2 boxes stacked, force applied to bottom box

In summary, the conversation discusses the setup of two boxes stacked vertically with friction between all surfaces. The masses of the two boxes are 15kg and 30kg, and a constant force of 200N and 400N is applied to the bottom box. The resulting acceleration of the bottom box is determined using equations, but there is confusion on setting up the equation correctly for the 400N force due to the different accelerations of the two boxes. The suggestion is to look at the top block separately and solve for its acceleration, then look at the bottom block with all friction forces and solve for its acceleration.
  • #1
johnsonc007
8
0
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 :( )
 
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  • #2
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.
 

Related to 2 boxes stacked, force applied to bottom box

1. What is the force applied to the bottom box?

The force applied to the bottom box is the weight of the top box plus any additional force applied by an external source.

2. How does the force applied to the bottom box affect the stability of the stacked boxes?

The force applied to the bottom box determines the stability of the stacked boxes. If the force is too strong, it can cause the boxes to topple over. If the force is too weak, the boxes may not be able to support the weight of the top box and may collapse.

3. What happens to the force when the top box is removed?

When the top box is removed, the force on the bottom box decreases. This can cause the stacked boxes to become unstable and potentially topple over if the remaining force is not enough to support the bottom box.

4. How does the weight of the boxes affect the force on the bottom box?

The weight of the boxes directly affects the force on the bottom box. The heavier the boxes are, the greater the force placed on the bottom box. This can impact the stability of the stacked boxes and the amount of force needed to keep them from toppling over.

5. Can the force on the bottom box be increased to support more weight on top?

Yes, the force on the bottom box can be increased by adding more weight on top or by applying an external force. However, it is important to ensure that the bottom box is able to support the increased force to prevent collapse or toppling over.

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