Physics HW help? Normal Force between two objects?

AI Thread Summary
The discussion revolves around a physics homework problem involving three blocks on a frictionless table, where a force is applied to the first block. The user successfully calculated the mass of the second block (m2) to be 11.29 kg using the equation F=ma. The main question is about determining the normal force between blocks 2 and 3, with an initial calculation yielding 40.8 N. Clarification was provided that this force is indeed relevant, and it was suggested that the normal force could be rounded to 41 N for significant figures. The conversation emphasizes the importance of understanding forces acting on the blocks in the context of Newton's laws.
VoNi
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Physics HW help!? Normal Force between two objects?

Hi, I've had trouble trying to gigure out this problem! I would really appreciate the help!

The Problem:

Three blocks rest on a frictionless, horizontal table (see figure below), with m1 = 7 kg and m3 = 12 kg. A horizontal force F = 103 N is applied to block 1, and the acceleration of all three blocks is found to be 3.4 m/s2.

A. Find m2
B. What is the normal force between blocks 2 and 3?
I coouldnt post the figure(picture) but I can explain it.. The three blocks are all lined up against each other. So block 1(m1) is first, then block two(m2) is right next to block1, then block 3(m3) is against block 2.. they are all lined up on top of the table.. side by side..
Relevant equation:

I found the mass of block 2 by doing F=ma. or M= F/A
7+12+m2= 103/3.4
19+m2= 30.29
m2= 11.29

So M2 is 11.29

My only problem is that I don't know how to find the normal force between blocks 2 and 3


Attempt at a Solution:

The way i tried to find the solution was figuring out what force is acting on block three so that it can accelerate 3.4 m/s^2.
I did:
F=m*a
F=12*3.4
F= 40.8

Now, is 40.8 the normal force between block two and three or is that just an irrelevant number?

Thank you so much for the help!
 
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VoNi said:
The way i tried to find the solution was figuring out what force is acting on block three so that it can accelerate 3.4 m/s^2.
very good.
I did:
F=m*a
F=12*3.4
F= 40.8
Yes, but don't forget, F is actually F_net (in Newtons, don't forget the units)
Now, is 40.8 the normal force between block two and three or is that just an irrelevant number?
It is very relevant. If F_net = 40.8 N, what forces act on that block in the x direction? Draw a free body diagram of block 3. Is F_net and and F_normal the same??
 


Thank you SO much! that was very helpful! and the normal force, is 40.8 N. Thanks for taking the time to help me out! =)
 


You are welcome. Note that the normal force should probably (I think) be rounded off to 41 N (it's a significant figure thing), but your answer is essentially correct. Wecome to PF!
 
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