A ladder of mass 39.0 kg and length 3.18 m is leaning against a wall at an angle θ. The coefficient of static friction between ladder and floor is 0.298; assume that the friction force between ladder and wall is zero. What is the maximum value that θ can have before the ladder starts slipping?
To determine the maximum value of θ before the ladder starts slipping, we need to consider the rotational equilibrium and the friction force acting on the ladder.
Let's start by analyzing the forces acting on the ladder. We have the weight of the ladder acting downward (W = mg), and the normal force (N) and friction force (F) acting upward on the ladder.
Since the friction force opposes the tendency of sliding, it can be calculated as the product of the coefficient of static friction (μs) and the normal force (F = μsN).
In this case, the normal force is equal to the vertical component of the ladder's weight (N = mgcosθ) and the friction force is equal to the product of the coefficient of static friction and the normal force (F = μsN = μsmgcosθ).
The ladder will start slipping when the friction force reaches its maximum value, which occurs when it equals the product of the coefficient of static friction and the normal force, i.e., μsmgcosθ. At this point, static equilibrium is lost, and the ladder begins to slide down the wall.
The maximum value for θ is achieved when the friction force reaches its maximum value. Therefore, we can set the friction force equal to its maximum:
μsmgcosθ = μsN
Substituting the expression for N:
μsmgcosθ = μsmgcosθ
Simplifying the equation, we find:
cosθ = 1
For the maximum value of cosθ (which occurs at θ = 0°), cosθ = 1.
Hence, the maximum value of θ before the ladder starts slipping is 0°.