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

 

[Nickie]

Yes, you are on the right track. To find the relationship between F, M, and m, we can use the concept of torque. Torque is defined as the product of force and the perpendicular distance from the point of rotation to the line of action of the force. In this case, the point of rotation can be taken as the point where the inclined plane meets the horizontal ground.

Let's start by considering the first situation, where F is acting up and along the inclined plane to support a mass M. We can resolve this force into two components - one perpendicular to the inclined plane and one parallel to the inclined plane. The perpendicular component will be balanced by the normal reaction from the inclined plane, while the parallel component will be balanced by the force of gravity acting down the inclined plane.

Now, since the particle is in equilibrium, the net torque acting on it must be zero. This means that the torque due to the force of gravity must be equal and opposite to the torque due to the parallel component of F. Mathematically, we can write this as:

Mg sinθ = Fsinθ

where θ is the angle of inclination of the slope. This equation is independent of the angle of inclination, as you can see, and it relates F and M.

Now let's consider the second situation, where the force of magnitude F acts horizontally to support a mass m on the same inclined plane. Again, we can resolve this force into two components - one perpendicular to the inclined plane and one parallel to the inclined plane. The perpendicular component will be balanced by the normal reaction from the inclined plane, while the parallel component will be balanced by the force of F acting in the opposite direction.

Similarly, for this particle to be in equilibrium, the net torque acting on it must be zero. This means that the torque due to the force of F must be equal and opposite to the torque due to the parallel component of the force of gravity. Mathematically, we can write this as:

F = mg sinθ

Combining this equation with the one we found earlier, we get:

F = Mgsinθ = mgsinθ

which shows the relationship between F, M, and m, independent of the angle of inclination. I hope this helps!