zeion
Feb28-09, 08:04 PM
Hello. This is my first time posting, I hope I have done everything correctly.
1. The problem statement, all variables and given/known data
A small box is resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.
1) What force causes the small box to accelerate horizontally?
2) If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.
3. The attempt at a solution
1)
Horizontal Fa on the lower box causes the top box to move, but Fa cannot > Fs between the two boxes or top will slip. But if Fs between two boxes is > than Fa then bottom cannot accelerate therefore top will not accelerate. Therefore Fa = Fs.
Fa causes the top box to move, but Fs between top and bottom causes top to not slip.
2)
\mu_s = \frac{F_s}{F_n}
It does not slip so F_A = F_s = ma
There is not vertical movement so F_n = mg
Therefore \mu_s = \frac{ma}{mg} = \frac{a}{g} = \frac{2.5m/s^2}{9.8m/s^2} = 0.25
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1. The problem statement, all variables and given/known data
A small box is resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.
1) What force causes the small box to accelerate horizontally?
2) If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.
3. The attempt at a solution
1)
Horizontal Fa on the lower box causes the top box to move, but Fa cannot > Fs between the two boxes or top will slip. But if Fs between two boxes is > than Fa then bottom cannot accelerate therefore top will not accelerate. Therefore Fa = Fs.
Fa causes the top box to move, but Fs between top and bottom causes top to not slip.
2)
\mu_s = \frac{F_s}{F_n}
It does not slip so F_A = F_s = ma
There is not vertical movement so F_n = mg
Therefore \mu_s = \frac{ma}{mg} = \frac{a}{g} = \frac{2.5m/s^2}{9.8m/s^2} = 0.25
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