# Force and Torque or Center of Gravity ^^

1. Nov 22, 2006

### twotaileddemon :uhh: I think I'm getting pretty good at my free body diagrams.. at least I hope so...
Well, in any case, I'm having trouble with my homework problem again X_X (specificially c, because it's new). I was wondering if someone could check my work because on problems that are assigned before we go over them in class, I can't seem to figure them out completely. (We just finished another chapter, were given this for homework with no background on torque/center of gravity. I looked it up in my textbook, but am a bit confused. Here is my homework problem:

A box of uniform density weighing 100 N moves in a straight line with constant speed along a horizontal surface. The coefficient of sliding friction is .4 and a rope exerts a force F in the direction of motion. (There is a diagram with a box and an arrow going right with an "F" by it.

a) On the diagram to the right (just a box), draw and identify all the forces on the box.

I have the "F" mentioned in the problem to the right as it was originally drawn, "Ff" to the left, "Fn" up, and "mg" down, which the weight = 100 N

b) Calculate the force F exerted by the rope that keeps the box moving with a constant speed.

Well... I said that F = Ff = uFn
And if Fn = mg, then uFn = umg
So.. F = umg = (.4)(100 N) = 40 N

c) A horizontal force F' is applied at a height 5/3 meters above the surface, is just sufficient to cause the box to begin to tip forward about an axis through point P. The box is 1 meter wide and 2 meters high. Calculate the force F'.
(The diagram is a rectangle - the top and bottom are 1m, the sides are 2m, and on the right (that is, where the F' is) F' is draw to the right 5/3 m above the ground.

I'm pretty sure this involves torque somehow.. if not that, then definitely center of gravity. The only thing is, I'm not sure if I use the forces I gathered in my free body diagram or not. If I do, then F' would have to be slightly above 40 N, but that doesn't seem like the right way to go because I'm given all this other information. Could I use the equation for center of gravity, perhaps? But I only have one mass.. Could someone possibly give me any hint on how to start this? I'm really having trouble getting started. Thanks for any help. I appreciate it immensely. 2. 3. Nov 22, 2006

### Doc Al ### Staff: Mentor

The force F' clearly exerts a torque about P that tends to tilt the box upward. What other force acts on the box tending to prevent that tilting? What torque does it exert about P?

4. Nov 22, 2006

### twotaileddemon I would say Ff prevents the tilting because Fn and mg cancel out. And.. that's 2 m.. Ff = 40
So perhaps..
2(40 N) = 5/3(F')
F' = 80/(5/3) = 48 N?

And.. 48 is greater than 40 N, so I think that would be able to tip the box over slightly.

Last edited: Nov 22, 2006
5. Nov 22, 2006

### Doc Al ### Staff: Mentor

Two problems with that idea: What torque does Ff produce about point P? When the box begins to tilt up off the surface, what happens to Fn and the torque it produces about point P?

6. Nov 22, 2006

### twotaileddemon The torque of Ff about point P... hmm.. >_<
Well torque by Ff itself is 80 I'm pretty sure.
So if it's about point p.. then maybe
80 N = 1mP = 80 N-m

7. Nov 22, 2006

### twotaileddemon Oh.. as for the second part, Fn = mg = 100.. so Fn = 100. If it starts to tilt.. then it becomes less in order to begin tilting?

To be honest, I'm still really confused and try to put what you said together, but I'm really new to torque and I'm trying to combine forces and torque = Fd with little knowledge n_n;

8. Nov 22, 2006

### Doc Al ### Staff: Mentor

Although you didn't define it, I assume point P is the leading edge of the bottom of the box. What's the definition of torque? The direction of the force makes a big difference in calculating torque. If you extend the line of force Ff, how close does it get to point P?

9. Nov 22, 2006

### twotaileddemon P is the edge at the bottom right of the box.. which is the leading edge, which is what you said, I think.
Torque is the magnitude of a force times the length between the forces...
If I extend Ff, which is to the left... I don't think it ever reaches P. Oh.. unless you mean just extend it to the right to get it in terms of height, then it would be.. 1 m. And there is a 1/3 difference between Ff and F' then..

10. Nov 22, 2006

### twotaileddemon And Ff is opposite F'.. so it's sort of like
1 m 40 N left, and 5/3 ? N right..with 1/3 space. am I getting closer to the solution, or am I just going off track?

11. Nov 22, 2006

### Doc Al ### Staff: Mentor

One way of defining torque is force times the perpendicular distance between the axis of rotation and the line of the force vector.

What height is Ff? (Friction acts where?) What height is point P?

12. Nov 22, 2006

### Doc Al ### Staff: Mentor

You're all over the map. First you need to understand what torque is produced by Fn and Ff just as the box starts to tilt up. Hint: They exert the same torque about point P.

