# 2.7 kg wood box

1. Oct 9, 2009

### gbedenba

1. The problem statement, all variables and given/known data
A m = 2.7 kg wood box slides down a vertical wood wall while you push on it at a θ = 35° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

2. Relevant equations

3. The attempt at a solution

I got 64.5766 N for the answer. Don't know how to do it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Oct 9, 2009

### Chewy0087

Do you know what the net force on the box is equal to if it's travelling at a constant speed?

What force is the box moving downwards with? (if your hands aren't there).

just some things to consider

remember that for it to not be accelerating up or down the force you push it up with (vertically) must be equal to the force it's coming down with. think about splitting the 35 degree force into vertical and horizontal forcess

3. Oct 9, 2009

### gbedenba

don't really understand that.... could you tell me what formulas to use? thanks

4. Oct 9, 2009

### Chewy0087

okay, do you understand that in order for something to be static, or travelling with a constant velocity, there must be no net force acting on it?

i think it's far more important to understand what to do here than simply use formulae.

also do you know how to resolve forces into vertical and horizontal?

5. Oct 9, 2009

### gbedenba

yeah I understand the no net force thing...and resolving vertical and horizontal forces...using sin and cos. yes...

6. Oct 9, 2009

### Chewy0087

Okay, so you must therefore realise that if it's moving at a constant speed (downwards) it's net force will be 0.

therefore the force of gravity down, and the vertical component of your pushing (upward) will be equal

your equation will look something like so;

mg - Fcos(theta) = 0

is friction mentioned at all?

Last edited: Oct 9, 2009
7. Oct 9, 2009

### gbedenba

I got 29.2799 for the final answer. it is wrong

8. Oct 10, 2009

### Chewy0087

what working out did you use to get that? if friction mentioned at all?

9. Oct 10, 2009

### jesh462

I have this exact same question. Friction is not mentioned at all. He stated the exact wording already. I need help on this question as well. Is it ok to hijack this thread, or do I have to start a new one?

10. Oct 11, 2009

### Chewy0087

No that's fine, it's the same question.

Hmm when i tried it I got 32.3N, is that the correct answer? If it is, see my post above, which says this

Okay, so you must therefore realise that if it's moving at a constant speed (downwards) it's net force will be 0.

therefore the force of gravity down, and the vertical component of your pushing (upward) will be equal

your equation will look something like so;

mg - Fsin(theta) = 0

if not tell me, the answer may be wrong

Last edited: Oct 11, 2009
11. Oct 11, 2009

### sArGe99

Actually its the sine component that's balancing the weight of the body. So your answer might be around 47 or so, I reckon.

12. Oct 11, 2009

### Chewy0087

Hm, would presume they mean 35 degrees from the vertical, not horizontal.

edit: oh right just saw the picture attached, my bad :P