Forces: Wedge and Block Problem + Pulley System Problem

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To solve the wedge and block problem, the horizontal acceleration of both the wedge and block must be equal to prevent sliding. The force required to keep the block stationary on the wedge can be derived from the equation mgsin18° = m1ax, but further steps are needed for clarity. For the pulley system, focusing on one block, particularly the 20N block, can simplify the analysis amidst the complexity of multiple pulleys. The equilibrium condition should be applied, balancing the forces acting on the blocks to find the tension. Overall, breaking down the problems into manageable parts and applying equilibrium principles is essential for finding the solutions.
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Homework Statement


9) A 15 kg wedge is pushed along a frictionless surface. There is no friction between the wedge and the block. What force, F, is required to keep the block from sliding up or down the wedge?

See attachment for diagram.

10) Find the tension, T.

See attachment for diagram.

Homework Equations



ƩF=ma

The Attempt at a Solution



9) So I know that in order for the block to not slide, the horizontal acceleration for both the wedge and the block have to be equal. To solve for the horizontal acceleration I used the horizontal component of block (m1) (where the xy axis is tilted).

mgsin18° = m1ax

I don't really know where to go from there because I do not feel that I am on the right track.

10) All I have figured out is if all three blocks were to be laid out on a flat surface, the three blocks would all move in one direction. I am unsure about what I'm supposed to do with all the pulleys. I was wondering if all I really needed to do was to focus on one block (the 20N one?). I am admittedly overwhelmed by how many pulleys there are.
 

Attachments

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For question 9) remembering that only the wedge will experience the force F, and therefore the acceleration. Since the wedge/block interface is smooth if F was very large the block would simply slide up until it fell of the top.

So you could equally consider the acceleration to be in the opposing direction, just for the block. Try drawing the block, with the acceleration from the wedge, and gravity.For question 10) I assume this is supposed to be equilibrium, in which case the forces must balance. So write an equilibrium statement of forces up and forces down. I think this is designed to look scary
 
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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