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:

1. the efforts in the two cases.
2. the potential energy gained by the loads in the two cases.
3. the efficiencies in the two cases.

Answer 1

Given: Load = 30 kgf

a. As the mechanical advantage (MA) of a machine is given by,

MA = `"Load"/("Effort" (E_1))`

For a single fixed pulley, the mechanical advantage (MA) is 1.

Then,

MA = `"Load"/("Effort" (E_1))`

⇒ E₁ = `"Load"/"MA"`

= `30/1`

= 30 kgf

For a single movable pulley, the mechanical advantage (MA) is 2.

Then,

MA = `"Load"/("Effort" (E_2))`

⇒ E₂ = `"Load"/"MA"`

= `30/2`

= 15 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 is given by,

Potential energy gained = Load × Height raised

If the effort displacement is the same:

In a fixed pulley, the load rises by the same distance as the effort.

In a movable pulley, the load rises by half the distance of the effort.

Let, for the fixed pulley, the height raised by the load be h.

Therefore,

Ratio of the potential energies in two cases is given by,

Ratio of PE = `"PE gained by the load in fixed pulley​"/"PE gained by the load in movable pulley​"`

= `(30 xx h)/(30 xx h/2)`

= `(30 h xx 2)/(30 h)`

= `2/1`

= 2 : 1

Hence, the ratio of the potential energies in two cases is 2 : 1.

c. In an ideal machine, efficiency = 100 % for both pulleys.

Then,

Efficiency ratio = 1 : 1

Hence, the ratio of the efficiencies in the two cases is 1 : 1.