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Torque about the origin is

  • Thread starter Cal124
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Homework Statement


a point particle is located in the position r=i+2j m . A gravitation (vertical) force of 1.5N acts on the particle. The torque about the origin is...

Homework Equations



thau = r X F = rFsin(theta)

The Attempt at a Solution



it's 0 right?

magnitude r = sqrt(1^2+2^2)m
mag(r).1.5N.Sin(180)=0
 

Answers and Replies

  • #2
MarcusAgrippa
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What does "vertical" mean in this problem? j or k? - I assume it is not i.
 
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  • #3
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I'm assuming its the k component, that's the vertical part of i,j,k, right?
Thanks for the reply, i just seem to be struggling understanding the question to be honest. If it is the k component then surely the cross product would give 0 as well?
 
  • #4
MarcusAgrippa
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Assume "vertical" means the k-direction. Then [itex] \vec{F} = F \hat{k}[/itex], where F = 1.5 N. Then
[itex] \vec{\tau} = \vec{r} \times \vec{F} = (\hat{i} + 2\hat{j}) \times (F \hat{k}) = F \underbrace{\hat{i} \times \hat{k}}_{=-\hat{j}} + 2 F\underbrace{\hat{j} \times \hat{k}}_{=\hat{i}} = 2F \hat{i} - F \hat{j} = (3.0 \hat{i} - 1.5 \hat{j} ) N.m
[/itex]

You should be able to see from this calculation what to do if your were wrong about the vertical direction. The answer would still not be zero, as your surmised.
 
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  • #5
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sorry you've lost me, how does that let me work out the torque in Nm? I have the possible answers -4.5, 4.5, 1.5, 3, 0
 
  • #6
MarcusAgrippa
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Ok. You were probably wrong about the direction of the vertical. Replace k by j and redo the calculation - you try it this time.

The magnitude of the above torque is calculated from [itex] |\vec{\tau}| = \sqrt{ (3.0 N.m)^2 + (-1.5 N.m)^2} [/itex], which does not give one of your answers, so you must have guessed the vertical direction incorrectly.
 
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  • #7
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F(i x j) + 2F(j x j) = Fk + 0 = (1.5k) Nm = 1.5 Nm
 
  • #8
MarcusAgrippa
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Correct. Understand it now? You had an incorrect angle in your first calculation.

If they give you vector data, calculate in vector form. You need to learn to use vectors, so do not try to bypass them. Vectors are a powerful mathematical tool, even if you do not see the point of them yet. So aim to master vectors as soon as you can.
 
  • #9
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Thanks a lot! yeah, I'm going to spend some much needed time on vectors. I'm not sure how i go about finding the direction as in questions like this?
 
  • #10
MarcusAgrippa
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Direction is specified by an unit vector in the direction of the given vector. Unit vectors are defined by dividing the vector by its magnitude:
[itex] \hat{A} = \frac{\vec{A}}{|\vec{A}|}[/itex]
The vector [itex]\hat{A}[/itex] is the direction of the vector [itex]\vec{A} [/itex].

This is a new way to specify direction. The unit vector contains all f the information that you need to convert it into any other format. It is therefore a complete specification of direction.

If you want to work out angles this direction makes with reference directions or lines, use the dot product. Thus the angle made with the x-axis can be found from
[itex] \cos \theta = \hat{A} \cdot \hat{i} [/itex]

The beauty of vectors is that you need never convert the information into the kind of information that you needed at school to do these same problems. The problems can be solved directly by vector methods.
 
  • #11
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Thanks for all your help, I don't suppose you could reccomend a book/website to learn vectors? I have a physics book by Feynman that has a section on vectors I plan to read. Or would I get a better understanding of vectors from a mathematical viewpoint?
 
  • #12
MarcusAgrippa
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Try Christie's Vector Mechanics. It is a bit old, but it does all the right things and in the right way.
 

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