A motor delivering 100 inch ounces of torque will force 1 0unce on the end of a lever arm 100 inches long attached perpendicular to the motor shaft. it will also force 2 ounces on the end of a 50 inch shaft or 4 ounces on the end of a 25 inch shaft. because 4*25=100 and 2*50=100.
OK?
Torque is about the turning force on an axis (sometimes that axis is a rotatable shaft). It is a force at a position that is displaced from the axis. The SI unit is newton meter (N・m), so is calculated as newtons of force x meters of displacement. Note that is not ounces per inch.
When used with a shaft that can rotate, the torque is equivalent to a force perpendicular to the radius from the axis of the shaft, at one unit of displacement. Unit displacement means at a distance of one unit (foot, meter etc.). When the force is applied to a lever to rotate it around an axis, the force may not be perpendicular to the lever so the angle of the force vector has to be considered.
If the applied torque is the same as the load torque, the torque is balanced and the shaft does not turn. If the torqe is not balanced, the shaft accelerates in the direction of rotation according to the greater torque.
Incidentally 100 ounce inches of load torque does not stall a motor specified as able to provide 100 ounce inches of torque. This is the condition of applied torque = load torque, so the motor continues at the same speed, does not accelerate or decelerate. The balancing of torque is more or less inherent with a typical electric motor operated normally. A change in load results in an almost instantaneous increase in applied torque, and a new balance is established, though with a non-ideal motor the speed changes for other reasons as a result of the changed load. To stall a motor the load has to be increased beyond the maximum torque the motor can produce at the speeds involved. This generally results in the motor attempting to draw a current beyond its rating, so motor protection involves an understanding of this situation. Many electric motors have a constant maximum torque rating according to maximum allowed current (wire sizes).
Torque = Force x Radius
Unit in SI = N.m
To explain in such a way that make you understand, consider a two-wheeler: To go fast, you increase the gear but with this high speed, you face difficulty in climbing slope. There you need to reduce the gear to increase the 'Torque' to climb the slop at the cost of speed. In first case, 'speed high & Torque low' & in second case, 'Torque is more & speed low'.
Irrespective of the gear the power output of your vehicle is same provided the engine output is constant!
Another practical example is screw-jack where you feel easy to lift weight if the handle is long. FOr constant Torque, you can reduce your effort (Tangentinal force you apply) of turning the handle by increasing the length of handle.
What has shaft length got to do with the torque to stall a motor ?
Torque = Force x Radius
Unit in SI = N.m
Torque is (Length)(Force). To stall a motor of 100 ounces /inch you need to apply a force of 100 ounces if the length of the shaft is a inch. If we decrease the length of the shaft to make it half so we need to apply a force of 200 ounces.
I understand how this will happen .
Pls explain with calculations,diagrams and videos if necessary
