Solving an Equilibrium Problem Involving Torque: Need Help!

[findley]

Hello all

I am working on some homework and I am having a problem with a question that involves torque. Here is the question.

A 62.5 kg painter stands 1.35 m from one end of a 45.0 kg scaffold that is 4.50 m long. He has a 11.0 kg supply of paint 1.00 m from the end he is nearest to. Find the reaction forces at the supports if the scaffold is supported at each end.



I have no idea how to set this problem up once I get the weights of each of the objects.

I understand that because the scaffold is in equilibrium that the sum of forces are equal to zero. I just have no idea where to go next.

Any help would be greatly appreciated.

Thanks a lot!

 

[Filip Larsen]

It may help to draw a diagram with all the forces involved, including the reaction forces and torques from the support. You should probably also think about how a single downward force between the two endpoints is distributed to two reaction forces at the two endpoints (hint: think about what torque the force gives at either end of the support, and how this torque then translates to a force at the other end (I am assuming that the support ends are free to rotate)).

 

[findley]

I made a diagram but I wasn't sure where to put the axis of rotation. I always thought the middle was best, but I have been told to use one end of the scaffold. I did that but I wasn't sure what to do next. Do I multiply the distances by the weights all the way down the 4.5m scaffold?Thanks.

 

[Kabbotta]

Put the axis of rotation at the contact point of any of the forces on the scaffold. I believe there are four in this problem, both supports, the paint supply, and the painter himself.

\tau = \textbf{r x F}

So, if \textbf{r} = 0 for one of your torques then it can be removed from the calculation.

After that add up the rest of the forces that cause a torque, any force that acts at a distance from your axis of rotation, and set them equal to zero since the bridge is in static equilibrium.

Also remember that you will get needed information from Newton's regular second law equation for the bridge. In the end you have three separate equations to help you solve this,

Newton's Second law y-direction
Newton's Second Law x-direction
Net Torque about axis of rotation

 

[findley]

I came out with an answer of 428N on the side furthest from the painter/paint can and 734N on the other side. Does that look like the right answer?

Thanks!

 

[Kabbotta]

It seems reasonable, you'd expect the side with the painter and his can would experience a much greater force relative to the other support.

A good way to check is to just return to one of those three equations, see if you when you plug all your numbers in it really equals zero. Otherwise, you'll know something would be moving in your painter's world ( i.e. a NOT static scaffold )

The only thing you might have missed is that if you did not pick the center of the scaffold as your axis of rotation then the force of gravity on the scaffold would also cause a torque, since r!=0 anymore for that force.

Hopefully that doesn't mess things up if you already got it figured out.

 

[findley]

Well I found the total force of the scaffold including the weight of the painter and paint can was 1162N which both reaction forces equal. I guess I'll find out tomorrow.

Thanks a lot for your help. It is grealty appreciated!

 

[Kabbotta]

Yep, you got it.