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A 5.00 kg mass rests on top of a 15.0 kg mass that is pulled to the right with a constant horizontal force of 125 N. What is the minimum coefficient of static friction between the 5.00 kg and 15.0 kg masses if the smaller one does not slide? Assume there is no friction between the floor and the 15 kg mass.

By Jason817 (Jason817) on Thursday, October 02, 2003 - 08:48 pm: Edit

bump

By Perry2006 (Perry2006) on Thursday, October 02, 2003 - 10:07 pm: Edit

F = ma
a = 125/15=8.333

Fs, max = 5*8.333
Fs, max = 41.6665 (That is the force that is imposed on the 5.00 kg object)

In order to make the small object (5.00 kg) not slide, the Us (Static friction) must be strong enough to resist the Fs, max described above.
Us * Fn = Fs, max
Us * 5(9.81) = 41.6665
Us = .84947
Us = .849

Correct me if I got it wrong.

By Jason817 (Jason817) on Thursday, October 02, 2003 - 10:28 pm: Edit

thanks. I dont know if its right or not...

Second one:

A large box (m=25.0 kg) is pushed to the right with an applied force of F across a frictionless table. A second, small box (m=4.00 kg) is in contact with the front of the box (not on the ground though). If the coefficient of friction between the two boxes is 0.710, calculate the minimum value of F so that the small box does not slide down the face of the larger box. Hint: The reaction force between the 4.00 kg and 25.0 kg boxes is the normal force for the 4.0 kg box

By Perry2006 (Perry2006) on Thursday, October 02, 2003 - 10:50 pm: Edit

Btw, how can the small box come in contact with the *front* of the box and is high in the air? (should it be "on top of"?)

This is my solution although I'm not 100% sure about it.

Us * Fn = Fs, Max
Fs, Max = 0.710 * 4 (9.81)

Fs, Max / m = a
so, a = 0.710 * 4 (9.81) / 4 = 0.710 (9.81) = 6.9651 m/s^2

Force for the large block is the Acceleration (which got from above) * mass of the large block.

Thus we have:
F = 6.9651 * 25
F = 174.135 = 174 N

By Jason817 (Jason817) on Thursday, October 02, 2003 - 11:14 pm: Edit

Btw, how can the small box come in contact with the *front* of the box and is high in the air? (should it be "on top of"?)

no, check my profile for the pic

By Perry2006 (Perry2006) on Thursday, October 02, 2003 - 11:34 pm: Edit

I c. My answer is wrong then.
I'm sorry I do not know the answer for this one.

By Jason817 (Jason817) on Thursday, October 02, 2003 - 11:53 pm: Edit

yeah this one was hella hard. thanks.