Find the magnitude of the torque

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    Magnitude Torque
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To find the torque about point P from a 68 lb force, the initial calculation used t = (4)(68)(sin30), resulting in 136. However, there is a suggestion to consider the distance as 4√2 and the perpendicular component of the force as F*cos15°. A critique highlights that the torque equation should be expressed in vector form using the cross product, rather than a scalar calculation. This approach would provide a more accurate torque value and its magnitude. Overall, the discussion emphasizes the importance of using vector calculations for torque to ensure precision.
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Homework Statement


Find the magnitude of the torque about P if a 68 lb force is applied as shown. Give your answer correct to the nearest integer.
12-4-40.gif

Homework Equations


t=r X F

The Attempt at a Solution


t = (4)(68)(sin30)
= 136
 
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Don't you want to consider the distance at which it is acting is 4*√2 ?

At the point of interaction then the ⊥ component of F looks to me like F*Cos15°
 


jonnejon said:

Homework Statement


Find the magnitude of the torque about P if a 68 lb force is applied as shown. Give your answer correct to the nearest integer.
12-4-40.gif



Homework Equations


t=r X F


The Attempt at a Solution


t = (4)(68)(sin30)
= 136


You wrote what must have been intended to be a vector equation for the torque, t = r x F, because the cross product only makes sense if t, r, and F are all expressed as vectors. But then you proceeded to do a scalar calculation. Why not make your torque calculation in vector form using the determinant expression for the cross product and giving the resulting torque as a vector? Then if you need the magnitude of that vector, dot it with itself and take the square root of the result.
 
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