Understanding Normal Force on an Inclined Plane: Pushing vs. Carrying a Mass

[Donald K]

Is there any difference in the force I need to apply between pushing and carrying a mass upward along an inclined plane with a constant velocity ?

I'm puzzled with the normal force of the mass at the inclined plane. When I push up a mas, the normal force of it is exerted by the inclined plane. If I carry it, the normal force is exerted by me.

Thanks!

 

[Doc Al]

Not sure I quite understand the question. If you're pushing a mass up a frictionless incline, then you only need to push a fraction of the weight (mgsinθ). If you are carrying it up an incline (with friction, otherwise you won't be able to walk), then you must support the full weight. OK. What's the issue?

 

[grzz]

The push of the plane on the mass is sometimes called the reaction of the plane on the mass. If there is no friction, this reaction will be normal to the plane. Hence it is usually called the 'normal force'. This is the force with which the plane pushes on the mass.

 

[Donald K]

Assume there is no friction.

When I push up a mass up the inclined plane ,the reaction force (mgcosθ) is offered by the plane. The force I need to apply is Mgsinθ, <mg.

If I carry the mass, (e.g. I put it above my head) the reaction force is provided by me.

The total force I need to provide is mgcosθ + Mgsinθ which is > mg.

Is it a paradox?

 

[Doc Al]

The total force I need to provide is mgcosθ + Mgsinθ which is > mg.

Those components are perpendicular; they must be added as vectors. They add to mg.

Is it a paradox?

No paradox. The work you need to do against gravity to raise the object a certain height is the same. With the incline, since you exert less force, the distance you have to push is greater. (Allowing you to exert less force is the benefit of using an incline.)

 

[Donald K]

Thanks!

I know the work done is path independent.

But how about the mgcosθ ( the reaction force) if I carry it (in my hands / put it on my head)

If I carry the mass in my hands, the reaction force exerted by the incline is ( my mass +m)cosθ.

The reaction force I need to keep the mass from falling down my hand is mgcosθ, is it right?

So the total force I need to spend (mgcosθ + mgsinθ) is greater at the incline if I carry the mass in my hands?

 

[Doc Al]

Thanks!

I know the work done is path independent.

But how about the mgcosθ ( the reaction force) if I carry it (in my hands / put it on my head)

If I carry the mass in my hands, the reaction force exerted by the incline is ( my mass +m)cosθ.

So both you and the object are on the frictionless incline now? (What's preventing you from sliding down?) If so, the normal force exerted by the incline will be (M + m)cosθ. So?

The reaction force I need to keep the mass from falling down my hand is mgcosθ, is it right?

If you are holding the mass in your hand, and it's not accelerating, then you must be exerting a force of mg on it to support it. If you are sliding down the incline as you hold the mass, then you only exert a force of mgcosθ on it.

So the total force I need to spend (mgcosθ + mgsinθ) is greater at the incline if I carry the mass in my hands?

I'm still not understanding your point. If you are holding the mass as you both slide down a frictionless incline, then the force you exert on the mass is as outlined above.

 

[Donald K]

Assume constant velocity. There is no acceleration and friction on the ramp.

The force needed to push up the incline is mgsinθ which is smaller than directly raising up , i.e., mg. This is mechanical advantage of the incline.



But how about the reaction force component mgcosθ if I hold the mass? If I hold it, it is offered by me rather than the incline surface.

It means the total force I need to spend is more. i.e. mgcosθ + mgsinθ

Hence, the ramp can save your force only if you pushing up the mass (the mgcosθ is done by the incline), not carrying it.

The force needed to handle a mass up a ramp varies on your lifting method. If you hold it
by your arms, you spend more. If you just push it up, it's saving your force.

 

[Doc Al]

Assume constant velocity. There is no acceleration and friction on the ramp.

Still confusing. Friction? Yes or no?

If the object is being slid up a frictionless incline at constant velocity, then something is pushing it along.

The force needed to push up the incline is mgsinθ which is smaller than directly raising up , i.e., mg. This is mechanical advantage of the incline.

OK.

But how about the reaction force component mgcosθ if I hold the mass? If I hold it, it is offered by me rather than the incline surface.

Describe this situation. You are now on the frictionless incline holding the mass? What's pushing you up the incline?

It means the total force I need to spend is more. i.e. mgcosθ + mgsinθ

Hence, the ramp can save your force only if you pushing up the mass (the mgcosθ is done by the incline), not carrying it.

The force needed to handle a mass up a ramp varies on your lifting method. If you hold it
by your arms, you spend more. If you just push it up, it's saving your force.

Are you comparing:
- the force needed to push an object up a frictionless incline (= mgsinθ)
to
- the force needed to carry an object as you walk up an incline (with friction) (= mg)

If so, then sure, carrying an object up a ramp requires more force.

(There's no conservation law for force.)

 

[Naty1]

of course: as log as the inclined plane is not vertical...
that's one reason the wheel (as "wagon wheel") is so useful.

push,carry,roll will each, in general, require a different force from you.