Answer to Question #215293 in Physics for Sera

Question

A 5 m uniform ladder weighing 120 N is placed against a smooth wall in such a way that it makes an angle of 60o with the horizontal. A man weighing 480 N climbs up the ladder and stays 1.3 m from its upper end. Find (a) the force exerted by the wall against the ladder and (b) the horizontal and vertical components of the force exerted by the floor on the lower end of the ladder.

Answer 1

(a)

$$\sum F_x=F_{fr}-F_{wall}=0,$$ $$F_{wall}=F_{fr}.$$ $$\sum \tau=0,$$ $$\tau_{wall}=\tau_{man}+\tau_{ladder},$$ $$F_{wall}Lcos\theta=(L-1.3)W_{man}sin\theta+\dfrac{L}{2}W_{ladder}sin\theta,$$ $$F_{wall}=\dfrac{(L-1.3)W_{man}tan\theta}{L}+\dfrac{1}{2}W_{ladder}tan\theta,$$ $$F_{wall}=\dfrac{(5\ m-1.3\ m)\cdot480\ N\cdot tan60^{\circ}}{5\ m}+\dfrac{1}{2}\cdot120\ N\cdot tan60^{\circ},$$ $$F_{wall}=719\ N.$$

(b)

$$F_{wall}=F_{fr}=719\ N.$$ $$\sum F_y=F_{floor}-W_{man}-W_{ladder}=0,$$ $$F_{floor}=W_{ladder}+W_{man}=120\ N+480\ N=600\ N.$$