A uniform ladder of mass m = 6.5 kg and length l = 2.1 m leans at an angle against a frictionless wall, Fig. 12-72. If the coefficient of static friction between the ladder and the ground is µs = 0.3, what is the minimum angle at which the ladder will not slip?

To find the minimum angle at which the ladder will not slip, we need to consider the forces acting on the ladder.

1. Weight: The weight of the ladder can be calculated as W = m * g, where m is the mass of the ladder and g is the acceleration due to gravity. In this case, assuming the acceleration due to gravity is 9.8 m/s^2, we have W = 6.5 kg * 9.8 m/s^2 = 63.7 N.

2. Normal force: The normal force is the force exerted by the ground on the ladder perpendicular to the contact surface. Since the ladder is not sinking into the ground, the normal force is equal in magnitude and opposite in direction to the weight of the ladder. So, the normal force is N = 63.7 N.

3. Frictional force: The frictional force acting between the ground and the ladder can be calculated as F_friction = µs * N, where µs is the coefficient of static friction. In this case, µs = 0.3 and N = 63.7 N, so F_friction = 0.3 * 63.7 N = 19.11 N.

4. Force due to the ladder's weight: The component of the ladder's weight parallel to the ground is given by W_parallel = W * sin(θ), where θ is the angle the ladder makes with the ground.

5. Force due to the ladder's weight along the wall: The component of the ladder's weight perpendicular to the ground is given by W_perpendicular = W * cos(θ).

For the ladder not to slip, the frictional force should be equal to or greater than the force due to the ladder's weight parallel to the ground, i.e., F_friction >= W_parallel.

Substituting the values, we have:

0.3 * 63.7 N >= 63.7 N * sin(θ)

19.11 N >= 63.7 N * sin(θ)

Dividing both sides by 63.7 N:

0.3 >= sin(θ)

To find the minimum angle at which the ladder will not slip, we need to find the angle whose sine is equal to or greater than 0.3.

Using the inverse sine function (sin⁻¹), we find:

θ >= sin⁻¹(0.3)

Using a calculator, sin⁻¹(0.3) is approximately 17.46 degrees.

Therefore, the minimum angle at which the ladder will not slip is approximately 17.46 degrees.

To find the minimum angle at which the ladder will not slip, we need to determine the maximum angle at which the static friction force can counteract the ladder's tendency to slip. We know that the static friction force is given by:

fs ≤ µs * N

Where fs is the static friction force, µs is the coefficient of static friction, and N is the normal force exerted on the ladder by the ground.

In this case, the normal force N is equal to the vertical component of the ladder's weight, which is mg, where g is the acceleration due to gravity.

N = mg

Now, we need to determine the maximum static friction force that can be exerted before the ladder slips. This maximum static friction force is when the ladder is about to slip, which means the ladder is in equilibrium. Therefore, the sum of all the forces acting on the ladder must be zero.

Taking the rotational equilibrium about the point where the ladder is in contact with the ground, we can write:

Στ = 0

Where Στ is the sum of the torques. In this case, torsional equilibrium means that there is no net torque acting on the ladder.

The only torque acting on the ladder is the torque due to the weight acting at the center of the ladder. The weight force can be represented as a point force acting downward at the center of mass, which is at a distance l/2 from the point of contact with the ground.

The torque due to weight can be calculated as:

τ = mg * (l/2) * sinθ

Where τ is the torque, m is the mass of the ladder, l is the length of the ladder, and θ is the angle the ladder makes with the ground.

Since there is no net torque acting on the ladder, the torque due to weight must be balanced by the torque due to the static friction force:

τ = fs * l * cosθ

Now, we can equate the expressions for torque due to weight and torque due to static friction:

mg * (l/2) * sinθ = fs * l * cosθ

Simplifying the equation, we find:

fs = (mg * sinθ) / (2 * cosθ)

Substituting the expression for fs from earlier:

(µs * N) = (mg * sinθ) / (2 * cosθ)

Substituting the expression for N:

(µs * mg) = (mg * sinθ) / (2 * cosθ)

Canceling out mg:

µs = (sinθ) / (2 * cosθ)

Rearranging the equation:

2 * µs * cosθ = sinθ

Taking the inverse sine of both sides:

θ = arcsin(2 * µs * cosθ)

By substituting µs = 0.3 into the equation, you can solve for the minimum angle at which the ladder will not slip.