# Weight/Normal Force of a block

AnkhUNC
[SOLVED] Weight/Normal Force of a block

## Homework Statement

If P = 1.98, M = 1, Theta = 45 degrees what is the weight of m in Newtons?

http://img261.imageshack.us/img261/3339/fig450ph7.gif [Broken]

## The Attempt at a Solution

I'm apparently doing this wrong. I was trying to solve for the normal force given that N = Mg/cos theta but this is just 9.8 so I'm doing something wrong. Any advice would be appreciated.

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Mentor
Any further description of the problem? Is the block sliding down the incline?

AnkhUNC
P (1.98) is such that the block is not moving.

Thats all I'm given :(

Mentor
Start by identifying the forces acting on each mass. Draw a free body diagram for each.

What can you say about the net force on the block?

Mentor
I was trying to solve for the normal force given that N = Mg/cos theta ...
How did you deduce this?

AnkhUNC
N cosθ – mg = ma(y). a(y) = 0 so N = Mg/cos theta?

Forces on triangle block are W (Mass pointed down), P (->), Normal force up mgcos(theta) opposite the angle of the block?

Small block is w pointed straight down and Normal force at an angle opposite (90d?)

Mentor
N cosθ – mg = ma(y). a(y) = 0 so N = Mg/cos theta?
Good. But don't mix M and m: N = mg/cos(theta).
Forces on triangle block are W (Mass pointed down), P (->), Normal force up mgcos(theta) opposite the angle of the block?
Only worry about horizontal forces on the triangle.
Small block is w pointed straight down and Normal force at an angle opposite (90d?)
OK

Keep going. Apply Newton's 2nd law to the horizontal direction and see what you can deduce. Hint: What net horizontal force acts on the block? On both masses?

AnkhUNC
P = (M + m) g tanθ?

But this gives me m = .2020408163 so * 9.8 just = 1.98 again which is wrong.

N = Nx i + Ny j = N sinθ i + N cosθ j

–Nsinθ = –mgtanθ

P – mg tanθ = Max

N sinθ = mg tanθ = max

P = (M + m) g tanθ

I guess I'm just having a hard time seeing what I need to get out of this.

F = Mgcos(45) - mgsin(45)? That doesn't even look right...

Mentor
P = (M + m) g tanθ?
No. Just apply Newton's 2nd law to M + m. (Where did you get the g tanθ?)

After you do that, apply it to the block alone.

AnkhUNC
Ah its equal to 1! Sweet! I don't know if I did it right or not though. I did N = Mg/cos(45) to find N then N = mg/sin(45) so m = 1!

Mentor
P = (M + m) g tanθ?
Actually, I take it back. This seems correct to me. (I just didn't see how you got. You gave your conclusion first. )

But your data doesn't seem OK.
But this gives me m = .2020408163 so * 9.8 just = 1.98 again which is wrong.
I don't see how you deduced this value for m.

Mentor
Ah its equal to 1! Sweet! I don't know if I did it right or not though. I did N = Mg/cos(45) to find N then N = mg/sin(45) so m = 1!
Where did you get N = Mg/cos(45) = mg/sin(45)? I thought we had established that N = mg/cos(45).

AnkhUNC
I was just trying to find a value for N without having m. Like I said I got the right answer but I'm sure I did it wrong

Mentor
What makes you think you got the right answer?

Homework Helper
N cosθ – mg = ma(y). a(y) = 0 so N = Mg/cos theta?

Hi AnkhUNC! waaah … you've left out the y-component of the poor little friction force. When you draw a diagram, you should always mark in all the forces

You need to take components in the normal direction, so that the friction component will be 0.

Then N = ? Mentor
..the poor little friction force. Don't cry, tiny-tim! Now I'm starting to cry... I would assume, lacking any statement to the contrary, that the surfaces are frictionless.

AnkhUNC
The answer checked as right but like I said I was sure I did it wrong. If its not moving then N is going to be equal to P?

Any websites you can recommend to help with these types of problems? :(

Mentor
How about the site you're on right now? Just attack it systematically, as I suggest in post #7. (I assume the surfaces are frictionless, correct?)

(I'd say that given the data you supplied, there is no correct answer. But you can solve for m symbolically in terms of P and M.)

AnkhUNC
Yeah there isn't a frictional force. I think the problem was just to get me thinking about the process and I spent way too much time on it :P

Mentor
Just do it step by step. Analyze the forces and apply Newton's 2nd law:
(1) To the block (vertical direction)
(2) To the block (horizontal direction)
(3) To the entire system (horizontal direction)

Don't cry, tiny-tim! Now I'm starting to cry... Presumably the whole system is accelerating, but the little block is not moving relative to the big block, so resolving vertically was correct after all. Thanks, Doc Al! 