QQuestion
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 cases is 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.
✓Answer
Given,
Load = 30 kgf
(a) Mechanical advantage (MA) of a machine is:
\mathrm{MA}=\frac{\mathrm{Load}}{\mathrm{Effort}}For a single fixed pulley, MA = 1.
\mathrm{E_1}=\frac{\mathrm{Load}}{\mathrm{MA}}=\frac{30}{1}=30\ \mathrm{kgf}For a single movable pulley, MA = 2.
\mathrm{E_2}=\frac{\mathrm{Load}}{\mathrm{MA}}=\frac{30}{2}=15\ \mathrm{kgf}Ratio of efforts = E1 : E2 = 30 : 15 = 2 : 1.
Hence, the ratio of the efforts in the two cases is 2 : 1.
(b) Potential energy gained = Load × Height raised.
If the effort displacement is the same:
- In a fixed pulley, the load rises through the same distance as the effort.
- In a movable pulley, the load rises through half the distance moved by the effort.
Let the load in the fixed-pulley case rise through height h. The load in the movable-pulley case rises through h/2.
Ratio of potential energies:
\frac{30\times\mathrm{h}}{30\times\frac{\mathrm{h}}{2}}=\frac{2}{1}=2:1Hence, the ratio of the potential energies gained by the loads is 2 : 1.
(c) In an ideal machine, efficiency = 100% for both pulleys.
Therefore, the ratio of their efficiencies is 1 : 1.
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