2 boxes stacked, force applied to bottom box

  • #1
2 boxes are stacked vertically, box A on top with box B below. Mass of box A is 15kg. Mass of box B is 30kg. There is friction between all surfaces; static coefficient of 0.40, and kinetic coefficient of 0.35. The blocks are stationary when the constant force F is applied to box B (bottom box). Determine the resulting acceleration of box B if (a) F=200N and (b) F = 400N.

For (a) i used the following equations:
sumFy = [N][/s] - [m][/A]g - [m][/B]g = 0
sumFx = F - [f][/B] = ([m][/A] + [m][/B])[a][/B]

This gave me the correct solution for the force applied of 1.01m/s^2. HOWEVER, for the 400N force it does not give me the correct solution. I am setting up the equation incorrectly and/or setting up my FBDs incorrectly. I don't thoroughly understand how to distribute the normal forces correctly. By simple intuition one knows that the upper box will fall off of the lower box as the force applied increases, however I have no idea to apply it. Also, I know that the friction of box A onto B and the friction of box B onto the floor surface both go to the left as the force is applied to the right.

Can anyone give my hints on setting this up properly?? Thanks.

(Not sure why the inserts for subscript and summation aren't working :( )
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
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In the first case, the blocks move together, with the same acceleration.. In the 2nd case, they each have different accelerations with respect to the ground, because the static friction force between the blocks is overcome. You must look at the top block separately, and set the kinectic friction force on the top block equal to its mass times acceleration, and solve for the acceleration of the first block. Then look at the bottom block with allfriction forces acting top and bottom, and solve for its acceleration.
 

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