text: rotating disk and torque? the 26 cm diameter disk in figure ex13.14 can rotate on an axle through its center. what is the net torque (counterclockwise) about the axle if f = 24 n?  figure ex13.14: http://i137.photobucket.com/albums/q208/infinitbelt/p13-14alt.gif   comments: i tried summing all of the torques together and i get 2.002 nm counterclockwise but that appears to be wrong.  any ideas?   thanks torque is not just force times lever arm length.  it is the lever arm length times the component of force that is perpendicular to the said lever arm.  also, counter clockwise torques will be positive and clockwise torques will be negative (right hand rule).  for instance, the torque caused by the 30n force that is located at 5 cm radius and is oriented at 45 degrees would be computed as follows:  torque1 = (0.05 m) * (30 n) * sine (45 degrees) = 1.0606 nm  this torque is positive because it tends to rotate the wheel in a counterclockwise direction.  notice that there is a force that is oriented radially outward from the disk.  this will have no torque associated with it since there is no component of that force that is perpendicular to the lever arm.  in this case, since the line of the force includes the center of the disk, this force will not cause the disk to rotate.  the other torques are pretty straightforward:  torque2  = (.26m) * (-30n) = -7.8 nm  this torque is negative since it tends to rotate the wheel in a clockwise direction.  torque3 = (.05m) * (24n) = 1.2 nm  now we just sum up all the torques acting on the disk to get the net torque:  torque1 + torque2 + torque3 = 1.0606 nm  -7.8 nm +1.2 nm = -5.53  nm