# AP Physics Problem

1. Oct 27, 2009

### swimchica93

As you all have seen, I am completely lost, but within the last two hours of being a member, I now know how to add and subtract vectors. Anyway!

I made the image, because there is a diagram, and I have some clue as in what to do. My head is still spinning from the 40 vector addition/subtraction problems.

http://i47.photobucket.com/albums/f162/mythologymaniac/image-4.png [Broken]

I really need to know how to do it, not the answer, I changedthe values, so answers won't help. :) THANK YOU IN ADVANCE!!

Last edited by a moderator: May 4, 2017
2. Oct 27, 2009

### longball4153

First step is drawing a free body diagram.

3. Oct 27, 2009

### swimchica93

Is that different than the diagram that I have? :)

4. Oct 27, 2009

### swimchica93

I am pretty sure I have no clue what I am doing.

It is beyond hard. Or I am missing some key piece of information. I have tried three things, and they are not right. At all.

http://i47.photobucket.com/albums/f162/mythologymaniac/phys2.png [Broken]

I really need to know how to do it, because I changed the values. :)

I really want an A in physics. >.<

Last edited by a moderator: May 4, 2017
5. Oct 28, 2009

### XanziBar

A free body diagram usually consists of an object (a dot) and the forces acting on it (yeah actually it is pretty close to what you have). Now that you are an expert at decomposing vectors...what part of the force is acting parallel to the floor?

6. Oct 28, 2009

### swimchica93

Is there a force parallel? I have no clue what I am doing. lol So far for what is acting on the box I have this:

http://i47.photobucket.com/albums/f162/mythologymaniac/phys2.png [Broken]

There wouldn't be friction if it is a frictionless floor, right?

Last edited by a moderator: May 4, 2017
7. Oct 28, 2009

### Staff: Mentor

Re: I am pretty sure I have no clue what I am doing.

Show us the 3 things that you've tried. We do not do your work for you.

Last edited by a moderator: May 4, 2017
8. Oct 28, 2009

### swimchica93

Re: I am pretty sure I have no clue what I am doing.

I think I figured out that the muk is kinetic friction.

I tried an equation with pulleys and that is definately wrong. I also tried one, but it involved two boxes, again wrong, and another one that I don't really know where I got it.

When the rope breaks I know that I will hae to use 9.8 m/s^2 but I hae no clue where.

I have two equations: Fgx= mg sin (angle) & Fgy = mg cos (angle)

I am not looking for answers, I changed the values so even if someone did give me answers I wouldn't use them. I have a test soon so not knowing how to do the problems might be detrimental to my grade. :)

9. Oct 28, 2009

### Staff: Mentor

(merged two threads with the same questions)

10. Oct 28, 2009

### swimchica93

They are both different. One is about a box and an incline, while the other is a box being pushed...

11. Oct 28, 2009

### Staff: Mentor

I'm pretty sure that the figures were the same for the two threads I merged. In any case, show us the 3 things you tried in the post where you said you tried 3 things... Showing your work is important here.

12. Oct 28, 2009

### swimchica93

Ok. For this one: http://i47.photobucket.com/albums/f162/mythologymaniac/phys2.png [Broken]

I tried this to no avail: Fgx=mg sin (angle) and Fgy = mg cos (angle)

The thing is, the examples in the book are nothing like this problem. That is for finding acceleration and I need force, and magnitude of acceleration. Which I am guessing magnitude of acceleration= acceleration. So it works for that.

This is where the example gets sticky.It uses a skier as the example and then they literally use greek symbols from then on. Sigma F y = ma y
Fn - mg cos (angle) = ma y = 0

That is where I get confused... It is all greek to me. (pun intended)

Then the example switches to speed.

All the other examples involving anything similar are in the next chapter, so wouldn't relate.

I also tried something involving a sled and pushing, so pushing isn't what I am looking for for that problem.

Then I tried another equation: Sigma Fy= Fn - mg = 0

I tried substitution and it didn't work, and turned out unbearbly long. I am guessing seeing the space to put it, should at least fit an eighth of the problem, I was on th wrong track.

I tried something else but scribbled it out.

On this: http://i47.photobucket.com/albums/f162/mythologymaniac/image-4.png [Broken]

I don't have any clue what to even try... I did change the values. I just am at a loss for both of them.

The main thing is, they are both due in 5 hours... -sigh-

Last edited by a moderator: May 4, 2017
13. Oct 28, 2009

### Feldoh

In http://i47.photobucket.com/albums/f162/mythologymaniac/image-4.png [Broken]

What forces are acting on the box, and what direction(s) are they? That's a good place to start.

Last edited by a moderator: May 4, 2017
14. Oct 28, 2009

### XanziBar

Let's talk about the ramp problem first. The capital sigma means "sum of", as in you add up all of the things you see after the sigma. The example in your book add up all the forces in the y direction, and set them equal to zero for equilibrium (everything is sitting still or moving at a constant velocity).

What I suggest you do, especially for the ramp problem, is to draw a free body diagram for the box, what forces are acting on it? Then, knowing that if you want a constant velocity in the direction parallel to the ramp set the sum of those forces equal to zero, just like in your text.

15. Oct 28, 2009

### swimchica93

I have this right now: http://i47.photobucket.com/albums/f162/mythologymaniac/phys2.png [Broken]

With no friction, and I added the force from the floor.

Last edited by a moderator: May 4, 2017
16. Oct 28, 2009

### swimchica93

I tried drawing a free body diagram, and came up with the pull of gravity, the force from the ramp, magnitude, and the force from the rope.

What do you mean by adding them together? This was never explained in class. Is tension at all a part of this. We went over that some. The book is completely useless...

17. Oct 28, 2009

### XanziBar

What book are you using?

Parallel to the ramp, you have the component of gravity which is parallel, you said a few posts ago this is mg*sin(theta), you also have the friction, and you have the tension from the rope. In the direction perpendicular to the ramp you have the normal force, and the perpendicular component of gravity. The sum of the forces (like force A +force B- force C) has to be equal to zero.

This is not so different from your vector problem from before.

18. Oct 28, 2009

### swimchica93

Thank you for your help. :) I got an extension for the problems. I figured out the box one.

I used Fpx=(force)(cos(angle))

then ax=(Fpx)/(m) which gave me acceleration

then v=(initial velocity which equals zero) + acceleration*force= speed

then I took that, and multiplied it by the time to get the distance. :)

Now I am working on the ramp problem... I drew a free body diagram, but it doesn't give me any idea what to do.

19. Oct 28, 2009

### XanziBar

Not to quibble, but why did you multiply acceleration and force? once you have the acceleration velocity=initial velocity+acceleration*time.

As for the ramp, what are the components of the forces that are acting parallel to the ramp (there are 3)?

20. Oct 28, 2009

### swimchica93

Sorry! I did v= v(knot) + at

Gravity, Friction, and the constant force from the rope?

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