Force Question -- Solidworks model of a pulley system

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The discussion revolves around simulating a pulley system in Solidworks, specifically addressing the need to double the force while adjusting the dimensions of the pulley. The hypothesis suggests that multiplying all dimensions by the cubed root of 2 will maintain consistent simulation results, as both force and volume are doubled. However, the simulation results did not align with expectations, prompting questions about torque calculations and the relationship between force and radius. Participants clarify that increasing dimensions affects torque, which increases significantly due to the changes in both force and radius. The conversation emphasizes the importance of accurately defining force and its implications on the system's performance.
Elijah Carstens
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Hello Everyone,

I'm currently making simulations in Solidworks and am confused.

I have to double the force on a pulley and change the dimensions from the original pulley, in order to withstand the new force.

My hypothesis:
If I multiply all the dimensions by the cubed root of 2, then my volume will double and the results of both simulations will remain the same, because I doubled the force and the volume of the second experiment.

Does that make sense?

I figured if the volume of the object doubles along with the amount of force, then the results should remain the same.

When i ran the simulation, it didn't go as planned.

Why?

Thank You so much!
 
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Elijah Carstens said:
My hypothesis:
If I multiply all the dimensions by the cubed root of 2, then my volume will double and the results of both simulations will remain the same, because I doubled the force and the volume of the second experiment.
Try to express what you're doing in terms of physics, for those who might not know anything about "Solidworks".
 
Sorry I seem to have accidentally deleted my first attempt to post this so...

Strength of materials isn't really my field but consider the torque calculation..

Torque = force * radius

You are doubling the force and increasing the radius by sqrt 2 so the torque increases by about 2.8.

Perhaps that doesn't matter?
 
CWatters said:
Sorry I seem to have accidentally deleted my first attempt to post this so...

Strength of materials isn't really my field but consider the torque calculation..

Torque = force * radius

You are doubling the force and increasing the radius by sqrt 2 so the torque increases by about 2.8.

Perhaps that doesn't matter?

Maybe clarify what you mean. My math was dividing the new amount of force (800N), by the original amount (400N).

That's how I got 2. I needed that number in order to find what to multiply the original dimensions by.

3√2 * (original dimensions of pulley)
 
Sorry I made a mistake in my post above. I should have used 3√2 instead of sqrt.

If you are increasing all dimensions by 3√2 then does that also includes the radius?

When you refer to "force" I assume you mean the difference in tension between the two sides of the belt (eg the force used to calculate the torque on the pulley)?

Putting these two together you can calculate that the torque is increasing by a factor of...

(New Radius/Old Radius)*(New force/Old force) = 3√2 * 800/400 = 2.5

Perhaps by "force" you mean something different?
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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