Applying Newton's laws to find forces

1. Feb 18, 2013

Mangoes

1. The problem statement, all variables and given/known data

"A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp. The cable makes an angle of 31 degrees above the surface of the ramp, and the ramp itself rises at 25 degrees above the horizontal.

(a) Find the tension in the cable.

(b) How hard does the surface of the ramp push on the car.

2. Relevant equations

Newton's laws of motion.

3. The attempt at a solution

The car's in equilibrium, so there is no net force. Two forces are acting on the car: the normal force of the surface and the cable's pull.

I decide to set my x-axis parallel to the surface of the ramp. This means the normal force will be perpendicular to the x-axis and will be go in the direction of the y-axis.

The car has a weight.

Drew a picture:

http://imgur.com/0Bs4rfV

Pretend the car isn't hovering over the ramp.

We know ƩFx = 0

ƩFx = Tcos31 - wcos245 = 0

Tcos31 = wcos245

But weight is (1130 kg)(9.8 m/s^2) = 11074 N

Plugging the value in,

T = (11074*cos245)/cos31 = -3343.76

But this is wrong, T is supposed to be 5460N.

I just can't see where I'm going wrong though.

I haven't done (b) yet, but I figure it would be set up as

normal force + y-component of tension - y component of weight = 0

Last edited: Feb 18, 2013
2. Feb 18, 2013

bossman27

In your first equation, it should be wsin(25), not wcos(255) or wcos(25). I'm assuming you meant 25. I think you just goofed up the components of weight.

3. Feb 18, 2013

voko

First, why 245?

Second, since you want to take the full angle, you should not have the minus sign before the weight.

4. Feb 18, 2013

Mangoes

I actually meant 245, why is taking the angle to be 245 wrong?

If the angle of weight is indeed 25, wouldn't it be in the first quadrant? Then the weight would be towards the sky?

EDIT:

I just realized I wrote 255 instead of 245 in both my original post and this one. Editted to fix it, but I still don't understand why I'd use 25 instead of 245. Sorry about that.

I redid the calculations with cos(245) and I now have the correct magnitude but the incorrect sign...

Last edited: Feb 18, 2013
5. Feb 18, 2013

bossman27

When finding components of the weight, you're concerned with the angle between the mg Force and the "y-axis" in this case. I see what you're saying, but if you want to do it that way, the angle should actually be 335, not 255. 255 is the angle between the mg force and the positive x-axis.

You can indeed do sin(335), but notice that sin(25)= -sin(335). You're correct that it's now pointing in "the right direction," but in such a simple problem I imagine you could match up the forces correctly either way. I was simply finding the magnitudes of the components and matching them up by common sense.

6. Feb 18, 2013

voko

The sign is incorrect because you have the incorrect sign in the weight term. You are taking the full angle from the horizontal. The angle already accounts for the negative direction of the weight, you don't need to account for it again.

It is conceptually simpler, as suggested by others, to take the angle from the downward vertical direction. Then you have sin, not cos, the angle is 25, and you have the minus sign in the weight term, just like one expects intuitively.

7. Feb 18, 2013

bossman27

There's a number of ways you could look at the angles here, so hopefully I can lessen the confusion.

If we define the x-axis as the surface of the ramp, the mg force is pointing down and to the left, at an angle of 245 from the positive x-axis, which is of course intuitive because you're probably used to unit circles.

But, notice that in matching up forces in the x direction, we want the component of mg in the x direction. We know that the angle between mg and the negative y-axis is 25 degrees "to the left."

Ways you can get the correct magnitude, for example:

*the component of w in the positive x direction: wsin(-25) = wsin(355) = wcos(245) = wcos(115) = -4680N
*the opposite of the w component in the negative x direction: - wsin(155) = -wsin(25) = -cos(65) = -wcos(295) = -4680N

The general point is that you can measure the angle and component any number of ways, but in a problem like this it doesn't particularly matter since you can obviously set them up in the correct way by common sense.

8. Feb 18, 2013

bossman27

And, as voko notes, the simplest way is just to take the angle between the negative y-axis and the mg force. A sketch will make it painfully obvious how simple it is to set up.

9. Feb 18, 2013

Mangoes

Glad you posted this, finally makes sense now. That sentence made it click.

I had seen throughout the book that the authors used the opposite trigonometric function I was expecting and I suspect it was because of this.

I had also noticed in previous problems that I'd get incorrect signs, but I checked back through the work and found the same error that voko pointed out in this one.

Thanks a lot for the help guys, really glad I posted this