# Homework Help: Newtons Laws Multiple Objects

1. Nov 14, 2009

### tigerwoods99

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. Nov 15, 2009

### rl.bhat

FF on m2 is wrong.
Net force = applied force - FF.
Acceleration = Net force/ mass

3. Nov 15, 2009

### tigerwoods99

Thanks! Thats what I tried mathematically but resulted with the wrong answer becuase I didn't multiply M2 by the resistance..

4. Nov 15, 2009

### tigerwoods99

And what about the contact forces?

5. Nov 15, 2009

### rl.bhat

If T1 is the contact force between M1 and M2, them
T1 - FF(M1) = M1a.
Similarly you can find T2.

6. Nov 15, 2009

### tigerwoods99

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.

7. Nov 15, 2009

### rl.bhat

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.

8. Nov 15, 2009

### tigerwoods99

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 im not sure.. Thanks!!

9. Nov 15, 2009

### tigerwoods99

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)
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10. Nov 15, 2009

### rl.bhat

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.