Particle on an inclined plane formula proof.

[Final_HB]

Homework Statement

A force of magnitude F acting up and along a smooth inclined plane, can support a mass M in equilibrium. If a force of the same magnitude acts horizontally, it can support a mass m on the same inclined plane in equilibrium.

Find a relationship between F, M and m which is independent of the angle of inclination of the slope.

Homework Equations

The Attempt at a Solution

Im lost with this I have no idea where to start. I guess you:
Take each force and particle separately.
Resolve the forces into horizontal/vertical components.
Let the sum of all these forces equal 0.
Do the same for the second particle and equate the two equations to each other through F.

Good so far?

 

[ehild]

Homework Statement

A force of magnitude F acting up and along a smooth inclined plane, can support a mass M in equilibrium. If a force of the same magnitude acts horizontally, it can support a mass m on the same inclined plane in equilibrium.

Find a relationship between F, M and m which is independent of the angle of inclination of the slope.

Homework Equations

The Attempt at a Solution

Im lost with this I have no idea where to start. I guess you:
Take each force and particle separately.
Resolve the forces into horizontal/vertical components.
Let the sum of all these forces equal 0.
Do the same for the second particle and equate the two equations to each other through F.

Good so far?

Yes, it is good in principle. Go ahead.

_{( You might consider to decompose the forces into components, parallel and perpendicular to the slope)}

ehild

 

[Final_HB]

when you say resolve perpendicular/parallel to the slope, do you mean like:

F=Fcosα + Fsinα

 

[ehild]

when you say resolve perpendicular/parallel to the slope, do you mean like:

F=Fcosα + Fsinα

Remember the force is vector. Its components are different if it is parallel with the slope, or horizontal. Only the magnitude is the same in both cases.
If it is a horizontal force, the parallel component is Fcosα and the normal component is -Fsinα.
The whole force vector is $\vec F = F\cos(\alpha )\hat i-F\sin(\alpha) \hat j$ where $\hat i$ and $\hat j$ are unit vectors along the slope (up) and perpendicular to it (out), respectively.


ehild