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Box on a Toboggan - Newton's Laws

by NKKM
Tags: laws, newton, toboggan
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NKKM
#1
Jul20-07, 01:16 PM
P: 8
1. The problem statement, all variables and given/known data
HI,
I am confused about how to approach this question.

A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on
top of the toboggan. The coefficient of static friction m
s between the block and the surface of the toboggan is 0.60, whereas the kinetic friction coefficient is 0.51. The block is pulled by a 30 N-horizontal force as shown. What are the magnitudes and directions of the resulting accelerations of the block and the toboggan?


2. Relevant equations

Fpull- Ffriction = Fnet = ma
Ffriction = uk X Fnormal
Fnormal = mg

3. The attempt at a solution

If I calculate the acceleration of the box as so: Fpull- Ffriction = Fnet = ma
using the kinetic friction coefficient and solve for acceleration. Does that make sense.. and how then do I approach the acceleration of the toboggan? Would it slide under the box?
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PhanthomJay
#2
Jul20-07, 03:30 PM
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Quote Quote by NKKM View Post
1. The problem statement, all variables and given/known data
HI,
I am confused about how to approach this question.

A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on
top of the toboggan. The coefficient of static friction m
s between the block and the surface of the toboggan is 0.60, whereas the kinetic friction coefficient is 0.51. The block is pulled by a 30 N-horizontal force as shown. What are the magnitudes and directions of the resulting accelerations of the block and the toboggan?


2. Relevant equations

Fpull- Ffriction = Fnet = ma
Ffriction = uk X Fnormal
Fnormal = mg

3. The attempt at a solution

If I calculate the acceleration of the box as so: Fpull- Ffriction = Fnet = ma
using the kinetic friction coefficient and solve for acceleration. Does that make sense.. and how then do I approach the acceleration of the toboggan? Would it slide under the box?
Yes, that makes sense, using the mass of the block for m, and your result is the acceleration of the block with respect to the ground. Now draw a free body diagram of the toboggan.....what net force in the horizontal direction acts on the toboggan? Then use Newton 2 again for the he toboggan to determine its acceleration with respect to the ground.
wizzle
#3
Feb19-09, 10:59 AM
P: 26
Hello,
I'm working on this same problem. I calculated the force of static friction between the block and the toboggan: Fn*Us = (2 kg*9.8)(0.60) = 11.76 N, so it would seem to me that the box will move since this is less than the 30 N pulling on it. I'm confused with the fact that you're saying it moves "with respect to the ground", since it would seem to me that since this is less than the maximum static friction between the box and toboggan, that the box will be moving with respect to the toboggan. Is that incorrect?

I calculated the net force on the box to be Fpull-(FnUk) = ma. (30)-(2.0*9.8)(0.51) = 2.0(a) = 10.002 m/s^2, and you're saying that this is with respect to the ground? For the toboggan, do I use {Fx = 30N-0(since frictionless) = ma, with m being the mass of the block and toboggan combined? I'm having trouble figuring out how to approach this part.

Thanks so much!


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