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Weight/Normal Force of a block

  • Thread starter AnkhUNC
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  • #1
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[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]

Homework Equations





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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Answers and Replies

  • #2
Doc Al
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Any further description of the problem? Is the block sliding down the incline?
 
  • #3
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P (1.98) is such that the block is not moving.

Thats all I'm given :(
 
  • #4
Doc Al
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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?
 
  • #5
Doc Al
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I was trying to solve for the normal force given that N = Mg/cos theta ...
How did you deduce this?
 
  • #6
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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?)
 
  • #7
Doc Al
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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?
 
  • #8
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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...
 
  • #9
Doc Al
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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.
 
  • #10
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Ah its equal to 1! Sweet! I dunno 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!
 
  • #11
Doc Al
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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. :wink:)

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.
 
  • #12
Doc Al
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Ah its equal to 1! Sweet! I dunno 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).
 
  • #13
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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
 
  • #14
Doc Al
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What makes you think you got the right answer?
 
  • #15
tiny-tim
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N cosθ – mg = ma(y). a(y) = 0 so N = Mg/cos theta?
Hi AnkhUNC! :smile:

waaah … you've left out the y-component of the poor little friction force. :cry:

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 = ? :smile:
 
  • #16
Doc Al
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..the poor little friction force. :cry:
Don't cry, tiny-tim! Now I'm starting to cry... :cry:

I would assume, lacking any statement to the contrary, that the surfaces are frictionless.
 
  • #17
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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? :(
 
  • #18
Doc Al
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How about the site you're on right now? :wink:

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.)
 
  • #19
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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
 
  • #20
Doc Al
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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)
 
  • #21
tiny-tim
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Don't cry, tiny-tim! Now I'm starting to cry... :cry:

I would assume, lacking any statement to the contrary, that the surfaces are frictionless.
Yes, looking at the diagram again, I think you're right!

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. :redface:

oh … I'm so much more cheerful now!

Thanks, Doc Al! :smile:
 

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