What is the magnitude of the contact force between the boxes?

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SUMMARY

The discussion revolves around calculating the contact forces between three boxes subjected to a pushing force of 7.50 N. The masses of the boxes are m1 = 1.30 kg, m2 = 3.30 kg, and m3 = 5.10 kg. To find the acceleration, the total mass of the system must be considered, leading to the correct formula a = F/m, where m is the sum of all box masses. The contact forces can then be determined using Newton's Second Law applied to each box individually.

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youxcrushxme
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I'm really confused about this problem -

As shown in Figure 5-22, a force of magnitude 7.50 N pushes three boxes with masses m1 = 1.30 kg, m2 = 3.30 kg, and m3 = 5.10 kg.

05-19alt.gif


(a) Find the magnitude of the contact force between boxes 1 and 2.
(b) Find the magnitude of the contact force between boxes 2 and 3.

So F=7.5N. First you need to find the acceleration right? So I got a=F/m and did 7.5/(1.3+3.3) for part a. That gave me 1.63. I multiplied that with 3.3 because that's the mass of the second box and got a contact force of 5.379N but it's telling me that it's not right. If anyone could help I'd really appreciate it
 
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youxcrushxme said:
So F=7.5N. First you need to find the acceleration right? So I got a=F/m and did 7.5/(1.3+3.3) for part a. That gave me 1.63.
What happened to the third mass?
 
I'm confused about how I work that in there...I was kind of doing the problem as if the third one wasn't there which is wrong. So I do 7.5/(1.3+3.3+5.1) to get a? And then once I have a how do I find the contact force?
 
You apply Newton's 2nd law: First to the entire 3-block system, to find a. Then to each mass separately, to find the contact forces. (Realize that the F in F = ma is the net force.)
 
Ahh, I figured it out, thanks
 

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