Holding a block of mass in equilibrium on a slope

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SUMMARY

The discussion focuses on determining the weight of a block held in equilibrium on a 30-degree incline by a horizontal force of 500N, while ignoring friction. The key equations involved are F = MA, Normal Force y = Fg*sin(theta), and Normal Force x = Fg*cos(theta). Participants emphasize the importance of correctly identifying the angles between the forces and the incline, particularly noting that the angle between the horizontal force and the incline is 90 - 30 degrees. The consensus is that the net force along the incline must be zero, leading to the equation 500*cos(30) = mg*sin(30).

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adam19325
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Homework Statement


A block of mass is held in equilibrium on an incline of angle 30 degrees by the horizontal force 500N. Determine the blocks's weight, ignore friction.

Homework Equations


F = MA
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)

The Attempt at a Solution


WlLTfvI.jpg

I'm just wondering if this solution is correct. I'm worried because I think the angle is not the same for the horizontal force as the force of gravity pulling it downward.
 
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It is the forces along the incline that need to balance. You need to get correct expressions for the component of the applied force and for the gravitational force component along the incline. There is no requirement for the forces perpendicular to the incline, because the incline will supply the necessary force to balance any applied and/or gravitational forces in this direction.
 
adam19325 said:

Homework Statement


A block of mass is held in equilibrium on an incline of angle 30 degrees by the horizontal force 500N. Determine the blocks's weight, ignore friction.

Homework Equations


F = MA
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)

The Attempt at a Solution


WlLTfvI.jpg

I'm just wondering if this solution is correct. I'm worried because I think the angle is not the same for the horizontal force as the force of gravity pulling it downward.
express correctly the angles (between the forces).
For example what is the angle between horizontal push and the inclined plane?
Secondly the block is not sliding so net force along the incline must vanish-
you may draw a free body diagram?[/QUOTE]
 
So would it be 500*cos(30) = mg * sin(30)
 
adam19325 said:
So would it be 500*cos(30) = mg * sin(30)

It seems to be correct - actually the angle made by F with perpendicular to the incline is (90 -30) degrees.
 
adam19325 said:
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)
I don't know what situation those equations apply to, but it's not this one. There is no value in remembering equations separately from the context in which they apply.
 

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