# Holding a block of mass in equilibrium on a slope

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1. Mar 15, 2016

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
F = MA
Normal Force y = Fg*sin(theta)
Normal Force x = Fg*cos(theta)

3. The attempt at a solution

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.

2. Mar 15, 2016

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.

3. Mar 15, 2016

### drvrm

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]

4. Mar 15, 2016

So would it be 500*cos(30) = mg * sin(30)

5. Mar 15, 2016

### drvrm

It seems to be correct - actually the angle made by F with perpendicular to the incline is (90 -30) degrees.

6. Mar 16, 2016

### haruspex

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.