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

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To determine the contact forces between the boxes, first calculate the total acceleration of the system using the total mass of all three boxes. The correct formula is a = F/m, where F is the total applied force of 7.5 N and m is the combined mass of 9.7 kg (1.3 kg + 3.3 kg + 5.1 kg). After finding the acceleration, apply Newton's second law to each box to find the individual contact forces. The initial confusion stemmed from neglecting the third box in the calculations. Ultimately, the correct approach involves considering all boxes to accurately compute the contact forces.
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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