Question

A rope supporting a mass passes over a horizontal tree branch. The mass exerts a force of 1000 N on the rope. The radius of the branch is 75 mm. The coefficient of static friction between the rope and all the surfaces it contacts is 0.3, and the angle of contacts is 300. What is most nearly the minimum force necessary to hold the mass in position?

Answer 1

The minimum force necessary to hold the mass in position can be calculated using the capstan equation with the coefficient of static friction, the angle of contact in radians, and the known load. By converting the contact angle to radians and applying the given values, we find the minimum force required.

Step-by-step explanation:

To determine the most nearly minimum force necessary to hold the mass in position, we consider the capstan equation, which relates the tensions on either side of the rope as it wraps around the branch. The equation is T1 = T2 * e^(θμ), where T1 is the tension on one side necessary to support the load, T2 is the tension on the other side (the load itself in this case, which is 1000 N), θ is the angle of contact in radians, and μ is the coefficient of static friction.

The angle of contact given is 300 degrees. We need to convert this into radians by multiplying by π/180 to use in the capstan equation. The angle in radians is hence 300 * π/180 = 5π/3 radians.

Now we can find the minimum tension T1 required to hold the mass in place using the capstan equation:
T1 = 1000 * e^(5π/3 * 0.3). By calculating this, we find T1 to be the minimum force necessary to hold the mass in position.