How to Calculate Normal Forces in a Uniform Board System

[astru025]

Homework Statement

Cliff and Will are carrying a uniform 2.0m board of mass 77kg. Will is supporting the board at the end while cliff is 0.6m from the other end as shown in the following figure. Cliff has attached his lunch to his end of the board, and the tension in the string supporting the lunch is 207N. Find the normal forces exerted by Cliff and Will.

Homework Equations

Are tension and force the same thing? So I would need to find tension exerted by cliff and will.. All the tensions must add up to equal 1000N. I don't know where to start with equations and such.

The Attempt at a Solution

Tried finding the torques but I don't believe that has anything to do with it. Any help would be much appreciated thanks!

 

[PhanthomJay]

The forces exerted by gentlemen on the board are usually called called normal forces. They can act up or down. The dowmward force exerted by the string on the Board (207 N) is called a tension force (tension forces always pull away from the object on which they act). You've got to sum torques about any point = 0. And the sum of all vertical forces must add to zero. Watch plus and minus signs.

 

[astru025]

So I must find the torques of cliff and will?

 

[PhanthomJay]

You can sum torques about any point, but it is easier to sum torques about either Cliff or Will. Try summing torques about Cliff. This way, Cliff exerts no torque about himself, so you can quickly solve for the force exerted by Will.

 

[astru025]

Okay can you help me get started with summing torques about cliff. I'm really struggling with this particular material.

 

[astru025]

I found the torque of the lunch. .6 m x 207 N = 124.2. Is this correct?

 

[PhanthomJay]

Thh torque of a force about a point is force times perpendicular distance, clockwise is a minus torque and counetrclockwise is a positive torque. There are two forces that exert torques about Cliff, one is the known tension force, and the other is the unknown force exerted by Will.

 

[astru025]

Okay so the torque to the right of cliff = 124.2 is that correct?

 

[PhanthomJay]

Okay so the torque to the right of cliff = 124.2 is that correct?

Yes, clockwise.

 

[astru025]

So that torque is -124.2. What do I do from here?

 

[PhanthomJay]

Well, the torques from all forces about Cliff must sum to 0, so what is the torque from Will about Cliff?

 

[astru025]

Positive 124.2?? Then what... I'm sorry I'm really struggling with this and this problem is due in an hour

 

[astru025]

So if I know these 2 torques how do I find the force exerted by will ?

 

[PhanthomJay]

Yes, counterclockwise. Will's force acts 1.4 m from Cliff. So the force exerted by Will must be how much? And up or down?

 

[astru025]

1.4 x 124.2? = 173.88 which is up. Is this correct?

 

[PhanthomJay]

1.4 x 124.2? = 173.88 which is up. Is this correct?

No, you are rushing too much and not paying attention. Torque is force times distance, so force = torque diviided by distance.
T = Fd
F =T/d
F = ?? Up or down?

 

[astru025]

F=t/d. F= 124.2/ 1.4 F=88.7 N. in the upward direction

 

[astru025]

My answer was incorrect. What am I doing wrong?

 

[PhanthomJay]

Oh darn it's me who is rushing because I am trying to hit the sack and watch the Sox lose. We forgot to include the 770 N weight of the board. It too exerts a torque about Cliff, the weight of the board acts down at its center , or 0.4 m left of Cliff. So the torque from the weight is 770(.4) = 308, ccw. So we have 124.2 cw, and 308 ccw, and the torque from Will.

-F(1.4) +308 -124.2 = 0
Solve F (Will) = 131, up

 

[astru025]

Wow yes just came up with that same answer. Then the 2 forces have to add to 1000N so the force exerted by Cliff is 869N! Thank you so much! Have a great night!

 

[PhanthomJay]

Well, the 2 forces have to add to 770 + 207 =977. So Chris force up is 846. I rounded a bit. Sorry for confusion (yawn).

 

[astru025]

Can't thank you enough!