Question

A ladder of length 10 is leaning against a wall. The vertical wall is frictionless while the horizontal surface is rough. The angle between the ladder and the horizontal surface is 50 degrees. If the mass of the ladder is 14 kg, what is the magnitude of the friction force (in N) exerted on the ladder in the point of contact with the horizontal surface

Answer 1

Answer:

105.1N

Explanation:

According to the newton second law

$\sum F_x = ma_x\\F_m - F_f = ma_x$

Since the acceleration is zero, then;

$F_m - F_f = 0\\Fm = F_f$

Since;

$F_m = mg sin \theta, \ hence \ F_f = mg sin \theta$

Given the following

mass of ladder m = 14kg

acceleration due to gravity g = 9.8m/s²

θ = 50°

Substitute

$F_f = 14(9.8)sin50\\F_f = 14(7.5072)\\F_f = 105.1N$

Hence the magnitude of the friction force (in N) exerted on the ladder in the point of contact with the horizontal surface is 105.1N