How to Solve a Basic Sliding Box Problem on a Frictionless Ramp?

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To solve the sliding box problem on a frictionless ramp, the applied horizontal force must counteract the gravitational force acting down the incline. The user initially calculated the horizontal force as a hypotenuse, leading to confusion about the normal force and gravitational components. After further analysis, a force of 468 N was determined, with a 220 N component potentially representing the normal force. It was confirmed that the component of the applied force must equal the gravitational force down the incline for the box to move at constant speed. Understanding the normal force is deemed unnecessary for this specific problem.
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Hi

A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
  1. when the ramp is frictionless.

My attempt at a solution.

I diagrammed the problem as in the file uploaded. To figure out the horizontal force required to counter the force of the box sliding down the incline I figured that the horizontal force would have to be the hypotenuse because it needs to be stronger than both Fn and Fg(sin28). However, it shows the normal force being larger than Fg(sin28) in terms of side lengths when it should be the opposite way around. I am confused on how to solve this problem even though it has been posted before on this website. Any help would be much appreciated.

Thanks
Brendan
 

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Using a right angle calculator I came up with a more logical answer of 468 N. But that leaves the opposite side of my triangle at 220 N which I am not sure how I would get. Or would that 220 N be the Normal force pushing back against my horizontal force? If it is then I guess in this question it is inconsequential to know the normal force of the box against the ramp.
 
Brendan Webb said:
I guess in this question it is inconsequential to know the normal force of the box against the ramp.

Indeed. The component of F directed up the incline just needs to be equal and opposite to the force of gravity directed down the incline.
I got pretty much the same answer as you. Now, if they had asked for the normal force on the box...
 
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Thanks for the verification!
 
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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