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. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) 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.

Answer 1

(a) The torque of the force exerted by the man on the ladder is $\tau=747.64 \approx 747 N-m$.

(b) The force exerted by the ground on the ladder is $R_1$ $=122.5 \mathrm{N}$.

Explanation:

Given:

Mass = 60 kg.  

The length of a ladder = 6.5 m,

The height of a wall = 6 m  

Hence, torque due to the body weight

(a) Torque of force exerted on the ladder by the man around the upper end of the ladder :

                              $\tau=600 \times \frac{6.5}{2} \sin \theta=i$

                             $\tau=600 \times \frac{6.5}{2} \times \sqrt{1-\left(\frac{6}{6.5}\right)^{2}}$

                             $\tau=600 \times \frac{6.5}{2} \times \sqrt{1-\frac{36}{42.25}}$

                             $\tau=600 \times \frac{6.5}{2} \times \sqrt{\frac{42.25-36}{42.25}}$      

                             $\tau=600 \times \frac{6.5}{2} \times \sqrt{\frac{6.25}{42.25}}$

                             $\tau=600 \times \frac{6.5}{2} \times \sqrt{0.147}$              

                            $\tau=300 \times 6.5 \times 0.383$

                             $\tau=1950 \times 0.383$

                             $\tau=747.64 \approx 747 N-m$

(b) Assuming the ladder's weight is negligible compared to the man and assuming the wall is smooth, find the force exerted on the ladder by the ground :

                               $\mathrm{R}_{2}=\mathrm{mg}$

                               $\mathrm{R}_{2}=60 \times 9.8$

                               $\mathrm{R}_{1}=\mu_{2} R_{2}$

                               $6.5 \mathrm{R}_{1} \cos \theta=60 \mathrm{g} \sin \theta \times \frac{6.5}{2}$

                                $\mathrm{R}_{2}=60 \mathrm{g} \tan \theta=60 g \times \frac{2.5}{12} \text { (because } \tan \theta=\frac{2.5}{6})$

                                $R_{1}=\frac{2 s}{2} g$

                                $R_{1}=\frac{25}{2} \times 9.8$

                                $\mathrm{R}_{1}=12.5 \times 9.8$

                                $R_1$ $=122.5 \mathrm{N}$.

Thus, the torque of the force exerted by the man on the ladder is  $\tau=747.64 \approx 747 N-m$ and the force exerted by the ground on the ladder is $R_1$ $=122.5 \mathrm{N}$.