Forces btwn Boxes: Find Acc & Magnitude of Contact Forces

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In this discussion, participants analyze a physics problem involving three boxes on a frictionless surface, where a 75 Newton force is applied to box 3. The total mass of the system is calculated as 85 kg, leading to an acceleration of 0.88 m/s² for all boxes. The contact force between box 3 and box 2 is determined to be 22 N, which is the net force acting on box 3. Further clarification is sought regarding the direction of contact forces, with emphasis on understanding that contact forces between boxes are not equal and must be considered in relation to the applied force direction. The discussion highlights the importance of free body diagrams and Newton's second law in solving for contact forces.
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Homework Statement


boxes 1,2 and 3 are touching each other. a force is applied to box 3. assume that the surface is frictionless. the masses are m1 20 kgs, m2 is 40 kgs, and m3 is 25kgs. the applied force to box 3 is 75 Newtons. find te acceleration of the blocks and the magnitude of there contact forces.


Homework Equations



Fnet=ma

The Attempt at a Solution

box 3 has applied force of 75N and and opposite normal force of 75N?? not sure what the value of the other horizontal forces are?? also not sure where to include the weight of the boxes??
 

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I think I've figured the problem out just not sure if the contact forces are all the same magnitude?
 
pb23me said:
I think I've figured the problem out just not sure if the contact forces are all the same magnitude?
Please show your work. You first need to look at the entire system of 3 boxes to solve for the acceleration using Newton 2. Then look at each box separately in a free body diagram, noting any contact normal or applied forces in the x direction, and use Newton 2 again for each box. They all move together, so each has the same acceleration. The contact forces are not the same.
 
i considered the whole system as one unit. added the masses to get mass 85kgs. Fnet= 75N then found acceleration = .88 m/s^2. i used the accleration and mass of box 3 to determine the force of box 3 acting on box 2 F3=22N now how do i find the other contact forces?
 
am i correct in saying that the force applied on box 2 from box 3 is 22N or is that the net force?
 
When you looked at block 3, the force you solved, 22 N , was the NET force acting on block 3. Since Block 3 has an applied force of 75 N acting left, then the normal force of 2 on 3 must be __?___ in which direction? Then try the same approach using Block 1.
 
idk the contact force is 53N so assume it would be -53N of 2 on 3.
 
What do you mean by the minus sign? If the 75 N force acts left, the 53 N force of 2 on 3 must act right. The net force (22N) is always in the direction of the acceleration (in this case, F_net acts left).
 
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