Torque- Vector cross product using both geometric and algebraic methods

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Homework Help Overview

The problem involves calculating the torque produced by a force acting on a lever in a Cartesian coordinate system. The lever is oriented along the y direction, and the force is given in vector form. The original poster attempts to calculate the torque using both geometric and algebraic definitions of the cross product.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster calculates torque using both the sine definition and the cross product, but their result differs from the lecturer's solution. Some participants question the orientation of the lever and the angle used in the calculations.

Discussion Status

The discussion includes attempts to reconcile the original poster's calculations with the lecturer's solution. Some participants express agreement with the original poster's answer, while others suggest there may be a misunderstanding regarding the lever's orientation or a possible typo in the lecturer's solution.

Contextual Notes

There is uncertainty regarding the angle used in the torque calculations and the orientation of the lever, which may affect the results. The original poster's calculations yield a torque of -1.5 Nm, while the lecturer's solution is -2.5 Nm.

garyd
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Homework Statement


A lever is orientated along the y direction in a Cartesian coordinate system. The length of the lever is 0.5m and one end of it is at the origin of the coordinate system. A (3i-5j)N force applied to the other end of the lever. Calculate the Torque produced by the force acting on the lever about the origin. Do calculation twice, firstly using the geometric definition of the cross product and secondly using the algebraic definition of a cross product

L=(0i+0.5j+0k)m
F=(3i-5j+0k)N

Homework Equations



T=LFsin(theta)

T=L × F


The Attempt at a Solution



theta= 90+59= -149°
mag F= √(3^2+-5^2)= 5.83N
mag L= .5m
LFsin(theta)= .5*5.83*sin(-149)=T=-1.5Nm

&

(0*-5) - (.5*3) =T=-1.5k Nm

My lecturer has a solution of -2.5Nm! Help appreciated

 
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I agree with your answer.
 
I think my lecturer must be putting the lever in the 'i' direction and using theta=-59
 
garyd said:
I think my lecturer must be putting the lever in the 'i' direction and using theta=-59

That's a good guess. There is some kind of typo.
 
Dick said:
That's a good guess. There is some kind of typo.
Thanks for your help.Much appreciated.
 

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