Undergrad Forces to move on the level versus up an inclined plane

[annamal]

If on a flat ground, we exert a force F to move forward, then we go to an incline plane of theta degrees. Why wouldn't the force F2 to move up the incline plane with respect to ground be F2*cos(theta) = F --> F2 = F/cos(theta) disregarding the effects of gravity?

 

[Baluncore]

On wheels, or friction between surfaces ?
How can you disregard gravity on an inclined plane ?
Is the force applied horizontally or in line with the plane ?

 

[berkeman]

on a flat ground

up the incline plane

disregarding the effects of gravity

The plane is inclined with respect to what? If there is no gravity, there is no "level"...

 

[Lnewqban]

If on a flat ground, we exert a force F to move forward, then we go to an incline plane of theta degrees. Why wouldn't the force F2 to move up the incline plane be F2*cos(theta) = F --> F2 = F/cos(theta) disregarding the effects of gravity?

A force is a vector quantity, it has both, magnitude and direction.
The direction of your walking force is originally horizontal.

After you go up onto an inclined plane of theta degrees, the direction of that vector changes to be parallel to that incline plane.

How do you compensate for the change in height, if you exert the same magnitude of vector force F to move forward as to move up the inclined plane?

Please, see:
https://www.engineeringtoolbox.com/inclined-planes-forces-d_1305.html

https://scienceblogs.com/startswithabang/2010/03/10/the-physics-of-an-inclined-tre

 

[annamal]

On wheels, or friction between surfaces ?
How can you disregard gravity on an inclined plane ?
Is the force applied horizontally or in line with the plane ?

no friction between surface. force applied in line with plane

 

[annamal]

The plane is inclined with respect to what? If there is no gravity, there is no "level"...

incline with respect to ground

 

[jbriggs444]

If on a flat ground, we exert a force F to move forward, then we go to an incline plane of theta degrees. Why wouldn't the force F2 to move up the incline plane with respect to ground be F2*cos(theta) = F --> F2 = F/cos(theta) disregarding the effects of gravity?

Let us change the problem description so that it makes a bit of physical sense.

We have a bead on a uniform, straight wire. Our lab is in space, far from any gravitational source. There is vacuum. No air resistance.

The bead grips the wire with a fixed and even pressure so that it resists motion along the wire. The maximum force of static friction is $F$. The force of kinetic friction is also $F$.

This force of friction is fixed. It remains the same as the bead moves. It remains the same if the bead happens to be subject to a lateral force.

We push on the bead, parallel to the wire with force $F$. It moves at a steady pace. The force is in balance with friction.

We now push on the bead, at an angle $\theta$ to the wire with force $F_2$ so that it moves at a steady pace. The force is in balance with friction.

What is $F_2$? How does it depend on $\theta$?

 

[annamal]

Let us change the problem description so that it makes a bit of physical sense.

We have a bead on a uniform, straight wire. Our lab is in space, far from any gravitational source. There is vacuum. No air resistance.

The bead grips the wire with a fixed and even pressure so that it resists motion along the wire. The maximum force of static friction is $F$. The force of kinetic friction is also $F$.

This force of friction is fixed. It remains the same as the bead moves. It remains the same if the bead happens to be subject to a lateral force.

We push on the bead, parallel to the wire with force $F$. It moves at a steady pace. The force is in balance with friction.

We now push on the bead, at an angle $\theta$ to the wire with force $F_2$ so that it moves at a steady pace. The force is in balance with friction.

What is $F_2$? How does it depend on $\theta$?

on straight line F - F = m*a = 0
on angle F2 - F = m*a = 0 F2 = F... not sure?

 

[jbriggs444]

on straight line F - F = m*a = 0
on angle F2 - F = m*a = 0 F2 = F... not sure?

Forces are vectors. Direction matters. You already used $\cos \theta$ in the original post. Surely you thought that it has some relevance?

Maybe something to do with components?

 

[annamal]

Forces are vectors. Direction matters. You already used $\cos \theta$ in the original post. Surely you thought that it has some relevance?

Maybe something to do with components?

Yes but F2 and F are vectors parallel to the incline plane. So why should the direction matter there?

 

[jbriggs444]

Yes but F2 and F are vectors parallel to the incline plane. So why should the direction matter there?

$F$ and $F_2$ are vectors respectively parallel to and at angle $\theta$ to the wire on which the bead rides.

There is no inclined plane since there is no gravity and no "horizontal".

 

[annamal]

$F$ and $F_2$ are vectors respectively parallel to and at angle $\theta$ to the wire on which the bead rides.

There is no inclined plane since there is no gravity and no "horizontal".

I don't know how this answers my original question:
If on a flat ground, we exert a force F to move forward, then we go to an incline plane of theta degrees. Why wouldn't the force F2 to move up the incline plane with respect to ground be F2*cos(theta) = F --> F2 = F/cos(theta) disregarding the effects of gravity?

 

[jbriggs444]

I don't know how this answers my original question:
If on a flat ground, we exert a force F to move forward, then we go to an incline plane of theta degrees. Why wouldn't the force F2 to move up the incline plane with respect to ground be F2*cos(theta) = F --> F2 = F/cos(theta) disregarding the effects of gravity?

Because, in the absence of gravity there is no reason to require any force at all. The modified scenario is intended to remove this impediment.

In the modified scenario, yes, $F_2 = \frac{F}{\cos \theta}$.