Forces to move on the level versus up an inclined plane

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Discussion Overview

The discussion revolves around the forces required to move an object on flat ground versus an inclined plane, particularly focusing on the relationship between the applied force and the angle of inclination. Participants explore theoretical scenarios, including cases where gravity is disregarded, and examine the implications of vector forces in these contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the force required to move up an incline can be expressed as F2*cos(theta) = F, leading to F2 = F/cos(theta), assuming gravity is disregarded.
  • Others challenge the assumption of neglecting gravity on an inclined plane, questioning the validity of the scenario without gravitational effects.
  • There are discussions about the direction of the applied force, with some participants emphasizing that the force must be considered as a vector quantity with both magnitude and direction.
  • A modified scenario is introduced where a bead on a wire in a gravity-free environment is analyzed, raising questions about the relationship between the forces involved and their dependence on the angle of application.
  • Some participants express uncertainty about the implications of using cosine in the context of vector forces and whether directionality matters when forces are parallel to the incline.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the treatment of gravity in the problem. While some agree on the mathematical formulation presented, others dispute the feasibility of the scenario without gravity, leading to multiple competing views on the topic.

Contextual Notes

Limitations include the assumption of no gravity, which raises questions about the definition of "level" and the applicability of vector components in the absence of gravitational forces. The discussion also reflects varying interpretations of the problem setup and the role of friction.

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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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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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"...
 
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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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
 
berkeman said:
The plane is inclined with respect to what? If there is no gravity, there is no "level"...
incline with respect to ground
 
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##?
 
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?
 
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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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?
 
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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}##.
 

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