A uniform ladder 10.0m long and weighing 50.0N 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 ground.
The coefficient of static friction is equal to the ratio of the normal force to the force of gravity.
Normal force = 50.0N * cos(50°) = 43.3N
Force of gravity = 50.0N * sin(50°) = 43.3N
Coefficient of static friction = 43.3N / 43.3N = 1.0
To find the coefficient of static friction between the ladder and the ground, we need to consider the forces acting on the ladder.
Let's assume:
- The weight of the ladder is acting downward, denoted as W.
- The normal force between the ladder and the ground is acting perpendicular to the ground, denoted as N.
- The frictional force between the ladder and the ground is acting parallel to the ground, denoted as F.
Since the ladder is on the verge of slipping, the maximum value of friction is given by:
F_max = μ_s * N
where μ_s is the coefficient of static friction.
To find the normal force:
N = W * cos(θ)
where θ is the angle made by the ladder with the ground.
N = 50.0 N * cos(50°)
N = 50.0 N * 0.6428
N ≈ 32.14 N
Now, we can substitute the value of N into the equation for the maximum frictional force.
F_max = μ_s * N
F_max = μ_s * 32.14 N
Since the ladder is on the verge of slipping, the maximum value of frictional force is equal to the maximum value of static frictional force:
F_max = μ_s * N
μ_s * 32.14 N = μ_s * 32.14 N
Therefore, the coefficient of static friction between the ladder and the ground is 1.0 (since both sides of the equation are equal for any value of μ_s).
To find the coefficient of static friction between the ladder and the ground, we can use the concept of equilibrium. When the ladder is on the verge of slipping, the forces acting on it must balance out. Let's analyze the forces involved:
1. Weight of the ladder (mg): The weight of the ladder is given as 50.0N. This force acts vertically downward.
2. Normal force (N): The normal force exerted by the ground on the ladder acts perpendicular to the surface of contact (i.e., vertically).
3. Frictional force (f): The static friction experienced by the ladder opposes its tendency to slip. It acts parallel to the surface of contact between the ladder and the ground.
To begin, let's resolve the forces acting along the ladder:
Horizontal forces:
The only force acting horizontally is the frictional force (f).
Vertical forces:
The vertical forces acting on the ladder are the weight (mg) and the normal force (N).
The angle the ladder makes with the ground is given as 50 degrees. Since the ladder is on the verge of slipping, the angle it makes with the wall will be 90 degrees to the ground.
Now, let's resolve the forces along the ladder:
Horizontal forces:
The horizontal component of the weight (mgcosθ) must be equal to the frictional force (f).
Vertical forces:
The vertical component of the weight (mgsinθ) must be equal to the normal force (N).
Since the ladder is in equilibrium, the sum of the moments about any point (let's take the point where the ladder touches the ground) must be zero.
Taking moments about the bottom of the ladder:
Clockwise moments: None because the only force acting at the bottom is along the ladder.
Counterclockwise moments: The frictional force (f) and the vertical component of the weight (mgsinθ) contribute to the counterclockwise moments.
Using the equation:
f * l - m * g * sinθ * l/2 = 0,
where
f = frictional force,
l = length of the ladder,
m = mass of the ladder,
g = acceleration due to gravity,
θ = angle the ladder makes with the ground,
we can rearrange the equation to solve for the coefficient of static friction (μs):
μs = (m * g * sinθ) / (f * l) = (50.0N * sin(50°)) / (f * 10.0m).
Finally, substitute the given values and solve for the coefficient of static friction (μs).