A uniform ladder of length (L) and weight 100 N rests against a smooth vertical wall. The coefficient of static friction between the bottom of the ladder and the floor is 0.5. Find the minimum angle theta , which the ladder can make with the floor so the ladder will not slip.
To find the minimum angle (θ) at which the ladder will not slip, we need to consider the forces acting on the ladder.
Let's analyze the forces acting on the ladder:
1. Weight (W) acting downward with a magnitude of 100 N.
2. Normal force (N) acting perpendicular to the floor and exerted by the floor on the ladder.
3. Frictional force (F) acting parallel to the floor and exerted by the floor on the ladder.
Since the ladder is at rest, the forces acting vertically must balance each other. Therefore, N = W.
The frictional force can be calculated using the formula: F = μN, where μ is the coefficient of static friction.
For the ladder not to slip, the frictional force F must be sufficient to oppose the tendency of the ladder to slip. This means that the frictional force needs to be equal to or greater than the horizontal component of the weight.
The horizontal component of the weight can be calculated as: Wh = W * sin(θ).
Now, let's equate the frictional force F to Wh and solve for θ. We have:
F = μN
μN = Wh
μW = W * sin(θ)
μ = sin(θ)
To find the minimum angle θ, we need to find the maximum value of the coefficient of static friction (μ). The maximum value of μ is given as 0.5 in the problem statement.
Using the equation μ = sin(θ), we can rewrite it as:
0.5 = sin(θ)
To find the minimum angle θ, take the inverse sine (sin^-1) of both sides of the equation:
θ = sin^-1(0.5)
Calculating θ using a calculator, we find that the minimum angle θ is approximately 30 degrees.
In order to solve this problem, we need to analyze the forces acting on the ladder and use the conditions for static equilibrium.
Let's consider the ladder in a vertical position, making an angle theta with the floor. The weight of the ladder (W) can be decomposed into two components: one parallel to the floor (W_parallel = W * cos(theta)) and one perpendicular to the floor (W_perpendicular = W * sin(theta)).
The static friction force (F_friction) acts at the bottom of the ladder and opposes the tendency of the ladder to slip. The maximum value of static friction force (F_friction_max) can be calculated using the coefficient of static friction (μ) and the normal force (N) exerted by the floor on the ladder's bottom.
The normal force (N) is equal to the perpendicular component of the ladder's weight (N = W_perpendicular).
In order for the ladder not to slip, the maximum static friction force (F_friction_max) must be greater than or equal to the parallel component of the ladder's weight (W_parallel). Mathematically, this can be expressed as:
F_friction_max ≥ W_parallel
Let's substitute the values and solve for theta:
F_friction_max = μ * N
W_parallel = W * cos(theta)
N = W * sin(theta)
Now we can write the inequality equation:
μ * N ≥ W * cos(theta)
Substituting the values:
0.5 * (W * sin(theta)) ≥ W * cos(theta)
Canceling out the common factor of W:
0.5 * sin(theta) ≥ cos(theta)
Dividing both sides by cos(theta):
0.5 ≥ tan(theta)
Now we can find the angle theta by taking the inverse tangent of both sides:
theta ≥ arctan(0.5)
Using a calculator, the minimum angle theta is approximately 26.57 degrees.
Therefore, the minimum angle theta, which the ladder can make with the floor so that the ladder will not slip, is approximately 26.57 degrees.
Set the moment about the wall contact point be zero. Since the wall is smooth, the vertical force floor component at the floor equals the weight, W = 100 N.
W*(L/2)costheta + mus*W*L*sintheta = W*L*costheta
(1/2)costheta = mus*sintheta
tantheta = 1/(2*mus)