Finding the Magnitude of a Horizontal Force

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Homework Help Overview

The problem involves a 2.0 kg block on a 60-degree incline held in equilibrium by a horizontal force, with a focus on determining the magnitude of this force and the normal force acting on the block, while disregarding friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about which equations to use, noting a lack of relevant information. Some participants suggest resolving the horizontal force into components and relate it to the weight of the block, while others question how to perform this resolution and discuss the components of gravitational force along and perpendicular to the incline.

Discussion Status

The discussion is ongoing, with participants exploring different methods to resolve forces and clarify the relationships between the components. Guidance has been offered regarding the resolution of forces, but no consensus has been reached on the approach to take.

Contextual Notes

Participants are working under the constraint of disregarding friction and are attempting to clarify the setup of the problem, particularly the components of forces involved.

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[tex]\oslash[/tex] an equation I would use?
 
GhostlyKiss said:
How do I resolve the horizontal force? Is Fn=mgcos[tex]\oslash[/tex] 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 :)
 

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