What is the magnitude of the torque in this scenario?

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SUMMARY

The torque generated by a force of F = (3.00 i - 2.00 j) N applied at the position r = (1.00 i + 2.00 j) m about an axis through the origin perpendicular to the xy plane is calculated using the cross product. The resulting torque vector is 0 i + 0 j - 8 k, leading to a magnitude of 8 N-m. This calculation confirms the correctness of the torque's magnitude as 8 N-m, aligning with standard physics conventions.

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Homework Statement



A force of F = (3.00 i - 2.00 j) N acts at a location r = (1.00 i + 2.00 j) m on an object. What is the magnitude of the torque that this force applies about an axis through the origin perpendicular to the xy plane?

Homework Equations



Torque is equal to the cross product of the position (vector r) and the Force (vector F).


The Attempt at a Solution



I took the cross product and obtained 0 i + 0 j - 8 k.

Taking the magnitude, I found positive 8 m*N. Is this correct?
 
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BIOPSYCH said:
Taking the magnitude, I found positive 8 m*N. Is this correct?
Looks good. (Although the units are traditionally written as N-m.)
 
the final equation that you've obtained was correct and also the magnitude of your vector is CORRECT. since the equation of getting the magnitude of the vector[the torque in particular] is:

T^2= (Tx)^2+(Ty)^2+(Tz)^2

so, by using the values you have there'

T^2= (0)^2+(0)^2+(-8)^2
T^2=(-8)^2
T=[squre root of](64)
T=8

good job.
 

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