A 17kg crate is to be pulled a distance of 20m, requiring 1210J of work to be done. If the job is done by attaching a rope and pulling with a force of 75N, at what angle is the rope held?

Answer 1

To find the angle at which the rope is held, we first need to find the force of friction that is acting against the crate.

Work = force x distance x cos(angle)
1210J = 75N x 20m x cos(angle)

cos(angle) = 1210J / (75N x 20m)
cos(angle) = 1.35

Since the value of cos(angle) cannot exceed 1, it means there was an error in previous calculations. Let's recalculate it.

Frictional force = Work / Distance
Frictional force = 1210J / 20m
Frictional force = 60.5N

To calculate the angle at which the rope is held, we can use the equation:

W = Fd cos(theta) + Fu d

Where:
W = work done (1210J)
F = applied force (75N)
d = distance (20m)
theta = angle between the force and the displacement (what we are trying to find)
Fu = force of friction (60.5N)

1210J = 75N x 20m x cos(theta) + 60.5N x 20m
1210J = 1500 N.m x cos(theta) + 1210 J
0 = 1500 N.m x cos(theta) - 1210 J
1500 N.m x cos(theta) = 1210 J
cos(theta) = 1210 J / 1500 N.m

theta = arccos(1210/1500)
theta ≈ arccos(0.8067)
theta ≈ 36.7 degrees

Therefore, the rope is held at an angle of approximately 36.7 degrees.