Equilibrium - Rotational Motion (Upward Force)

Question

A 5.77 kg rod is 0.625 m long, and attached at one end to a wall with a hinge, so it can be rotated up and down. How much upward force must you exert at the end of the rod to hold it horizontal?

Answer 1

т= p. F• sin(0)

where:

1. r is the distance from the pivot (hinge) to the point where the force is applied,
2. 1 is the force applied,
3. is the angle between the lever arm and the force.

For a horizontal rod, 0 = 90°, and sin(90°) = 1. So the torque becomes:

T=p.F

r.F=p•W

Now, solve for the force F:

Given the mass of the rod (m = 5.77 kg), the

length of the rod (r = 0.625 m), and the

acceleration due to gravity (g= 9.8 m/s,

you can substitute these values into the formula to find the upward force F which is 5.77kg

Answer 2

mgl/2 = Fl; F = mg/2 = 5.77kg·9.81m/s²/2 = 28.3 N