Peyton E.

asked • 03/06/15

A 50-kg crate is being pushed across a level, frictionless floor by a force F = 200 N.

A 50-kg crate is being pushed across a level, frictionless floor by a force F = 200 N. A 10-kg box is placed initially at rest against the crate as shown at the moment the force is applied. The coefficients of friction between the crate and the box are μ_"k" =0.4 and μ_s=0.5.
Draw free-body diagrams of (i) the system, (ii) the crate by itself, and (iii) the box by itself.
Determine the horizontal acceleration of the system.
Does the box remain in its initial position with respect to the crate?
What is the minimum value of F that will keep the box from sliding down?

1 Expert Answer

By:

Michael J. answered • 03/07/15

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5 (5)

Effective High School STEM Tutor & CUNY Math Peer Leader

Brad J.

I think Micheal's calculations on the initial frictional force summation is incorrect! Nu = m1g x u = 50 x 9.8 x .4 = 196N.  I dont know how he got 16.2 N???
 
If I am correct, that would mean his acceleration calculation is wrong as well. The net force should be 200n-196n = 4n.
 
Somebody else wanna comment on this?
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03/14/15

Brad J.

Also, there is no friction between the floor and the system. So wouldn't the net force on the system in the X-Direction be 200n???
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03/14/15

Michael J.

I should have tried using system of equations that applies the static equilibrium concept.  Equation 1 could be the 
∑Fx = 0 and equation 2 could be the ∑Fy = 0.
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03/14/15

Brad J.

Yay! I did get the same answer as Micheal for the last part, 49N is the lowest value of F that would keep the smaller box from falling.
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03/15/15

Brad J.

But, that is not what the original question asked for. So, if the big box is exerting a force of 49 on the small box to keep it from falling, you need to work backwards to calculate that "F" value. Find the A of the system via F=MA: 49N=10kg x A --> 4.9m/s^2. 4.9m/s^2=Fnet/Mass total --> 4.9=F/60 --> F=294N
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03/15/15

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