The diagram below shows a pulley used to lift an iron beam of length 3 m and weight 100 kgf, lying on the ground.

(a) Calculate the minimum effort needed to just lift the beam off the ground. [Assume no loss of energy]

(b) When the beam is being lifted from the ground, with the other end touching the ground, state with a reason whether the effort needed keeps on increasing, decreasing or remains constant.

Given,

Weight of iron beam (W) = 100 kgf

Length of iron beam (X) = 3 m

The minimum effort needed to just lift the beam off the ground:

Since the beam is uniform, its weight is evenly distributed over its length.

To just lift one end of the beam, we treat the beam like a uniform rod being pivoted at one end (the other end still touching the ground). The centre of gravity (CG) lies at the midpoint, i.e., at 1.5 m from the lifting end.

To lift one end, the torque due to effort must balance the torque due to the weight of the beam acting at the CG.

Let the effort E be applied vertically upwards through a pulley at one end.

Let Y be the minimum torque due to effort required to lift the beam.

Torque due to weight = 100 x 1.5 = 150 kgf m

Torque due to effort = Y x 3

By principle of levers:

Y x 3 = 150 kgf m

⇒ Y = $\dfrac{150}{3}$ = 50 kgf

M.A. = $\dfrac{\text {Load}}{\text {Effort}}$ ……………(i)

M.A. = Total number of pulleys in both the blocks = 4 ……………(ii)

From equation (i) and (ii)

4 = $\dfrac{50}{\text E}$

E = $\dfrac{50}{4} =12.5 \text{ kgf}$

(b) Effort needed remains the same.
Reason: Ratio between the effort arm and load arm is the same when the beam is being lifted.