# Torque or force

1. Apr 20, 2005

### NoHeart

the question: a board 2 ft long with weight 50 lb is pressed against a wall. what force, applied at 30 degrees, should be applied to keep it horizontal?

i'm not sure how to go about this, so i started by converting the weight of 50 lb to 22.67 kg, then getting a force of mass at 2.31 N
then i tried using the pythagorean theorem to find the magnitude of applied force, which would be the hypotenuse, but it's not making sense. i get 0.37 using 2.31 N and i get 0.017 using 50 lb. what am i leaving out? does torque play a part?

2. Apr 20, 2005

### gnome

Is there a picture to go along with this problem? Did you leave out any details, like:
where is the force applied?
at 30 degrees relative to what?
what's going on where the board meets the wall? is it attached? is friction involved?

3. Apr 20, 2005

### OlderDan

You have to assume the wall will have sufficient friction to keep the board from slipping. The 30 degrees is a bit vague, but my guess is they mean 30 relative to the horizontal. You could think of this force as being applied by a rope attached to the end of the board and to the wall to make a 30-60-90 degree right triangle, but any other source for this force would work the same. Then you have a typical equilibrium problem.

4. Apr 20, 2005

### gnome

That's a pretty big assumption. Where it's attached makes all the difference in the world.

5. Apr 20, 2005

### gnome

Maybe it's attached by a hinge, so friction is not an issue.
Maybe the coeff. of friction is so high that "any" normal force produces enough frictional force to keep it from slipping (your assumption). Essentially this makes it the same problem as the hinge problem; friction is not an issue.

Or maybe the coefficient of friction is given, and then you have to be sure that the applied force results in a sufficient normal force to keep the board from slipping.

6. Apr 20, 2005

### NoHeart

friction is not mentioned at all
the force is applied at an angle of 30 upward from the end of the board, just like older dan's rope suggestion.
i still don't get it...

7. Apr 20, 2005

### gnome

There's no reason to convert to metric. That just makes more work, & more opportunity to make mistakes.

If no friction is mentioned, treat it as if attached to the wall by a hinge and let that point be your axis. Form your equation to equate the torque produced by the weight of the board (remember, that one is applied at the center of mass) to the torque applied by the vertical component of your force applied at the end of the board.

PS: you can't solve this using forces alone (i.e. without considering torques) because you don't know how much upward force is applied by the wall.

Last edited: Apr 20, 2005
8. Apr 22, 2005

### NoHeart

okay i'm starting to get it- so if the force downward is 50 lb, then the torque of the force applied will equal the torque of the force downward, because it is at equilibrium? yes?

9. Apr 22, 2005

### OlderDan

Yes. I assume you are calculating torques about the contact point of the board with the wall. Just make sure you get your moment arms right.

10. Apr 22, 2005

### NoHeart

since the weight is from the center of gravity, would the moment arms be 1 ft, not 2?

11. Apr 22, 2005

### NoHeart

if the moment arm is 1, then the applied force would be 50 lbs as well, but don't i need to figure the 30 degrees in somewhere?

12. Apr 22, 2005

### NoHeart

sorry about my manic posting here, but if my brain is working properly then the moment arm of the force acting downward is 1 ft and the moment arm of the applied force is 2 ft. that would make the magnitude of the applied force 100 lbs. please make my brain stop spinning

13. Apr 22, 2005

### OlderDan

You are half correct, on second though maybe 1/3 . The moment arm of the board's weight is 1 foot. The moment arm of the applied force is the shortest distance between the line of that force and the point of rotation. That is not 2 feet.

When you used 2 feet your answer should not have been 100. Can you see why? What would it have been if the moment really were 2 feet?

Last edited: Apr 22, 2005
14. Apr 22, 2005

### minger

There was another post just a day or two ago that was very very similar involving a ladder. Anyways, I will give the same advice that I gave in that post. You are clear that you will need to sum moments (torques as your referring to it). Let's call the point where the board meets the vertical wall as A. We sum moments around point A. We can disregard any forces at point A since any force through the point of rotation will not cause a moment (no moment arm). The only forces that will cause a moment are the weight of the board, and the force at the bottom of the board.

Now what you were getting to about the angle. The angle is very important, because you only count the component of the force that is perpendicular to the board. For example, if someone is pulling axially on the board (as in like in the direction of the board) you can imagine that this would not "spin" the board around point A. Conversly, that same force applied perpendicular to the board will make it much easier to "spin." In this case, we will use Cosine(30)*Weight to find the component perpendicular to the board.

The other moment will again come from the bottom of the board. When you sum moments, you will get a force that is applied at the bottom of the board. You are not done yet though. Again, that force that you solved for is perpendicular to the board. To find the horizontal component, you will need to sin(30)*F, F being the force calculated from the moment equation.

Given this information if you wanted to find the reactions at our initial point A, you would sum forces in the x and y directions to get those components reacting at the top of the board.

Edit: this is of course assuming that there is no friction and the force applied will be at the end of the board, where the board meets the floor.

Answer whited out: 10.825 lbf

Last edited: Apr 22, 2005
15. Apr 22, 2005

### OlderDan

I think it has been resolved that the 30 degrees is measured from the horizontal rather than the vertical. An alternative to breaking the tension force into components is to find the shortest distance between the line of the force and the point of ratation. Either approach brings the same trigonometric ratio factor into the problem.

16. Apr 22, 2005

### NoHeart

thank you very much for all of your help.
is the moment arm of the applied force then the line coming from the center of the board to a point on the line of force, perpendicular to the line of force? this would make the moment arm 0.5ft (sin 30*1ft)
that would make the answer 25 lb
if it is the point where the board meets the wall, it is 1ft (sin 30*2ft), making the answer 50lb. i believe this one is correct, but with all the mistakes i've made so far i need some confirmation...thanks again

17. Apr 24, 2005

### NoHeart

seriously, anyone?

18. Apr 24, 2005

### OlderDan

The first part is not correct. You have been calculating the torques about the point of contact between the board and the wall. The distance between the line of force and the point of rotation you are using is sin 30*2 ft. You cannot use the middle of the board for this calculation unless you include the unknown force of friction as a torque producing force.

In the second part, you got it. You could also get the same result by following the suggestion of breaking the applied force into vertical and horizontal components. The vertical component would be Fy = F*sin 30 and the horizontal component would be Fx = F cos 30. The horizontal component would be cancelled by the normal force at the wall. The vertical component would have a moment arm of 2 feet, and so 2*Fy = 1*W = 50lb leading to Fy = 25lb. Then F = Fy/(sin 30) = 50lb. Now that you know Fy you can find the force of friction, f. Fy + f = W = 50lb, so f = 25lb. And you can find the normal force Fx = 50lb*cos 30 = 43.3lb

Last edited: Apr 24, 2005