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Net force on an object and contact force between objects

by kathyt.25
Tags: contact, force, object, objects
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kathyt.25
#1
May27-09, 06:43 PM
P: 49
1. The problem statement, all variables and given/known data
"Three blocks (m1, m2 and m3) are in contact with each other on a frictionless, horizontal surface. A horizontal force is applied to m1.
http://i4.photobucket.com/albums/y11...sb-pic0556.png
What is the net force on block 1 and block 2?
What is the magnitude of the contact force between blocks 1 and 2?"



2. Relevant equations
Fnet = ma


3. The attempt at a solution
I found acceleration by calculating the Fnet = m(total)*a
a = 17.9 / (1.61 + 3.39 + 4.16) = 1.95m/s/s

Just for curiosity, I tried the Fnet (m1) = (1.61kg)(1.95m/s/s) = 3.14N and it was correct. I have no idea why the applied force, 17.9N is not considered in this... OR the force of m2 on m1 (according to Newton's 3rd law)

I also found the Fnet (m2) = (3.39)(1.95) = 6.61N. Why is it that the net force on each mass is simply equal to that object's mass * acceleration?

Also, I tried finding the magnitude of the contact force between m1 and m2. I figured that since the two forces are going in opposite directiosns, I should subtract Fnet(m2) - Fnet(m1) = 6.61 - 3.14 = 3.47N... but that is wrong!

Also for the contact force between m3 and m2, would the force acting on m2 be the applied force of 17.9N, or the net force that m1 exerts?
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jgens
#2
May27-09, 07:12 PM
P: 1,622
I can't view the picture you attached so I'm not positive about the orientation of the blocks. However, assuming I'm visualizing the correct thing, the net force on each mass is equal to that object's mass * acceleration essentially because it is defined that way. You first found the acceleration of the system which tells you that each block must be accelerating at that same rate as well (assuming I've visualized this correctly). Since, Fnet = ma, it follows that the net force acting on a specific block is ma. Hopefully this all makes sense.

To find the force of 1 on 2, we know that we applied 17.9 N of force to the first block, however, the net force acting on it was only 3.14 N. What does this suggest about the force of 2 on 1?

Again, since I haven't seen the picture, if none of this makes sense, just disregard it. Hope this helps.
Doc Al
#3
May27-09, 07:20 PM
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P: 41,471
The link doesn't work.

Just for curiosity, I tried the Fnet (m1) = (1.61kg)(1.95m/s/s) = 3.14N and it was correct. I have no idea why the applied force, 17.9N is not considered in this... OR the force of m2 on m1 (according to Newton's 3rd law)

I also found the Fnet (m2) = (3.39)(1.95) = 6.61N. Why is it that the net force on each mass is simply equal to that object's mass * acceleration?
That's what Newton's 2nd law tells you: Fnet = ma. If you know the mass and the acceleration, you can immediately find the net force.

To use the net force to find the individual contact forces, you have to find the net force in terms of the individual forces. For example: F1on2 + F3on2 = Fnet_2.

Also, I tried finding the magnitude of the contact force between m1 and m2. I figured that since the two forces are going in opposite directiosns, I should subtract Fnet(m2) - Fnet(m1) = 6.61 - 3.14 = 3.47N... but that is wrong!
Don't subtract the net forces. Subtract the contact forces (on m2, say) and set that equal to the net force on m2.


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