Finding the Net Torque of a Circular Object

[DrunkApple]

Homework Statement

A circular-shaped object of mass 9 kg has an
inner radius of 7 cm and an outer radius of
20 cm. Three forces (acting perpendicular
to the axis of rotation) of magnitudes 12 N,
27 N, and 16 N act on the object, as shown.
The force of magnitude 27 N acts 29 below
the horizontal. Find the magnitude of the net torque on
the wheel about the axle through the center
of the object.
Answer in units of N · m

Homework Equations

t1 = r * F
t2 = r * F
t3 = r * F

The Attempt at a Solution

net torque = r1F1 + r1F2 + r3F3
= (.2)(12) + (0.2)(16) + (0.07)(27)
= 7.49 N * m
But it's wrong... why...

 

[LawrenceC]

You mention 'as shown'. I don't see where it is shown.

 

[DrunkApple]

here is the picture that i draw. i hope it's good enough.

ok so i think i got it but check it for me.

If torque is going clockwise, it's negative. If it's counterclockwise, it's positive.

So two torques are negative because if I pull the radius, the circle will turn clockwise.
12(0.2) and 16(0.2) are negative torques which is -5.6.

But the third one is positive because it is when I pull the radius, it will spin clockwise.
27(0.07) which is 1.09

So net torque is -5.6 + 1.09 is -3.71.

But since it is asking for magnitude, I can put absolute value on net torque which is 3.71. Is that right?

 

[LawrenceC]

Yes, think of magnitude as an absolute value.

 

[asteriatic]

I would like to point out that the equation used to calculate net torque is incorrect. The correct equation is net torque = r1F1sinθ1 + r2F2sinθ2 + r3F3sinθ3, where θ is the angle between the force and the radius. In this case, the angle for the force of magnitude 27 N would be 29 degrees below the horizontal, which would result in a different value for the net torque. Additionally, the units for torque should be N*m, not N*m^2. I recommend double checking the calculations and using the correct equation to find the net torque.