Determining the Net Force on a Wedge and Block System

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The discussion revolves around calculating the net force on a wedge and block system where a block remains stationary on a frictionless wedge. Given the masses (M = 20.0 kg, m = 4.00 kg) and the angle (theta = 34.0 degrees), participants emphasize the application of Newton's second law of motion. One user expresses confusion about the reference frame and the normal force in their free body diagram. The importance of providing detailed calculations and clear attempts at solutions is highlighted, as vague descriptions are insufficient for assistance. Clarity in the approach is essential for solving the problem effectively.
PAstudent
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Homework Statement


[/B]
For the situation shown in the diagram, the block of mass m remains in place on the sloping front of the accelerating wedge, even though that sloping front is frictionless. Given these values, M = 20.0 kg, m = 4.00 kg, and theta= 34.0, calculate the net force acting on the wedge plus block (M+m) system.

Force problem.png


Homework Equations


Newton's second law of motion

The Attempt at a Solution


At first, I tried turning the frame of reference so my normal force for the block was on the y-axis on my free body diagram. However, when I was looking at it more, I was confused on what I was actually trying to find with the wedge.
 
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PAstudent said:

Homework Statement


[/B]
For the situation shown in the diagram, the block of mass m remains in place on the sloping front of the accelerating wedge, even though that sloping front is frictionless. Given these values, M = 20.0 kg, m = 4.00 kg, and theta= 34.0, calculate the net force acting on the wedge plus block (M+m) system.

View attachment 88942

Homework Equations


Newton's second law of motion

The Attempt at a Solution


At first, I tried turning the frame of reference so my normal force for the block was on the y-axis on my free body diagram. However, when I was looking at it more, I was confused on what I was actually trying to find with the wedge.
A vague description of an attempted method does not qualify as a posted attempt. We need to see your working.
 
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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