Question

A uniform ladder 10 m long weighing 50 n rests against a smooth vertical wall. the ladder starts to slip away when its inclination angle is 50° with the floor. what is the coefficient of static friction at the floor?

Answer 1

To calculate the coefficient of static friction for a 10 m ladder weighing 50 N inclined at 50°, we use the equilibrium equations for forces and torque. After solving, we can find the coefficient of static friction that prevents the ladder from slipping.

Step-by-step explanation:

To determine the coefficient of static friction when a 10 m long ladder weighing 50 N slips at an inclination angle of 50°, we need to apply the concepts of equilibrium and static friction.

Two reaction forces are acting on the ladder: the normal force (N) from the ground and the normal force from the wall. Since the wall is smooth, there is no frictional force at the wall's contact point, but there is a frictional force (f) at the base of the ladder on the ground. The weight of the ladder (50 N) acts at the center of the ladder. To find the coefficient of static friction (μs), we will use the following equations for equilibrium:

• ∑Fx = 0: f - Fwall = 0, where f = μsN
• ∑Fy = 0: N - w = 0
• ∑τ = 0: w * (L/2) * cos(β) = μsN * L * sin(β)

Using these equations, we can calculate N and then substitute into the torque equation to solve for μs.

By setting up the free-body diagram and applying the rules for static equilibrium, we find that:

N = w = 50 N (since the ladder is in equilibrium, the vertical forces must balance).

Thus, μsN = w * (L/2) * cos(β) / (L * sin(β))

Substituting the values, we get:

μs = (50 N * (10 m / 2) * cos(50°)) / (50 N * 10 m * sin(50°))

After performing the calculations, we obtain the coefficient of static friction required to prevent the ladder from slipping at a 50° angle with the floor.