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Force multiplier

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electrodacus:
Problem with real numbers.

Wheels in diagram (a) are not allowed to slip but the red box can slip relative to ground gear ratio 2:1.
coefficient of static friction us = 0.5
coefficient of kinetic friction uk = 0.3
Vehicle mass 5kg with a 50% distribution so 2.5kg on the right and 2.5kg on the left wheel.
Red box is 1kg

A) What is the minimum force F1 needed for the red box to start moving ?
B) What is F2 in relation to F1 if F1 is less than needed for the red box.
C) What is F2 in relation to F1 if F1 is larger than needed for the box to start accelerating and in what direction will the red box move.
D) What is F2 in relation to F1 when box speed is constant and what is that speed of the red box relative to treadmill speed.

Consider the treadmill infinitely long so no worry about the wheel falling of the treadmill same for the red box.

PlainName:

--- Quote ---There are only two points of contact if you understand what that means.

1. Left wheel on the red block.
2. Right wheel on the treadmill.
--- End quote ---

Well something is moving the treadmill, and to do that it has to be pushing (or pulling) against some part of your setup, otherwise nothing will move1. Perhaps missing that is the source of your mistake.

---
[1] Actually, if something drives the treadmill and it's not attached to anything else, the treadmill will just disappear off to the right. But it can't, because it is attached to the ground to which the left block is also attached. Thus it's another point of contact, as you call it, and completes the circuit.

electrodacus:

--- Quote from: PlainName on February 08, 2023, 11:49:58 pm ---
--- Quote ---There are only two points of contact if you understand what that means.

1. Left wheel on the red block.
2. Right wheel on the treadmill.
--- End quote ---

Well something is moving the treadmill, and to do that it has to be pushing (or pulling) against some part of your setup, otherwise nothing will move1. Perhaps missing that is the source of your mistake.

---
[1] Actually, if something drives the treadmill and it's not attached to anything else, the treadmill will just disappear off to the right. But it can't, because it is attached to the ground to which the left block is also attached. Thus it's another point of contact, as you call it, and completes the circuit.

--- End quote ---

Hey this is very good.  Maybe this is where the confusion comes from.
Treadmill is attached to ground so the force it applies to the right wheel is relative to ground.
There is an electric motor inside the treadmill with the stator connected to ground and rotor connected to the treadmill belt.
Is this helpful ?

PlainName:

--- Quote from: james_s on February 08, 2023, 10:55:07 pm ---Seriously? Another one of these threads? Why do you guys waste time arguing with a troll?

--- End quote ---

It's like trying to bang a nail in with a hammer, but whatever you do and however you hold the hammer, the nail always manages to be slightly somewhere else at the critical moment. Keep coming back because it's one of those things you can't quite believe unless you see it  ::)

PlainName:

--- Quote ---Is this helpful ?
--- End quote ---

You tell me. I reckon I know what you'll say, but say it anyway.

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