# This problem on Rotational motion is bugging me.

## Homework Statement

A stepladder of negligible weight is constructed as shown. A painter of mass 70 kg stands on the ladder 3 meters from the bottom. Assuming that the floor is frictionless, find (a) the tnesion in the horizontal bar connecting the two halves of hte ladder, (b) the normal forces at A and B, and (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. (Suggestion: Treat the ladder as a single object, but also treate each half of the ladder seperately)

This is the picture associated.

## Homework Equations

Well, mg = Na + Nb, I can say that much.
sum of Torque = 0
Torque = F*d*sin(theta)

## The Attempt at a Solution

So far, I have part b successfully completed. Since mg = Na + Nb, I can take the entire ladder as a single entity, and use torque as an equation, setting the distance at Na equal to 0.

By doing that, and then filling in the blanks, I get Na = 429.18, and Nb = 257.5. I don't know what to do for part A. I've moved my reference frame upwards towards the hinges, rather than the entire ladder, and did it with just the distance of two meters.

I can say that the force exerted towards the right has to equal the force exerted towards the left. But I can't figure out how to get the forces on the left or the right.

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