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

Click For Summary

Homework Help Overview

The discussion revolves around the application of Newton's laws to a scenario involving three crates with specified masses and forces. Participants are analyzing the acceleration of the crates and the contact forces between them, while also addressing frictional forces at play.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the acceleration of the system and the contact forces between the crates. There are discussions about the net force, frictional forces, and the application of Newton's second law. Some participants express confusion regarding the calculations and the assumptions made about the forces involved.

Discussion Status

The discussion is ongoing, with participants providing various calculations and questioning each other's reasoning. There is an acknowledgment of potential errors in the calculations, particularly concerning friction and the application of forces. Some participants are exploring different interpretations of the problem setup.

Contextual Notes

There are indications of missing information regarding the direction of applied forces and the arrangement of the crates, which may affect the calculations of contact forces and tensions. Additionally, the presence of friction coefficients is noted, which influences the dynamics of the system.

tigerwoods99
Messages
99
Reaction score
0

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
 
Physics news on Phys.org
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 because 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.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K