Force of 3 Boxes: Acceleration of System (m1,m2,m3)

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To determine the acceleration of the system consisting of three blocks (m1, m2, m3) each with a mass of 12.5 kg, the total mass must be considered, leading to the equation F = (m1 + m2 + m3) * a. Given the applied force of 91.4 N, the acceleration is calculated as a = F / (m1 + m2 + m3), resulting in a value of 2.43 m/s². Additionally, to find the contact force that block m1 exerts on block m2, Newton's second law can be applied to block m1 alone, factoring in the acceleration of the entire system. This approach clarifies the interaction between the blocks and the resultant forces acting on them. Understanding these principles is crucial for solving similar problems in physics.
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Three blocks of mass 12.5 kg each, on a frictionless horizontal surface are in contact with each other as shown in the figure below.



A force F = 91.4 N is applied to block 1 (mass m1). Determine the acceleration of the system (in terms of m1, m2, and m3).

I did 91.4 = 12.5a
91.4/12.5 = 7.312 but something is missing to that equation. how do I do it in terms of m1, m2, and m3)?
 

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Since all three blocks are in contact you are attempting to accelerate all three of them - i.e. the combined mass.
 
That was an obvious answer that I should've gotten, thank you.

Could you give me a pointer on this one too, having to do with the same problem?
Calculate the force of contact that block m1 exerts on m2.
 
strugglin-physics said:
Calculate the force of contact that block m1 exerts on m2.
Apply Newton's 2nd law to block m1 alone.
 

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