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?
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.