Okay...I have a question.In the case described in the picture, the medium between the two forces, which would normally cause stationary or uniform circular motion, is not straight..The moment arm of each force is equal, and the forces are also equal...Now my question is, would the sum of torques in this case also be 0, or not???....Like in the sense that, if these forces were unequal (with equal distance from origin), there would be some clockwise or anticlockwise torque, and THEN if one of the forces was made equal to the other, then would the system's motion be uniform circular motion (Sum of torques=0)....
P.S: I intentionally did not write the angle between moment arms, as I thought that would not affect the answer

Doc Al
Mentor
In the case described in the picture
You forgot the picture.

jbriggs444
Homework Helper
Okay...I have a question.In the case described in the picture, the medium between the two forces, which would normally cause stationary or uniform circular motion, is not straight..The moment arm of each force is equal, and the forces are also equal...Now my question is, would the sum of torques in this case also be 0, or not???....Like in the sense that, if these forces were unequal (with equal distance from origin), there would be some clockwise or anticlockwise torque, and THEN if one of the forces was made equal to the other, then would the system's motion be uniform circular motion (Sum of torques=0)....
P.S: I intentionally did not write the angle between moment arms, as I thought that would not affect the answer
There is no picture. But if you are describing something like the bent arm on a phonograph then the forces at the endpoints are equal, opposite and act along a common line. The bent arm will be under a bending stress in addition to the pure longitudinal stress.

Angular momentum is conserved. Internal torques always sum to zero.

Heres the picuture...I apologize for inconvenience

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jbriggs444
Homework Helper
Now my question is, would the sum of torques in this case also be 0, or not???....Like in the sense that, if these forces were unequal (with equal distance from origin), there would be some clockwise or anticlockwise torque, and THEN if one of the forces was made equal to the other, then would the system's motion be uniform circular motion (Sum of torques=0)....
For equal forces at right angles to equal moment arms, one moment arm 90 degrees clockwise from its force and one 90 degrees counter-clockwise, yes. The sum of the torques would be zero.

If you changed one of the forces to get the system spinning and then made the forces equal again, yes. The rotation rate of the system would be constant.

• Kaneki123
CWatters
Homework Helper
Gold Member
Okay...I have a question.In the case described in the picture, the medium between the two forces, which would normally cause stationary or uniform circular motion, is not straight..The moment arm of each force is equal, and the forces are also equal...Now my question is, would the sum of torques in this case also be 0, or not???....

Yes. The two torques have equal magnitude but opposite direction. Typically you would decide that clockwise (or anti-clockwise) was positive making the other negative. So they sum to zero.

Like in the sense that, if these forces were unequal (with equal distance from origin), there would be some clockwise or anticlockwise torque, and THEN if one of the forces was made equal to the other, then would the system's motion be uniform circular motion (Sum of torques=0)....
P.S: I intentionally did not write the angle between moment arms, as I thought that would not affect the answer

It doesn't. The two forces could even act on the same arm eg zero angle between them.

I note that both of the forces in your diagram have a downward component. If the object is to move in a circular manner around the pivot/origin and not accelerate downwards as well there must also be a reaction force on it at the pivot/origin. As drawn that must have an upward component so the net force is zero. In fact, if it only moves in a circle and doesn't have linear acceleration, at all times the net linear force must be zero. Eg the reaction force at the origin may rotate.

• Kaneki123
I thank you all for your help....Much appreciated...