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Two bodies of masses M1 and M2 are on a frictionless surface and are pushed with a force F, causing them to accelerate. The force between the bodies can be calculated using the equation f = FM2/(M1 + M2). This result is derived by applying Newton's second law to both masses and recognizing they share the same acceleration. A free-body diagram is essential for visualizing the forces acting on each mass. The discussion concludes that the problem is appropriate for a grade 11 physics level.
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Homework Statement



Two bodies of masses M1& M2 resting on frictionless surface. Bodies touch each other without any friction.
What will be the force between the bodies, If they are pushed from one end with force F and both are accelerating towards other end?
Answer given is (M2 x F)/(M1+M2).
How is it arrived at?

Homework Equations



F= G M1XM2/ distance squared

The Attempt at a Solution



I am stumped. I cannot arrive at this with above equation.
 
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You really need to draw a diagram with the forces and accelerations to solve this one - a free-body diagram.
 
You shouldn't be using Newton's Law of Universal Gravitation here.

When you draw a free body diagram of M1, you'll recognize that there is a resisting force (f) on it from the other mass.

1)

ƩFx = M1a

F - f = M1aThen draw a free body diagram of M2, which has no resisting force.

2)

ƩFx = M2a

F = M2a

You should understand the fact that both masses will have the same acceleration.

Knowing that detail, you can manipulate your equations in order to solve for the missing force "f."

Start by solving the 2nd equation for acceleration:

a = F/M2

Plug that back into the 1st equation and you should have:

f = FM2/(M1+M2)
 
Thank you. It was bugging me.

It looks like a grade 11 question to me.
 
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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