This problem on Rotational motion is bugging me.

[Orriofeden]

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.

http://img253.imageshack.us/img253/3585/physicsstepladderiy5.jpg

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.

 

[anathema]

Thank you for your post. I understand your confusion regarding part A of this problem. Let me guide you through the solution step by step.

First, let's consider the ladder as a single object. We know that the painter has a mass of 70 kg and is standing 3 meters from the bottom of the ladder. This means that the force of gravity acting on the painter is 70 x 9.8 = 686 N. This force is acting downwards, so we can represent it as a vector pointing downwards from the center of mass of the ladder.

Next, let's look at the horizontal bar connecting the two halves of the ladder. This bar is exerting a force on each half of the ladder, keeping them together. Since the ladder is in equilibrium, the sum of all forces acting on it must be equal to 0. This means that the force exerted by the bar on the left half of the ladder must be equal in magnitude and opposite in direction to the force exerted by the bar on the right half of the ladder.

Now, let's look at the torque equation. We know that the torque is equal to the force multiplied by the distance from the pivot point (in this case, the hinge at C). Since the distance from the pivot point to the center of mass of the ladder is 3 meters, and the force of gravity is acting downwards, we can say that the torque due to the force of gravity is 686 x 3 = 2058 Nm. This torque is acting in a clockwise direction.

To balance this torque, the bar must exert a force in the opposite direction, with a magnitude of 2058/2 = 1029 N on each half of the ladder. This is the answer to part A of the problem.

For part C, we can treat each half of the ladder separately. For the left half, the reaction force at the hinge C is equal in magnitude and opposite in direction to the force exerted by the bar, which we have already calculated to be 1029 N. This reaction force is acting upwards. For the right half of the ladder, the reaction force at the hinge C is equal to the force of gravity acting on the painter, which we have already calculated to be 686 N. This reaction force is also acting upwards.

I hope this explanation helps you understand the problem better. Let me know if you have any further questions. Good luck with

 

[Moetasim]

First of all, it's great that you're approaching this problem by breaking it down into smaller parts and using the relevant equations. It's also good that you've already solved part b successfully.

For part a, you need to consider the forces acting on the horizontal bar connecting the two halves of the ladder. Since the painter is standing 3 meters from the bottom, you can use the equation F = ma to find the horizontal force exerted by the painter on the bar (remember to use his weight, 70 kg, as the mass). This force must be balanced by the tension in the bar, so you can set F = T and solve for T.

For part c, you can treat the left half of the ladder as a separate object and use the equation F = ma to find the horizontal and vertical components of the reaction force at the hinge C. Then, you can use the fact that the total reaction force at C must be equal to the sum of the forces from both halves of the ladder to find the remaining component of the reaction force.

Remember to draw free body diagrams for each part of the ladder and apply Newton's laws to find the relevant forces. Good luck!