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General Question about Solving Torque

  • Thread starter shawn87411
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This isn't a specific problem, I just can't remember how to solve problems like these.

I attached a brief picture.

The dots on the top are screws that hold the box in place. There's torques on both the shafts in the clockwise direction (shaft coming out towards you is 30 and the one going horizontal is 20) and I know the distance between the screws. Theres only vertical forces on the screws. I just can't remember how to go about solving it - I know I do the summation of torques is 0 but I forgot how it works in a 3D plane.

I'm guessing you do the sum of the torques in the direction of one of the torques is 0 and then do the same for the other direction which gives 2 equilibrium equations. I'm missing where the other 2 come from?
 

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nrqed
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This isn't a specific problem, I just can't remember how to solve problems like these.

I attached a brief picture.

The dots on the top are screws that hold the box in place. There's torques on both the shafts in the clockwise direction (shaft coming out towards you is 30 and the one going horizontal is 20) and I know the distance between the screws. Theres only vertical forces on the screws. I just can't remember how to go about solving it - I know I do the summation of torques is 0 but I forgot how it works in a 3D plane.

I'm guessing you do the sum of the torques in the direction of one of the torques is 0 and then do the same for the other direction which gives 2 equilibrium equations. I'm missing where the other 2 come from?

[tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]

The magnitude is [tex] \tau = r F \sin \phi [/tex]. Draw a section of the box let's say in the xy plane. (I take z to be the axis coming out at you). Then only two of the screws will exert a torque along z (because the other are aligned with the z axis).

Then you have a two dimensional drawing. You can figure out the distance r from the axis of rotation to each screw and you can figure out the angle phi with a simple right angle triangle.
 
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[tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]

The magnitude is [tex] \tau = r F \sin \phi [/tex]. Draw a section of the box let's say in the xy plane. (I take z to be the axis coming out at you). Then only two of the screws will exert a torque along z (because the other are aligned with the z axis).

Then you have a two dimensional drawing. You can figure out the distance r from the axis of rotation to each screw and you can figure out the angle phi with a simple right angle triangle.
Well the forces aren't known, this is a general design problem. All that is given is the torques on the shafts and I need to find the torques on the screw. Also the screws are on the edge of the box for each one with the length of the box being 300mm and the width 450mm so those are considered the distances apart for each screw. So if you look at the XY plane here you'd have a 30 N-m torque in the clockwise direction so are you saying only 2 of the screws feel that torque? If thats true shouldn't the torques be equal on the two screws that feel it since they are equidistant from the torque
 
nrqed
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Well the forces aren't known, this is a general design problem. All that is given is the torques on the shafts and I need to find the torques on the screw. Also the screws are on the edge of the box for each one with the length of the box being 300mm and the width 450mm so those are considered the distances apart for each screw. So if you look at the XY plane here you'd have a 30 N-m torque in the clockwise direction so are you saying only 2 of the screws feel that torque? If thats true shouldn't the torques be equal on the two screws that feel it since they are equidistant from the torque

I am not sure what you mean by "equidistant from the torque". What matters is the distance from the axis of rotation.

The screws won't apply the same force because one screw is applying a torque clockwise and the other screw is applying a torque counterclockwise . So the one applying a torque counterclockwise has to cancel the torque of the other screw plus the external torque
 
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I am not sure what you mean by "equidistant from the torque". What matters is the distance from the axis of rotation.

The screws won't apply the same force because one screw is applying a torque clockwise and the other screw is applying a torque counterclockwise . So the one applying a torque counterclockwise has to cancel the torque of the other screw plus the external torque
Sorry I attached a more specific picture. Distance between D and B is 300mm and the distance between A and C is 450mm. If you draw something in just the XY plane you have something like the second picture attached, correct (the 20 is into the page)? I dont see how you can find the force on the screw if you don't know the torsion on each individual screw? Obviously the sum of them equates to the 30 and 20 in the respective directions
 

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PhanthomJay
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There are no torques on the screws themselves, just up or down vertical forces. They don't sum to 20 or 30. Look at each applied torque separately, and solve for the screw reactions using statics. One applied torque will create vertical positive or negative forces in the outer screws, the other, similarly, on the other 2 screws.
 
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There are no torques on the screws themselves, just up or down vertical forces. They don't sum to 20 or 30. Look at each applied torque separately, and solve for the screw reactions using statics. One applied torque will create vertical positive or negative forces in the outer screws, the other, similarly, on the other 2 screws.
Ahhh I see so the screws only have the force up but because the distance the torque is equated for. Now is it correct that the screws A and C won't compensate for the 30 N-m torque and the screws B and D don't compensate for the 20 N-m torque because of their locations?
 
PhanthomJay
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Ahhh I see so the screws only have the force up but because the distance the torque is equated for. Now is it correct that the screws A and C won't compensate for the 30 N-m torque and the screws B and D don't compensate for the 20 N-m torque because of their locations?
Yes, that is correct. But don't forget that not all the screws have a force 'up'. Some will have a force 'down'.
 

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