What is the force exerted by one crate on the other?

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SUMMARY

The discussion focuses on calculating the force exerted by one crate on another when a man pushes a larger crate with a force of 400 N. The larger crate weighs 50.0 kg and the smaller crate weighs 25.0 kg, with a friction force opposing the motion equal to 40% of their combined weight. The acceleration of the crates is determined to be 1.41 m/s². The force exerted by the larger crate on the smaller crate is calculated to be 200 N, correcting the initial misunderstanding of the net force calculations.

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Homework Statement



Two crates rests on the floor, one next to the other. A man pushes with a horizontal force of 400 N on the larger (50.0 kg) crate, which, in turn, pushes the other (25.0 kg) so that both slide as he walks along. Each crate experiences a friction force opposing the motion equal to 40% of its weight.

1.Determine the acceleration of the crates.
Answer in m/s^2

2.What is the force exerted by one crate on the other?
Answer in Newtons


Homework Equations

and

The Attempt at a Solution



1) total friction force Ff = 40 x (50+25) x 9.8/100 = 294 N
=>F(net) = F(applied) - Ff
=>F(net) = 400 - 294 = 106 N
=>By F = ma
=>106 = 75 x a
=>a = 106/75 = 1.41 m/s^2 => CORRECT

2)F(net) = 106 N => INCORRECT. Why? Please help!
 
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the answer is 200N. i did i have take one crate and i calculated the force.according to the force exerted by one is exactly opposite to athor
 
the answer is 200N. i did i have take one crate and i calculated the force.according to the force exerted by one is exactly opposite to anthor
 
the answer is 200N. i did i have take one crate and i calculated the force.according to the force exerted by one is exactly opposite to anothor
 
The force exerted on one crate to the other is:

A = 50 kg; B = 25 kg

F = ma
400 = 75.0 kg x a
a = 5.3 m/s^2

F + F(BA) = m1 x a
F + F(BA) = 50 kg x 5.3 m/s^2
F + F(BA) = 265 N

F + F(BA) = 265 - F
F(BA) = 265 - 400
F + F(BA) = -135 N (But put in 135 because that's the correct answer)
 
Thanks! I got it!
 

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