a uniform ladder of mass m and length l leans against fricitonless wall angle theta. If the coefficient of static friction between the ladder and the ground is u, find the formula for the minimun angle such that the ladder won't slip
Arctan(1/2u)
To find the minimum angle at which the ladder won't slip, we need to determine the maximum value of the angle theta for which the frictional force can prevent slipping.
Let's start by drawing a free body diagram of the ladder. The two forces acting on the ladder are the weight (mg) acting vertically downwards and the normal force (N) acting vertically upwards. Additionally, there is a frictional force (F) acting parallel to the ground.
Given that the ladder is in equilibrium, the sum of the forces in the vertical direction is zero:
N - mgcos(theta) = 0 (equation 1)
To prevent the ladder from slipping, the frictional force must provide a clockwise torque about the point of contact with the ground. The torque due to the frictional force is given by:
τ = F * l * sin(theta) = 0 (equation 2)
Since the ladder is frictionless against the wall, there is no torque about that point. Therefore, the net torque about the point of contact with the ground must be zero.
Now, for equation 1, let's solve for N:
N = mgcos(theta)
Substituting this back into equation 2, we get:
F * l * sin(theta) = 0
Since mass cancels out, we can rewrite the equation as:
F * l * sin(theta) = mgsin(theta)
Next, we want to find the maximum value of the static frictional force, Fs (given by Fs = uN), that can prevent slipping. Substituting the expression for N, we have:
u * mgcos(theta) = mgsin(theta)
Simplifying, we obtain:
ucos(theta) = sin(theta)
Dividing both sides by cos(theta):
u = tan(theta)
The minimum value of theta at which the ladder won't slip is the angle for which the coefficient of static friction is equal to the tangent of the angle:
theta = arctan(u)
Therefore, the formula for the minimum angle is:
theta = arctan(u)