Applying Torque at an angle

  • Thread starter Kamikaze
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  • #1
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Homework Statement


A Force F = 76.0N is applied to a box with dimensions y = 2.69m, x = 1.04m. The force is applied at an angle theta with respect to the horizontal as shown. For theata = 24.0 deg, calculate the magnitude of the torque produced by F about the point O.
2je4dad.png



Homework Equations


T=r(Fperpendicular)
Fperpendicular = Fsin(theta)


The Attempt at a Solution


I did pythagorean theorem (a2 + b2 = c2) to find the diagonal (r) at 2.88404m
I then found Fperpendicular by using Fperpendicular = 76Nsin(24deg) = -68.823956N . I then multiplied this by the r I found earlier to find the magnitude of Torque at -198.4911991N. Somehow this is wrong and I don't know how else to do it.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
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Welcome to PF

Hi Kamikaze! Welcome to PF :smile:

(have a theta: θ and a degree: ° and try using the X2 tag just above the Reply box :wink:)
I then found Fperpendicular by using Fperpendicular = 76Nsin(24deg)
wrong angle! :redface:

useful tip: always draw your diagrams fully … if you'd put in the extra lines, you'd see it isn't θ.
 

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