Forces to move on the level versus up an inclined plane

In summary, the conversation discusses the relationship between force and motion on an inclined plane. It is suggested that the force required to move up the inclined plane with respect to ground can be calculated using the formula F2 = F/cos(theta), disregarding the effects of gravity. However, further discussions explore the role of friction and the importance of considering the direction of forces, particularly when dealing with vectors on an inclined plane.
  • #1
annamal
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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?
 
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  • #2
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 ?
 
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  • #3
annamal said:
on a flat ground
annamal said:
up the incline plane
annamal said:
disregarding the effects of gravity

The plane is inclined with respect to what? If there is no gravity, there is no "level"...
 
  • #4
annamal said:
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

:cool:
 
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  • #5
Baluncore said:
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
 
  • #6
berkeman said:
The plane is inclined with respect to what? If there is no gravity, there is no "level"...
incline with respect to ground
 
  • #7
annamal said:
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##?
 
  • #8
jbriggs444 said:
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?
 
  • #9
annamal said:
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?
 
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  • #10
jbriggs444 said:
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?
 
  • #11
annamal said:
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".
 
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  • #12
jbriggs444 said:
##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?
 
  • #13
annamal said:
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}##.
 

1. What is the difference between forces needed to move on a level surface versus an inclined plane?

The main difference is the direction of the force. On a level surface, the force needed to move an object is applied horizontally, while on an inclined plane, the force is applied at an angle.

2. How does the angle of the inclined plane affect the force needed to move an object?

The steeper the angle of the inclined plane, the greater the force needed to move an object. This is because the component of the force acting against gravity increases as the angle increases.

3. What role does friction play in the forces needed to move on a level surface versus an inclined plane?

Friction acts against the motion of an object and is present on both a level surface and an inclined plane. However, on an inclined plane, the force of friction is reduced due to the component of the force acting against gravity. This means that less force is needed to move an object on an inclined plane compared to a level surface.

4. How does the weight of an object impact the forces needed to move on a level surface versus an inclined plane?

The weight of an object is directly related to the force needed to move it. The heavier the object, the more force is needed to overcome its weight and move it. This means that a heavier object will require more force to move on both a level surface and an inclined plane.

5. Can you explain the concept of work in relation to forces needed to move on a level surface versus an inclined plane?

Work is the product of force and displacement. On a level surface, the displacement is in the same direction as the applied force, so the work done is equal to the force multiplied by the distance moved. On an inclined plane, the displacement is along the surface of the plane, while the force is applied at an angle. This means that the work done on an inclined plane is less than on a level surface, as the force is not acting in the same direction as the displacement.

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