- #1

Curious3

- 7

- 0

**Figure:**

1. Homework Statement

1. Homework Statement

We know system is in equilibrium and we are given the mass m, and the distances d, r1, and r2.

We assume that the rotational inertia of the bar is negligible.

What are the forces at points a and b?

## Homework Equations

Torque = radius × Force

Force = mass * acceleration

ΣFi = 0

ΣTi =0

where i = {x,y,z}

## The Attempt at a Solution

The force of the mass is F=mg, therefore the torque produced is T = d×(mg) = d*mg*sin(pi / 2) = dmg. (Where g is 9.81 m/s^2.)

Since we are in equilibrium we know r1*Fa+r2*Fb=T. Where Fa and Fb are the forces on the bar at the points a and b respectively.

Now how is the opposing torque shared between the two points?

Perhaps the torques r1*Fa and r2*Fb are equal, or perhaps Fa and Fb are equal?