Torque (rotation on a pivot)

  • #1

Homework Statement


You have one meter stick that weights 100g. Calculate the torgue on the stick at 20 cm mark for
A. whole stick weight when the center is at 50cm
B. the left and right weight of the stick are separate, "with their masses located at the halfway points between the 20cm mark and the end on that side"

665025d318767f40132aabb6ca0c7874.png

the pan mass (left side) is unknown and the slider (right) is 10g

Homework Equations


t=F*D

The Attempt at a Solution


for A. I get the torque on both side by multiply the [mass] time gravity (9.8) times the distance (of how far they are from the 20 cm mark, so left side d would be 20cm and right side is 80cm
B. I don't fully understand what they mean by the mass being separated. Does it mean the mass of that side from the pivot point? so the right side would weight 80g and the left 20g?

Is this correct? if not why?
 
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Answers and Replies

  • #2
haruspex
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The question statement is incomplete. What is the orientation of the stick? What is attached to it? There was mention of a pivot. Is that at 20cm? About what axis is the torque to be found, the pivot?
Was there a diagram?
 
  • #3
The question statement is incomplete. What is the orientation of the stick? What is attached to it? There was mention of a pivot. Is that at 20cm? About what axis is the torque to be found, the pivot?
Was there a diagram?
updated, but I didnt add the masses because isn't it asking just for the ruler?
 
  • #4
haruspex
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multiply the force time gravity
Multiply mass by gravity to get force
so left side d would be 20cm and right side is 80cm
Those are the distances from pivot to end of stick. At what point in each section of stick should you consider gravity as acting?

Your work for A is actually what you need for B, calculating the torque separately from each side.
For A, you can simply add them (they will have opposite signs). Or you could have just taken the whole weight as acting at the centre of the stick instead of treating the parts separately.
 
  • #5
Multiply mass by gravity to get force
that what I mean, sorry.

Those are the distances from pivot to end of stick. At what point in each section of stick should you consider gravity as acting?
the whole system?

Your work for A is actually what you need for B, calculating the torque separately from each side.
For A, you can simply add them (they will have opposite signs). Or you could have just taken the whole weight as acting at the centre of the stick instead of treating the parts separately.
But I dont fully understand what b is asking.
For A are you saying to do the torque at center then shift the distance +/-20cm?
 
  • #6
haruspex
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For A are you saying to do the torque at center then shift the distance +/-20cm?
Not sure what you mean.
For the stick as a whole, where should you take its weight as acting? What torque does that have about the pivot?
 
  • #7
Not sure what you mean.
For the stick as a whole, where should you take its weight as acting? What torque does that have about the pivot?
where gravity is acting on the whole stick? anywhere that isnt the pivot point? I'm not sure why section of it wouldn't be affect by gravity
 
  • #8
haruspex
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where gravity is acting on the whole stick? anywhere that isnt the pivot point? I'm not sure why section of it wouldn't be affect by gravity
Gravity acts equally on all parts of the stick, but for a rigid body you can generally take it all as acting at one particular point. That point is therefore known as the centre of gravity. (It gets more complicated in non-uniform gravitational fields.). Where is the centre of gravity?
 
  • #9
Gravity acts equally on all parts of the stick, but for a rigid body you can generally take it all as acting at one particular point. That point is therefore known as the centre of gravity. (It gets more complicated in non-uniform gravitational fields.). Where is the centre of gravity?
The Center of mass? or middle of the stick?
 
  • #10
haruspex
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The Center of mass? or middle of the stick?
Right. So what is its torque about the pivot?
 
  • #11
Right. So what is its torque about the pivot?
F=(.100kg*9.8)(.50m-.20m)=.294?
 
  • #12
haruspex
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F=(.100kg*9.8)(.50m-.20m)=.294?
Right.
Now follow the same procedure but only considering the portion of stick to the right of the pivot.
 

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