Calculating Force Exerted by Man on 380kg Piano Down an Incline

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To calculate the force exerted by a man on a 380kg piano sliding down a 27-degree incline, it's essential to resolve the forces acting on the piano into components parallel and normal to the incline. The gravitational force acting on the piano can be split into two components using trigonometry, with one component causing the piano to slide down the slope. The effective coefficient of friction is 0.40, which also plays a crucial role in determining the total force opposing the motion. Since the piano is not accelerating, the forces must balance, allowing for the calculation of the man's force by equating the downhill gravitational force and friction with the man's push. Understanding these concepts and utilizing a free body diagram can clarify the problem-solving process.
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Homework Statement


A 380kg piano slides 3.5m down at 27 degree incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of friction is 0.40. Calculate the force exerted by the man


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The Attempt at a Solution


Im confused with this because i think i have to use F =ma but I am not sure because i don't know if i have to use the angle of incline as well to work out the force. Can someone please give me some hints or help
 
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A good place to start is resolving the forcing acting into their components that are parallel and normal to the incline. A free body diagram always helps.
 
what do you mean by resolving the forces acting into their components that are parallel and normal to the incline
 
resolving the force means making all the components of various forces on that object..since there's no acc u can take the force to b zero and thus equate the forces in opp direction..gud luck!
 
If the man was not pushing against the piano, it would accelerate down the slope. The direction of this acceleration (down the slope) is usually the best choice to choose your coordinate system.

In this question, the coordinate system is therefore slightly skewed in relation to the 'ground'.

The gravitational force acting on the piano however is directed straight 'into the ground' and is therefore not aligned with the coordinate system.
Whenever you see this, you can 'split' the force into two components, one along the y-axis and one along the x-axis.
The component of the gravitational force that acts along the line of the slope will be the force that actually makes the piano move.

You can find the size of this force by using some trigonometry (you know the angle of the slope).

Finally, don't forget that there is also friction and the force of the man pushing back that is stopping the piano.
 
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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