# Homework Help: Box on a frictionless ramp.

1. Mar 8, 2005

### Poobel

Fro some reason cant figure this one out:

A box is sitting at a frictionless ramp, with a force of 150 N acting on the box, perpendicular to the ground. The ramp is at 25 degrees towards the normal.
a)Determine the force parallel to the surface of the ramp, that would keep the box from moving, i.e. force of friction.
b) Determine the tension of the rope fastened at the middle of the top of the box, that pulls at a 45 degree angle to the top surface of that box and keeps it from sliding.

It seems that I have a problem figuring out what the hell is this question answering. The wording is confusing (at least to me)

As far as I understand, both a) and b) are asking for the same thing - force required to keep the box from moving. then is it 150/sin25 for a)? becuase the number seems to be too high?

2. Mar 8, 2005

### dextercioby

Not really,pay attention to the congruent angles...

Daniel.

3. Mar 8, 2005

### Poobel

then I should use 65 degrees? Im confused by the part that says perpendicular to the ground. Is it perpendicular to the ramp or the ground ground?

4. Mar 8, 2005

### dextercioby

It would have said "on the ramp",trust me...

Daniel.

5. Mar 8, 2005

### Poobel

ok then I can use 150/cos25 to find the friction, and thus answer a)?

6. Mar 8, 2005

### dextercioby

Nope,u'd have to use 150 sin 25°...

Daniel.

7. Mar 8, 2005

### Poobel

yeah, thats what I said in the beginning, but the force then is approx 355N is that actually correct?

then for part b) it would be 355N/sin 45?

8. Mar 8, 2005

### BobG

The 150N is perpendicular to the ground because it is the force due to gravity.

Your answer to a) is almost correct (at least you used the right parameters). However, it should be 150 * sin 25, not division (this is probably why your answer seemed to high).

b) kind of asks the same thing. a) had the force counteracting the sliding of the box parallel to the box, so the total force was equal to the force being directed down the ramp. In b), the force is at a different angle - 45 degrees relative to the ramp, so the total force will have to be greater (some of it is directed perpendicular to the ramp as well as parallel to the ramp).

Exactly how you solve these depends on what makes it easier to visualize what's happening. Some people find it easier to use the ground as a reference, some the gravity force as the reference, some the ramp's surface as the reference.

Personally, just make the surface of the ramp my 'x-axis' with the bottom being the positive direction. Then the force of gravity is pushing the box at a 65 degree angle (that's okay, because the cosine of 65 is the same as the sine of 25). The rope pulls the box back along the x-axis with the same force that gravity pushes, so I know the magnitude of the x component of the rope's force. The rope is at a 45 degree angle. The tension of the rope times the cosine of 45 degrees equals the x component (T * cos 45 = 150 * cos 65).
If I divide both sides by the cos 45, the tension is equal to the x-component divided by the cosine (150 * cos 65)/cos 45.

The important thing is to draw this out if you have problems understanding what's going on. It makes it easier to figure out which sides of which triangles you're dealing with.

Edit: I guess if I'm going to use cos 65, then I should actually use it. Like I said, it's better to draw it if you want to keep things straight. :rofl:

Last edited: Mar 8, 2005
9. Mar 8, 2005

### Poobel

Thanks a lot BobG. I did try to sketch.. more than once, but I guess Im just kinda slow today Thanks again.

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