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Net torque acting on square metal plate

A square metal plate 0.18m on each side is pivoted about an axis at its centre and perpendicular to the plate. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1= 18N, F2=26N and F3= 14N.
I've tried using τ=rFsinϕ\tau = rFsin\phi with r as 0.09m to get T1 as 1.62Nm, T2 as 2.34Nm and T3 as 0.89Nm, then working out the sum of the torques as 2.34Nm + 0.89Nm - 1.62Nm = 1.61Nm, but the correct answer is 2.5Nm, and I cannot figure out how to reach this.
I've attached my poor recreation of the diagram in the book.
(edited 9 years ago)
Untitled.png1.8KB
r=0.09m not 0.9m for F1 and F2

What value of r have you used for F3
(edited 9 years ago)
Reply 2
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Original post by Stonebridge
r=0.09m not 0.9m for F1 and F2

What value of r have you used for F3


Ah damn yeah, all the values are out by a factor of 10.
I used rsin(45) for F3.
If F3 is acting at the corner then its perpendicular distance d from the centre is the hypotenuse of a right angle triangle with 2 sides equal to r. That distance is not r sin 45.
Sin45 = r/d
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Texx
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Original post by Stonebridge
If F3 is acting at the corner then its perpendicular distance d from the centre is the hypotenuse of a right angle triangle with 2 sides equal to r. That distance is not r sin 45.
Sin45 = r/d


Thankyou! I see it now. :smile:

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