Calculating Net Torque on a Circle: 4.0cm Diameter, 20N and 30N Forces

[bigsaucy]

Hello all, I am new to the forum so excuse any errors in my posting format

The question is as follows:

There is a circle with a 4.0cm diameter. There is a force of 20 N pulling downwards on the left of the circle and on the right of the circle there is a force pulling down at 30N. The question is to find the net torque about the axle.

I reasoned that the net force acting would be 10N on the right hand side of the circle so therefore using the equation

T = rF

T = (0.02m)(10N) = 0.2 N m

which i converted to -0.2 N m because it is acting clockwise the back of the book however says that the answer is -20 N m

PLEASE HELP, THANKS!

 

[tiny-tim]

welcome to pf!

hello bigsaucy! welcome to pf!

your reasoning and answer are correct …

the book must have got confused between cm and m ​

 

[PhanthomJay]

Looks like a book error, your answer is correct, the net torque is (- 0.2 N-m) or (- 20 N-cm).

However, although your answer is correct, do not calculate net torques in the manner you have done. Sum torques individually, that is, net torque = 20(0.02) -30(0.02) = - 0.2 N-m.

 

[bigsaucy]

thanks guys, you lot are great!

 

[bigsaucy]

and phantom, thanks for the advise, ill be sure to follow your format from now on. thanks heaps!