Calculating Contact Forces- Three objects

[student2013]

Homework Statement

If you have a system with three boxes, two side by side and one on top of the first box and there is a horizontal force acting on the boxes, do you have to account for the third box in calculating the contact force between two of them? Say if box 1, box 2, and box 3 (three is on top of box 1) had masses of 10 kg, 5 kg, and 6 kg, respectively. And the horizontal force was equal to 20 N and you were asked to calculate the contact force between boxes 1 and 2...would you have to divide 20N by the sum of all THREE masses or just the masses of ONE and TWO in order to get the acceleration to calculate contact force?

Homework Equations

You would need to use F = ma (a = F/m) to get the acceleration in the horizontal direction and then to get the contact force.

The Attempt at a Solution

I thought that you would have to take the box on top into account so a = 20 N / (10 kg + 5 kg + 6 kg) = 0.95 m/s^2

And then to get the contact force between boxes 1 and 2, you would multiply 0.95 m/s^2 by 5 kg to get 4.76 N.

 

[PhanthomJay]

Yes you are correct

 

[CWatters]

Yes correct but...

If the force is applied to the lower boxes then you may need to look at the friction force between upper and lower boxes. If you try to accelerate the lower boxes too fast they will "leave the top box behind". The top box can only accelerate as fast as the friction force/it's mass. That might be less than the applied force/total mass.

 

[PhanthomJay]

Yes correct but...

If the force is applied to the lower boxes then you may need to look at the friction force between upper and lower boxes. If you try to accelerate the lower boxes too fast they will "leave the top box behind". The top box can only accelerate as fast as the friction force/it's mass. That might be less than the applied force/total mass.

Yes, thanks, I overlooked this, you include all thee masses in your calc for the acceleration of the 3-block system only if all three move together. As an example, if the bottom blocks are on a frictionless surface, and if there is also no friction between the top and bottom block, and a force of 20 N is applied to the lower block on the left, then the top block will not move at all and the acceleration of the lower blocks is determined without considering the mass of the upper block. The whole problem changes if the 20 N force is applied to the upper block, so you have to be more specific about values of friction coefficients and point of application of the applied force.

 

[Nickie]

This would be the contact force between the two boxes since they are the ones directly experiencing the force.

Yes, you are correct in your approach. In this scenario, the force of 20 N is acting on the entire system of three boxes, so you would need to take all three masses into account when calculating the acceleration. This is because all three boxes are connected and will experience the same acceleration.

To calculate the contact force between boxes 1 and 2, you would need to use the individual mass of box 2 (5 kg) since it is the one in direct contact with box 1. Multiplying the acceleration (0.95 m/s^2) by the mass of box 2 would give you the contact force between the two boxes (4.76 N). The mass of box 3 (6 kg) would not be relevant in this calculation since it is not in direct contact with box 1.

In summary, when calculating contact forces between two objects, you would only need to consider the masses of those two objects, but when determining the overall acceleration, you would need to take into account the masses of all objects in the system.