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

A. 80 N

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

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles