PS 4.3 – Torque (Rotational Equilibrium) Problems
1. A crane supports a 3.0 kN weight as shown. The crane's boom is 8.0 m long and is at an angle of 30˚ from the horizontal. If the hydraulic support is attached to the boom 2.0 m from the bottom of the boom, then: a) what is the compressional force in the hydraulic?
b) what is the magnitude and direction of the force on the pin at the bottom of the boom?
a) ∑Torque = 0
F cos 30˚ x 2.0 = 3000 cos 30˚ x 8.0
F = 24 000 ÷ 2.0
F = 12 000 N
b) ∑Fy = 0 (the sum of the forces up must equal the sum of the forces downward)
V + 3000 = 12000
V = 9000 N down
2. A crate is 8.0 m tall and 3.0 m wide. Its coefficient of friction to the floor is 0.70. What is the maximum distance from the floor a horizontal force may be used to slide the crate without tipping it?
∑T = 0 ∑Fx = 0 F x d = mg(1.5) F = Ffr
F = µmg µmgd = mg(1.5)
d = 1.5 ÷ 0.7
d = 2.1 m
Sample Problem 1: One mass on a See-Saw
A 3.0kg mass is place 2.00m to the right of the pivot point of a see-saw. What is the the magnitude and the sign of the torque applied?
This problem looks like the figure
The force exerted by the mass is due to gravity and is found from F=mg. The distance between the force and the pivot point is r=2.00m. We can find the magnitude of the torque by
If the mass is to the right of the pivot point, the rotation will be in a clockwise direction, so the torque is negative: . As always, note the use of significant figures. The distance r was given to three significant figures, but the mass (and therefore the force) is only known to two significant figures. Thus the torque must have only two significant figures. EXAMPLE PROBLEM ON TORQUE: The Swinging Door
In a hurry to catch a cab, you rush through a frictionless swinging door and onto the sidewalk. The force you extered on the door was 50N, applied...