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Question
A 6⋅5 m long ladder rests against a vertical wall reaching a height of 6⋅0 m. A 60 kg man stands half way up the ladder.
- Find the torque of the force exerted by the man on the ladder about the upper end of the ladder.
- Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.
Solution
Given
Mass of the man = m = 60 kg
Ladder length = 6.5 m
Height of the wall = 6 m
(a) We have to find the torque due to the weight of the body about the upper end of the ladder.
\[\tau = 60 \times 10 \times \frac{6 . 5}{2}\sin\theta\]
\[ \Rightarrow \tau = 600 \times \frac{6 . 5}{2} \times \sqrt{\left( 1 - \cos^2 \theta \right)}\]
\[ \Rightarrow \tau = 600 \times \left( \frac{6 . 5}{2} \right) \times \sqrt{\left\{ 1 - \left( \frac{6}{6 . 5} \right)^2 \right\}}\]
\[ \Rightarrow \tau = 740 N - m\]
(b) Let us find the vertical force exerted by the ground on the ladder.
\[N_2 = mg = 60 \times 9 . 8 = 588 N\]
Vertical force exerted by the ground on the ladder = \[\mu N_2 = N_1\]
As system is in rotational equilibrium, we have
\[\tau_{\text{net}} = 0 ............\left(\text{about O} \right)\]
\[\Rightarrow 6 . 5 N_1 \cos\theta = 60g \times \frac{6 . 5}{2}\sin\theta\]
\[\Rightarrow N_1 = \frac{1}{2}60g\tan\theta\]
\[= \frac{1}{2}60g \times \left( \frac{2 . 5}{6} \right) ...........\left[\text{using, }\tan\theta = \frac{2.5}{6} \right]\]
\[ \Rightarrow N_1 = \frac{25}{2}g\]
\[ \Rightarrow N_1 = 122 . 5 N \approx 120 N\]
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