13. Nov 22, 2006

### twotaileddemon In terms of height, P is 0.. immagine a rectangle, and to the bottom right corner.. that's where P is.
The height for Ff was not given, but I assumed it acted in the middle. However, perhaps it would be best to say that it argues exactly opposite F'? Which would mean both are 5/3 m?

Lets see.. Torque = F times perpendicular distance between axis of rotation and the line of the force vector...
perpendicular distance....
perpendicular distance...
I should account for Fn then..
Fn would be.. 2 m up from P..
F' would be 5/3 m up (but to the right) from P
Ff should be opposite F'.. I think
>_< >_<; Uh.. first lets see.. F' is 5/3 m from P
Then F' goes right.. the "axis of rotation" is the "5/3" m I'm assuming
The line of the force vector is "F'" which is unknown.. but if, perhaps, it's opposite the Ff, which I'm assuming is at the same height.. so 40..

Well.. let me think again
Fn is one force.. that's acting up from P..
Fn = 100... it's torque might be 2(100) = 200
Then I may be able to set that to F's torque.. which may be
(5/3)F'
F' = 200/(5/3) = 120..?

14. Nov 23, 2006

### Doc Al ### Staff: Mentor

Friction is an interaction between two surfaces, in this case the bottom of the box and the floor. So friction acts on the bottom of the box.

15. Nov 23, 2006

### twotaileddemon Friction acts on the bottom of the box.. okay
Ff = uFn = umg = 40 N up times 2 m, = 80 N-m (torque)
Then.. going to the right for F' we dont know the distance or Force value.. or do we just use 1m, making it 40 N?

(Also, I wont be able ot respond for three days because I will be without computer access, but I'll keep trying)

Last edited: Nov 23, 2006
16. Nov 23, 2006

### Doc Al ### Staff: Mentor

The question is: What is the torque produced by Ff about the point P? (Nothing to do with F'.) Read up on torques and lever arms here: http://hyperphysics.phy-astr.gsu.edu/hbase/torq.html" [Broken]

Last edited by a moderator: May 2, 2017
17. Nov 27, 2006

### twotaileddemon I went to the site you suggested, and I already knew everything there. I guess I just have trouble applying it.

Oh, sorry about waiting so long to respond. I got back late sunday from my weekend with the family and had school the next day.. =(

So.. the torque produced by Ff about point P. Torque is Fd.
I think Ff is 1m in terms of height. P is 0 m. If Ff = 100 N (because it's at the bottom of the box, right? so it'd be the same as weight. And F' is 5/3 meters, I assume there is a greater force needed to tip the box over, so (5/3)(100) = 166 N = F'?

18. Nov 27, 2006

### Doc Al ### Staff: Mentor

Make up your mind: Is Ff at a height of 1m or is it at the bottom of the box?

The force of friction is parallel to the bottom of the box; the line of force passes through point P, so its distance is zero.

To solve this problem: Think about what happens just as the box starts to tilt up about point P. There's only one force acting to pull it down. What force is that? That's the force who's torque F' must overcome in order to tilt the box.

19. Nov 27, 2006

### twotaileddemon The force pulling the box down has to be mg... that makes the most logical sense to me.
What I intended to say with Ff is that it is at the bottom, but isn't the bottom a length of 1 m?
So F' has to overcome the 100 N from gravity. That means it must be greater than 100 N in order to tip the box. If F' must overcome it, then
(100 N) (1 m) <- at the bottom of the box = F' (5/3) m
F' = 100 N-m/(5/3 m)
F' = 60 N

Now, I don't think I'm supposed to get a small answer like that. So when accounting for the distance of the weight, do I use its distance from point P, or from force F'?

20. Nov 28, 2006

### twotaileddemon I don't think the distance of mg can be 0, and 5/3 is the distance of the F' from point P.. so I think using 1 m would be correct.

21. Nov 28, 2006

### Doc Al ### Staff: Mentor

This is true.
When finding the torque due to a force, it's not just the distance that counts, but also the angle that the force makes. Sometimes the distance is called the "lever arm" or the perpendicular distance from the line of force to the pivot point. You need to review that link I gave you.

For example: Imagine a long wrench attached to a bolt. The wrench has length L. If you exert a force F at the end of the wrench, what torque do you exert on the bolt? Answer: It depends on the angle that you apply that force! If you apply it in the usual way--sideways to the wrench--then the torque is FL. But what if you push straight in--or pull straight out--what torque would you exert then?

No, it means that the torque exerted by force F' (about P) must be greater than the torque exerted by gravity (about P). Evalute those torques just as the box begins to tilt.

Note: The weight of the box acts at the center of mass. How far is the center of mass from point P?