Question

A 3.00-m-long ladder, weighing 200 N, leans against a smooth vertical wall with its base on ahorizontal rough floor, a distance of 1.00 m away from the wall. The ladder is not completelyuniform, so its center of gravity is 1.20 m from its base. What force of friction must the floor exerton the base of the ladder to prevent the ladder from sliding down

Answer 1

Answer:

28.3 N

Explanation:

W = weight of the ladder = 200 N

L = length of the ladder = AC = 3 m

AD = distance of base of ladder from wall = 1 m

AB = distance of center of gravity from base = 1.2 m

F = force from the wall

f = frictional force acting between the ladder and floor

In triangle ACD

Cosθ = $\frac{AD}{AC}$

Cosθ = $\frac{1}{3}$

Sinθ = $\sqrt{1 - \left ( Cos\theta  \right )^{2}}$

Sinθ = $\sqrt{1 - \left ( \frac{1}{3} \right )^{2}}$

Sinθ =  $\frac{2.83}{3}$

Using equilibrium of torque about the base of ladder

W Cosθ (AB) = F Sinθ (AC)

(200) ($\frac{1}{3}$) (1.2) = F ($\frac{2.83}{3}$) (3)

F = 28.3 N

Using equilibrium of force along the horizontal direction

f = F

f = 28.3 N