Numerical Problem 8, Vectors and Equilibrium … msa

Problem 8: A 4.0 m long ladder with weight of 120 N leans against a wall making 70° above a cement floor as shown in the figure. Assuming the wall is frictionless but the floor is not determine the forces exerted on the ladder by the floor and by the wall.

Solution

Given

Weight of ladder, W = 120 N

Length of ladder, ℓ = 4 m

Center of gravity of ladder = 2 m

Required

Forces exerted on the ladder by the wall and floor.

So, to find the forces of the wall and floor, we apply the conditions of equilibrium. From 1^st condition,

∑F_x = 0 ⇒ F_x – F_wall = 0 ⇒ F_x = F_wall … (A)

∑F_y = 0 ⇒ F_y – W = 0 ⇒ F_y = 120 N … (B)

Apply second condition of equilibrium, ∑τ = 0.

Now to calculate torques due to different forces,

AC × W = ACWsinθ = τ _clockwise

Now from simple geometry rules, we see θ = 20°. (Otherwise, see here).

Therefore,

ACWsin20° = τ_clockwise

AB × F_wall = ABF_wallsin70° = τ_{counterclockwise}

Thus

ACWsin20° = ABF_wallsin70° … (C)

Now put the values,

2 × 120 × 0.342 = 4 × F_wall × 0.939 ⇒ 80.08 = 3.576 × F_wall ⇒ F_wall = (80.08/3.576) = 21.32 N

This is the reaction of the wall.

To find reaction of the floor, F, we have,

Since, F_x = F_wall = 21.32 and F_y = W = 120, therefore,