 1.
\(\eta_{\text{suhas}} = 100%\)
\(\eta_{\text{radha}} = 100%\)
Ratio = \(1 : 1\).
Step 3: Final Answer:
The ratio of the efficiencies is 1 : 1.