Force Question -- Solidworks model of a pulley system

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Discussion Overview

The discussion revolves around a Solidworks simulation involving a pulley system where participants explore the relationship between force, dimensions, and torque. The focus is on how to adjust the dimensions of a pulley to accommodate a doubled force while maintaining simulation results.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant hypothesizes that multiplying all dimensions by the cubed root of 2 will double the volume and keep simulation results consistent when doubling the force.
  • Another participant suggests considering torque calculations, noting that doubling the force and increasing the radius by the square root of 2 results in a significant increase in torque.
  • A later reply clarifies the calculation of torque and questions whether the radius is included in the dimensional changes, proposing a different factor for torque increase based on the new dimensions and forces.
  • There is a correction regarding the use of 3√2 instead of sqrt for the dimensional increase, indicating a potential misunderstanding in the calculations presented.

Areas of Agreement / Disagreement

Participants express differing views on the implications of changing dimensions and forces, particularly regarding the calculations of torque and the assumptions about how these changes affect simulation outcomes. The discussion remains unresolved with multiple competing views on the correct approach.

Contextual Notes

Limitations include potential misunderstandings in the definitions of force and torque, as well as the assumptions made about the relationships between dimensions and simulation results.

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?
 

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