Question

Consider a bar, of length 5.5 m, shown in the figure, being acted on by three forces. The magnitudes of the first two forces are 16 N and 29 N, and the first force is acting on the end of the bar at an angle of 29° as shown. 

Part (a) What is the torque, in newton-meters, due to F1 on this bar relative to the left end? Use a coordinate system with positive directed out of the screen. 

Part (b) What is the torque, in newton-meters, due to F2 on this bar relative to the left end, if this force is acting at the midpoint of the bar? 

Part (c) What is the magnitude of the force F3, in newtons, if the force is a distance 0.95 m from the left end and the bar is not rotating?

Answer 1

Given,

The first two forces are, F₁ = 16 N and F₂ = 29 N

The length of a bar, L = 5.5 m

The angle, = 29⁰

Part a:

The torque, ₁ = F₁ L sin

= 16 * 5.5 * sin 29

= 42.66 N m

Part b:

The torque, ₂ = F₂ * (L / 2)

= 29 * (5.5/2)

= 79.75 N m

Part c:

Given,

The distance, d = 0.95 m

We know,

₁ +  ₂ - ₃ = 0

Where, ₃ = F₃ * d

₁ +  ₂ -  F₃ * d = 0

42.66 + 79.75 - F₃ * 0.95 = 0

0.95 F₃ = 122.41

F₃ = 128.85 N