Abstract question- solving for acceleration up an inclined plane

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The discussion revolves around calculating the acceleration of a safe being pushed up an inclined plane with friction. The net force equation is established as Fnet = Fcosθ - mgsinθ - μmgcosθ, leading to the expression for acceleration a = (Fcosθ/m) - gsinθ - μgcosθ. A participant struggles to reach the expected acceleration of 1.777 m/s² using given values for force, mass, angle, and coefficient of friction. Another member emphasizes the importance of including the vertical component of the applied force, which affects the normal force and, consequently, the frictional force. Properly accounting for this component is crucial for arriving at the correct acceleration value.
will5656
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Homework Statement



Burglars are pushing, with a horizontal force Fpush, a safe of mass m and coefficient of kinetic friction μk up a slope of angle θ. What is the safe's acceleration (in abstract terms)?

Homework Equations



as= +/-gsinθ (natural accl down a slope)
friction on a slope= μmgcosθ


The Attempt at a Solution



Fnet= Force up the slope- natural accl down - kinetic friction
Fnet= Fcosθ - mgsinθ - μmgcosθ
ma= Fcosθ - mgsinθ - μmgcosθ
a= Fcosθ/m - gsinθ - μgcosθ

I thought this was right but we have numbers to plug into check for correctness:
fpush= 4000N
m= 500kg
θ= 20°
coeff of kf= 0.20

and acceleration is supposed to be 1.777 m/s2

I can't seem to arrive at this answer.
 
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Hi will5656, Welcome to Physics Forums.

When you draw your free body diagram for the safe, be sure to include the portion of the applied force Fpush which is adding to the normal force...

You've accounted for the "uphill" portion of the force with Fcosθ, but another component of that force is acting to push the safe against the slope. There will be consequences...
 
Remember that the vertical component of the pushing force is increasing the safe's normal force!
 
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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