# Static using Torque

1. Nov 29, 2008

### phys1618

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

a uniform beam 5 meters long and weighing 60N carries a load of 80N at one end and 180N on the other end. if it is to beheldhorizontal while resting on a wooden saw horse 1.3 meters from the heavier laod, where would you have to hold it if you are exerting 100N of force?

2. Relevant equations
Sum Torque counterclockwise= Sum of Torque clockwise
Torque= Fd

3. The attempt at a solution

from this problem, I've drew a diagram, which showed 5 different forces. Fsmall-end=80N,Fbig-end=180N, Fme=100N,Fbeam=60N,Fwooden-saw-horse=is unknown but is 1.3m from the big end. The 2 Forces that going up is 110N and Fwoodensaw-horse. The 3 Forces going down is 80N, 60N, and 180N. My teacher mentioned that if we use the Torque equations then we would have to choose a Fulcrum. I chose the wooden saw horse point as a Fulcrum, since the F is unknown there. So I did the whole problem using the Torque equation and the answer came out to be d=1.34m. That means that I, exerting a force of 100N would have to hold the beam at 1.34m from the small end to keep the beam horizontal(balanced). Is my answer correct? or am i missing something and did it wrong? thank you for the help!

Last edited: Nov 29, 2008
2. Nov 30, 2008

### horatio89

Ok, I see you've got the idea that for the board to remain at equil., the moments about any point must be zero, and you've aptly chosen the sawhorse point where there's an unknown force acting to eliminate the unknown force. So far, so good...

I believe your confusion begins here. You've applied the equation correctly, and found that you must exert a moment of 134N.m on the board. Since you are exerting 100N, by simple math, the distance is 1.34m. But, 1.34 m from where, you ask?

When considering moments, you chose the sawhorse point as the fulcrum. Recall that the definition of moment is the product of force and the perpendicular distance from the fulcrum to the force. Hence to exert a moment of 134N.m, you would have to exert a force of 100N, 1.34m from the fulcrum, i.e the sawhorse. And so you have it... the answer. (In particular, it is 1.34m from the fulcrum, in the direction of the small end, a quick look at the diagram you've drawn will reveal the reason)

3. Nov 30, 2008

### phys1618

thank you for your help. greatly appreciates it!

Yes, you are right, i wasn't sure where the 1.34m is from but now that you said its from the fulcrum. it made a lot of sense!!! you made it much clearer for me.
the question i have is, the 134N.m. how did you know that first?? is there another method? Im curious, because any other opinion or method will help!
Also, if calculating and diagraming everything correctly, would the force of the horse be 220N??

Again, thank you for the help!

4. Nov 30, 2008

### horatio89

Ah, recall that for rotational equlibrium, for any point, total clockwise moments = total anticlockwise moments. Generally, upward forces to the left of the fulcrum and downward forces to the right of the fulcrum exert a clockwise moment about the point. The converse holds true for counterclockwise moment.

Taking moments for the known forces about the sawhorse, the difference between the clockwise moments and counterclockwise moments is 134 N.m, which implies that our unknown force must exert a moment of this value. This method is useful, because it gives the moment, which could be used in both the cases where force is unknown or distance is unknown.

And, yes, the force exerted by the sawhorse is indeed 220N. Interestingly, that force can be found without diagramming. The other condition for equil. is that the sum of the force in any one direction is zero. Hence, upward F = downward F

Downward forces = 180 + 80 + 60 = 320, and upwards forces = F_saw + 100 = 320

5. Nov 30, 2008

### phys1618

thank you thank you!!

i greatly appreciates your help and it's clearly explained!
thanks