Contact Force with two masses given

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A horizontal force of 32 N is applied to a 3.0 kg box, pushing it against a 5.0 kg box on a smooth floor. The total mass of both boxes is 8 kg, resulting in an acceleration of 4 m/s². The contact force between the two boxes can be calculated using the formula F = ma, leading to a contact force of 20 N. The discussion highlights the importance of free body diagrams in solving such problems. Ultimately, the correct answer for the contact force is 20 N.
lexikobie
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Homework Statement


A 3.0 kg and a 5.0 kg box rest side by side on a smooth, level floor. A horizontal force of 32 N is applied to the 3.0 kg box pushing it against the 5.0 kg box, and, as a result, both boxes slide along the floor. How large is the contact force between the two boxes?

a) 32N b)24N c) 0N d)20N e) 12N

Homework Equations


F=ma

The Attempt at a Solution


I found the total mass which is 8kg. And then using the F=ma equation, I found my acceleration which is 4 m/s^2. And now...I'm completely stuck on what to do. In class we weren't given an example like this, so I don't know where to go from here...
 
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lexikobie said:

Homework Statement


A 3.0 kg and a 5.0 kg box rest side by side on a smooth, level floor. A horizontal force of 32 N is applied to the 3.0 kg box pushing it against the 5.0 kg box, and, as a result, both boxes slide along the floor. How large is the contact force between the two boxes?

a) 32N b)24N c) 0N d)20N e) 12N

Homework Equations


F=ma

The Attempt at a Solution


I found the total mass which is 8kg. And then using the F=ma equation, I found my acceleration which is 4 m/s^2. And now...I'm completely stuck on what to do. In class we weren't given an example like this, so I don't know where to go from here...
Can you show us your free body diagram (FBD) for each of the two boxes? :smile:
 
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You know mass and acceleration of the 5 kg block.

But I'm not sure if the blocks are really supposed to be frictionless. The answer does not depend on that question, but the points you get for the answer could if there are points for intermediate steps.
 
berkeman said:
Can you show us your free body diagram (FBD) for each of the two boxes? :smile:
Yes lol I do but there's no way for me to upload my image from my smartphone since my laptop won't allow me to connect it anymore. Basically I can't upload an image, but I did draw a free body diagram, with two boxes sitting on a horizontal line with the force of 32N being pushed to the right (----->) like this, upon the 3.0kg box. Dunno if that really...explains it, but that's what I drew...
 
I finally figured it out! F_2 = m_2 x acceleration. Therefore I get 20 as my answer! Iol I at least appreciate the time you guys used to reply to my question. Many thanks.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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