How Do Newton's Laws Apply to Multiple Objects in Motion?

AI Thread Summary
The discussion focuses on applying Newton's laws to three crates in motion, specifically calculating acceleration and contact forces. The applied force of 910 N and friction coefficient of 0.277 are given, leading to an initial calculation of acceleration at 9.85 m/s². Participants clarify the correct frictional forces acting on each crate and the equations for determining contact forces between the crates. The conversation also touches on a second problem involving tension in ropes connecting different masses, emphasizing the importance of force direction and mass arrangement. Accurate calculations for both scenarios are essential for understanding the dynamics of multiple objects in motion.
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Homework Statement



Three crates are arranged. The masses of the crates are m1= 29.8kg, m2= 15.3kg, and m3=27.3kg. The applied force of 910 N is directed horizontally, and the coefficient of friction is u= .277. what is the acceleration of the three masses and the magnitude of the contact force between m1 and m2, and m2 and m3

Homework Equations





The Attempt at a Solution



Weight of m1= 292.04
normal force on m2 = 150
FF on m1 = 80.9
FF on m2 = 150
Ff on m3= 74.1
 
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FF on m2 is wrong.
Net force = applied force - FF.
Acceleration = Net force/ mass
 
Thanks! Thats what I tried mathematically but resulted with the wrong answer becuase I didn't multiply M2 by the resistance..
 
And what about the contact forces?
 
If T1 is the contact force between M1 and M2, them
T1 - FF(M1) = M1a.
Similarly you can find T2.
 
rl.bhat said:
If T1 is the contact force between M1 and M2, them
T1 - FF(M1) = M1a.
Similarly you can find T2.

ok so FF of M1= 80.89
FF of M2= 41.5
and the acceleration would equal 9.85m/s/s

so T1- 122.4(29.8)= (29.8)(9.85) ? That seems way to ghih.
 
From where did you get 122.4?
It should be
T1 - 80.89 = 29.8*9.85.
In the formula FF(M1) is the frictional force on mass 1.
 
so T1 would equal 374.42 The question asks what is the contact force between m1 and m2. 374.42 does not answer any of these questions. I have a feeling it has something to do with the friction source but I am not sure.. Thanks!
 
Additionally, I have another question regarding a different problem.
Three crates are connected with ropes and arranged. The masses of the crates are m1= 20kg m2= 16.2kg and m3= 26.6kg. The crates accelerate to the right at 4.05m/s/sand the fiction is 0.279.

what is the magnitude of the tension force on the rope between m1 and m2 = 293.22
T= 2MA
T= 2(16.2)(4.05
T= 293.22N

What is the magnitude of the tension force in rope 2 between m2 and m3 ( m2====m2=====m1)
?
 
  • #10
In the first problem it is not clear to which mass the force is applied. Direction of the force decides the contact force.
In the second case how did you get T = 2MA?
Show the arrangement of the masses and the direction of the force clearly.
 
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