Question

A uniform plank 4 m long is pivoted about its center to an angle of 30° with the vertical. What magnitude of force

must be applied 0.5 m from the pivot to maintain equilibrium with a 4 kg mass hanging from one end?
A. 80 N
B. 20 N
C. 40 N
D. 60 N
E. 100 N

Answer 1

A. 80 N

Step-by-step explanation:

Since the uniform plank is pivoted at its center at an angle of 30 to the vertical, then it is 90° - 30° = 60° to the horizontal.

Let L be the length of the plank = 4 m, the perpendicular distance of the 4 kg mass form the pivot point is Lcos60°/2. Then the moment of the 4 kg mass about its pivot point is mgLcos60/2 where m = mass = 4 kg and g = acceleration due to gravity = 10 m/s²

Let F be the force applied at 0.5 m from the pivot point. The perpendicular distance of F from the pivot point is 0.5cos60°. Thus the moment of this force about the pivot point is F × 0.5cos60° = 0.5Fcos60°

Since both moments are the same for equilibrium,

mgLcos60/2 = 0.5Fcos60°

F = mgL/(2 × 0.5 m)

F = 4 kg × 10 m/s² × 4 m/1 m

F = 80 kgm/s²

F = 80 N