A ladder 10-m long and weighing 50 N rests against a smooth vertical wall. If the ladder is just on the verge of slipping when it makes a 50 degree angle with the ground, find the coefficient of static friction between the ladder and the ground.
To solve this problem, we need to use the concept of equilibrium and the condition for an object to be on the verge of slipping.
Let's break down the forces acting on the ladder:
1. Weight (W): The weight of the ladder acts vertically downwards and is given as 50 N.
2. Normal Force (N): The normal force acts perpendicular to the surface the ladder is resting on (the ground). Since the ladder is not sinking or going through the ground, the normal force is equal in magnitude but opposite in direction to the weight.
3. Frictional Force (F): The frictional force acts parallel to the ground (opposite the direction of potential slipping). This is what we want to find.
Now, let's analyze the forces acting on the ladder in equilibrium:
1. Vertical Forces: The sum of the vertical forces must be zero since the ladder is not sinking into the ground nor floating away. Therefore, the vertical components of weight and normal force must cancel each other out:
N - W*cosθ = 0
2. Horizontal Forces: The horizontal component of the weight and the frictional force are in equilibrium (no slipping):
F - W*sinθ = 0
where θ is the angle between the ladder and the ground, which is given as 50 degrees.
Now we can solve the equations:
From the vertical forces equation, we can solve for N:
N = W*cosθ
= 50 N * cos(50°)
≈ 32.26 N
From the horizontal forces equation, we can solve for F:
F = W*sinθ
= 50 N * sin(50°)
≈ 38.26 N
The frictional force F is also equal to μ*N, where μ is the coefficient of static friction. Substituting the values, we have:
μ*N = F
μ * 32.26 N = 38.26 N
μ ≈ 1.19
Thus, the coefficient of static friction between the ladder and the ground is approximately 1.19.