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Slope Friction Problem

by Fusilli_Jerry89
Tags: friction, slope
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Fusilli_Jerry89
#1
Oct18-06, 07:45 PM
P: 162
a 2.0 kg brick rides on a 20kg sled and is dragged upa22 degree slope by a rope witha tension of 1000 N. Theropeis parallel totheslope. The friction coefficient(both static and kinetic) between the brick and sled is 0.30 and between the sled and the ground is 0.15. Predict each mass's acceleration. I got 0.8 m/s/s for thebrick which i'm sure is right, and for the sled sled I got 40 m/s but the teacher says it is 45 m/s/s.

Teacher's method:
1000-20(9.8)sin22-0.15(22)(9.8)sin22-5.4=20a
a=45.5m/s/s

My method:
1000-81-30-5.4=22a
a=40m/s/s
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Doc Al
#2
Oct18-06, 08:26 PM
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Quote Quote by Fusilli_Jerry89
Teacher's method:
1000-20(9.8)sin22-0.15(22)(9.8)sin22-5.4=20a
a=45.5m/s/s

My method:
1000-81-30-5.4=22a
a=40m/s/s
Since you realize that the brick and sled do not accelerate together, why do you use the total mass in your equation?

(FYI: Getting useful help will be easier if you post your work using symbols, not just numbers.)
Fusilli_Jerry89
#3
Oct18-06, 08:40 PM
P: 162
well doesn't the brick on top add to the force of gravity, which in turn would increase the force of friction and the other force which directs down the slope?

OlderDan
#4
Oct18-06, 11:03 PM
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Slope Friction Problem

Quote Quote by Fusilli_Jerry89
well doesn't the brick on top add to the force of gravity, which in turn would increase the force of friction and the other force which directs down the slope?
You need to look at the sled and the brick separately and consider all the forces acting on each object. Draw a free body diagram for each of them. Correct the things highlighted by Doc Al

In your last equation, the mass that multiplies a is just the mass of the sled, not the mass of sled plus brick. In your "81" you have included a force that is acting only on the brick as if it were acting on the sled.

The teacher's method is correct, but the numerical value seems a bit off.

I'd like to see how you calculated the 0.8m/s^2 acceleration of the brick. It would be interesting to calculate the total force acting on the brick if the sled were not moving.
Fusilli_Jerry89
#5
Oct19-06, 11:20 AM
P: 162
For the brick: 0.30(19.6)cos22-19.6sin22=2a
Doc Al
#6
Oct19-06, 11:52 AM
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Quote Quote by Fusilli_Jerry89
For the brick: 0.30(19.6)cos22-19.6sin22=2a
Looks good. How did you get 0.8 m/s^2 from that?
gunblaze
#7
Oct19-06, 11:55 AM
P: 186
where does the 5.4 come from?
Fusilli_Jerry89
#8
Oct20-06, 05:09 PM
P: 162
Quote Quote by gunblaze
where does the 5.4 come from?
from 0.30(19.6)cos22 which is friction
Fusilli_Jerry89
#9
Oct20-06, 05:15 PM
P: 162
shouldn't it be this for the sled:

1000-20(9.8)sin22-0.15(22)(9.8)cos22-5.4=20a
a=52.6m/s/s?
So the teacher is wrong?
Doc Al
#10
Oct20-06, 06:42 PM
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Quote Quote by Fusilli_Jerry89
shouldn't it be this for the sled:

1000-20(9.8)sin22-0.15(22)(9.8)cos22-5.4=20a
This looks right.
a=52.6m/s/s?
Check your arithmetic.


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