Solving Block on Incline Physics Problem

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The discussion focuses on solving a physics problem involving a block on an incline. The user presents their free body diagram (FBD) and equations but receives feedback indicating errors in the representation of forces and the use of equations. Specifically, it is pointed out that the gravitational force (Fg) should be resolved into its components without retaining the downward force in the FBD. Additionally, there is a correction regarding the order of gravitational components in the equations. The user is advised to make these adjustments to achieve the correct solution.
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Can someone tell me if I'm doing this problem right?

Homework Statement


Here is the picture of the problem:
http://i210.photobucket.com/albums/bb23/Kyashii628/block.jpg

In my FBD, I have Fg pointing down, Fgy perpendicular to the incline and pointing down, Fgx parallel to incline pointing down the ramp (to left), kinetic friction pointing up the ramp (to right), and normal force pointing opposite from Fgy.

Homework Equations


Fnet= -ma
fk- Fgx= -ma
u(Fn)= -ma
u(345.0615986)= 648.9664808 -240
u= 1.182397583
 
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Maiia said:
Can someone tell me if I'm doing this problem right?

Homework Statement


Here is the picture of the problem:
http://i210.photobucket.com/albums/bb23/Kyashii628/block.jpg

In my FBD, I have Fg pointing down, Fgy perpendicular to the incline and pointing down, Fgx parallel to incline pointing down the ramp (to left), kinetic friction pointing up the ramp (to right), and normal force pointing opposite from Fgy.
good, but remember in your FBD, that once you resolve Fg into its x and y components, you should 'cross out' the Fg downward force.

Homework Equations


Fnet= -ma
fk- Fgx= -ma
good
u(Fn)= -ma
I'm sure you meant u(Fn) = fk
u(345.0615986)= 648.9664808 -240
you have your mg(sin theta)and mg(cos theta) reversed.
u= 1.182397583
Correct your error and your result should be fine.
 
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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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