Question

A uniform ladder of length 13m and weighing 250N is placed against a smooth vertical wall with its lower end 5m from the wall. The coefficient of friction between the ladder and floor is 0.3. Show that the ladder will remain in equilibrium in this position. What is the frictional force acting on the ladder at the point of contact between the ladder and the floor? 

Answer 1

Since the ladder is placed against a smooth vertical wall, therefore there will be no friction at the point of contact between the ladder and wall. 

Resolving all the force horizontally and vertically. 

∑H = 0, F_r – R₂ = 0 ...(i) 

∑V = 0, R₁ – 250 ...(ii) 

From the geometry of the figure, BC = 12m 

Taking moment about point B, 

R₁ × 5 – F_r × 12 – 250 × 2.5 = 0 

F_r = 52N

For equilibrium of the ladder, Maximum force of friction available at the point of contact between the ladder and the floor = µR 

 = 0.3 × 250 = 75N 

Thus we see that the amount of the force of friction available at the point of contact (75N) is more than force of friction required for equilibrium (52N). Therefore, the ladder will remain in equilibrium in this position.