Balancing Forces | Min Force to Hold Rod at Point A

[saltine]

Homework Statement

What is the minimum force required to hold the rod at point A?

Homework Equations

The Attempt at a Solution

If no one is holding the rod, point B would get the force mg from the rod. So the normal force from the hinge to the rod is mg, pointing up. This normal force would create a torque τ, which will cause the rod to rotate.

τ = (L/2)(mg cos θ)

The force that holds the rod in place counters this torque by having the component

F_τ = mg cos θ

that is perpendicular to the rod, in the clockwise direction. However, now the three forces acting on the rod--gravity, the normal force from the hinge, and the force from the hand--no longer sum up to zero. What is wrong?

- Thanks

 

[turin]

Forget about B. Just use torques, and take for granted that B is the pivot point. Note: there is a range of forces in different direction that will hold the rod at A. You need to find the minimum.

 

[saltine]

Ok, suppose there is a force F_a at A, and we some the torque about B:

(L/2)(mg cosθ) - LF_a = 0

So the tangential component of F_a must be 0.5 mg cosθ. Suppose I let F_a to have no radial component and let the hinge do the rest, then:

F_bx = - F_asinθ

F_by = mg - F_acosθ

Since the torque was summed at B, the force at B won't change the net torque. This shows that the x and y direction forces are also balanced. So F_a must be 0.5 mg cosθ, and it is tangent to the rod in clockwise direction.