Finding the Magnitude of a Horizontal Force

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To find the magnitude of the horizontal force on a 2.0 kg block on a 60-degree incline, resolve the horizontal force into two components: one along the incline and one perpendicular to it. The weight of the block also needs to be resolved into components, with one being mgcosθ (perpendicular) and the other being mgsinθ (along the incline). The normal force can be calculated using the equation Fn = mgcosθ. By identifying the directions of these forces, the required results can be determined. The discussion emphasizes the importance of breaking down forces into components to solve the problem effectively.
GhostlyKiss
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Homework Statement


A 2.0 kg block on an incline at a 60 degree angle is held in equilibrium by a horizontal force.
A) Determine the magnitude of this horizontal force (disregard friction)
B) Determine the magnitude of the normal force on the block.

I have no idea which equation to use. All the equations I can find don't give me enough information to start the problem. The only equations that I can find include friction and other things not present in the problem.
 
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Hi GhostlyKiss, welcome to PF.
Resolve the horizontal force into two components. One along the inclined plane ant the other perpendicular to the inclined plane. Similarly resolve the weight of the block. From these you can get the required results.
 
How do I resolve the horizontal force? Is Fn=mgcos\oslash an equation I would use?
 
GhostlyKiss said:
How do I resolve the horizontal force? Is Fn=mgcos\oslash an equation I would use?
One component of mg is mgcosθ. What is the other component of mg along the inclined plane? Similarly find Fcosθ and Fsinθ. Identify their directions.
 
Ok thank you :)
 
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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