Calculate Torque on a Particle with Given Force and Position | Homework Solution

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SUMMARY

The torque on a particle subjected to a force of F = -11j N at a position r = (8i + 5j) m about the origin is calculated using the cross product formula τ = r × F. The correct calculation yields τ = (0i, 0j, -88k), indicating a torque of -88 Nm. The confusion arose from the incorrect application of the cross product, which is not commutative, necessitating the correct order of vectors in the calculation.

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Homework Statement


Force F⃗ =−11j^N is exerted on a particle at r⃗ = (8i^+5j^)m.
What is the torque on the particle about the origin? Express your answer using two significant figures. Enter coordinates numerically separated by commas.

Homework Equations

The Attempt at a Solution



F: 0i, -11j, 0k
r: 8i, 5j, 0k

F X r = (0-0)i - (0-0)k + (0 - -88)j
= 0, 0, 88

this is wrong because it should be -88, but I don't see why? It was my understanding that I should plug values into formula i - j + k. Thanks for any help.
 
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sp3sp2sp said:

Homework Statement


Force F⃗ =−11j^N is exerted on a particle at r⃗ = (8i^+5j^)m.
What is the torque on the particle about the origin? Express your answer using two significant figures. Enter coordinates numerically separated by commas.

Homework Equations

The Attempt at a Solution



F: 0i, -11j, 0k
r: 8i, 5j, 0k

F X r = (0-0)i - (0-0)k + (0 - -88)j
= 0, 0, 88

this is wrong because it should be -88, but I don't see why? It was my understanding that I should plug values into formula i - j + k. Thanks for any help.

What's the definition of torque?
 
sp3sp2sp said:

Homework Statement


Force F⃗ =−11j^N is exerted on a particle at r⃗ = (8i^+5j^)m.
What is the torque on the particle about the origin? Express your answer using two significant figures. Enter coordinates numerically separated by commas.

Homework Equations

The Attempt at a Solution



F: 0i, -11j, 0k
r: 8i, 5j, 0k

F X r = (0-0)i - (0-0)k + (0 - -88)j
= 0, 0, 88

this is wrong because it should be -88, but I don't see why? It was my understanding that I should plug values into formula i - j + k. Thanks for any help.

In future, please avoid using symbols like j^N; in typewriter math, this means ##j^N##, which is not what you intend. Just use j N, or perhaps enclose the "N" in parentheses, so write either F = - 11j N or F = -11j (N). Better still, use LaTeX to typeset your formulas and equations, so you could write ##F = - 11\, j## N or ##F = -11\, j## (N). You could even make them look like vectors by using a bold font, or an arrow on top, like this: ##\mathbf{F} = -11 \, \mathbf{j}## N or ##\vec{F} = -11 \vec{j}## N.

Anyway, back to your question. Is torque equal to ##\mathbf{r \times F}## or is it ##\mathbf{F \times r}##?
 
Thanks for the replies . Torque is not communicative so torque = r X F
Its the cross product of the Force and the distance from pivot.
 
OK I think i get it now. it should be in form rXF which when you apply the i-j+k gives 0i, 0j, -88k
thanks for help
 
sp3sp2sp said:
when you apply the i-j+k
I am not familiar with such a formula. I assume it is some way to remember how to get the signs right, but if so I dislike it. It loses the symmetry.
It is really quite simple. The result is the next one around in cyclic order:
ixj->k
jxk->i
kxi->j
If you have to switch the inputs to get them in cyclic order, switch the sign too:
jxi->-k
kxj->-i
ixk->-j
 
Yup that's exactly what it was .. its not a formula, just way I was remembering it because it was confusing me. thanks for the help
 

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