# Finding tension/force

1. Oct 12, 2013

### astru025

1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

3. 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!

2. Oct 12, 2013

### 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.

3. Oct 12, 2013

### astru025

So I must find the torques of cliff and will?

4. Oct 12, 2013

### 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.

5. Oct 12, 2013

### astru025

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

6. Oct 12, 2013

### astru025

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

7. Oct 12, 2013

### 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.

8. Oct 12, 2013

### astru025

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

9. Oct 12, 2013

### PhanthomJay

Yes, clockwise.

10. Oct 12, 2013

### astru025

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

11. Oct 12, 2013

### PhanthomJay

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

12. Oct 12, 2013

### astru025

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

13. Oct 12, 2013

### astru025

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

14. Oct 12, 2013

### 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?

15. Oct 12, 2013

### astru025

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

16. Oct 12, 2013

### PhanthomJay

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?

17. Oct 12, 2013

### astru025

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

18. Oct 12, 2013

### astru025

My answer was incorrect. What am I doing wrong?

19. Oct 12, 2013

### 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

20. Oct 12, 2013

### 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!

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