CWatters
Science Advisor
Homework Helper
Gold Member
- 10,546
- 2,324
Thought it might be worth checking that the torque really do sum to zero. If clockwise is +ve then...
Summing the torques about the pivot point gives..
(FD * 1.2) - (FS* 0.5) = ?
(1.2 * 1.2) - (2.88 * 0.5) = ?
1.44 - 1.44 = 0
So that's ok.
However if it's not accelerating in rotation you can actually sum the torques about any point and it will be zero. eg It doesn't have to be a real pivot.
Suppose we pick the sensor and sum the torque about that point. The equation is..
(FD*1.2) + (FL*0.5) +(FY*0.5) = ?
FX and FS don't feature because they pass through the point where the sensor is located (eg tangential distance = 0).
Substitute values..
(1.2*1.2) + (1.75*0.5) + (-4.63*0.5) = ?
1.44 + 0.875 - 2.315 = 0
This ability to pick any point about which to sum torques is handy. For example suppose there is an unknown force somewhere. If you pick the point where that force acts you eliminate it from the equation at a stroke.
Summing the torques about the pivot point gives..
(FD * 1.2) - (FS* 0.5) = ?
(1.2 * 1.2) - (2.88 * 0.5) = ?
1.44 - 1.44 = 0
So that's ok.
However if it's not accelerating in rotation you can actually sum the torques about any point and it will be zero. eg It doesn't have to be a real pivot.
Suppose we pick the sensor and sum the torque about that point. The equation is..
(FD*1.2) + (FL*0.5) +(FY*0.5) = ?
FX and FS don't feature because they pass through the point where the sensor is located (eg tangential distance = 0).
Substitute values..
(1.2*1.2) + (1.75*0.5) + (-4.63*0.5) = ?
1.44 + 0.875 - 2.315 = 0
This ability to pick any point about which to sum torques is handy. For example suppose there is an unknown force somewhere. If you pick the point where that force acts you eliminate it from the equation at a stroke.