a. Is it true to consider ‘torque’ as a ‘force’? Give its explanation.
b. A torque that causes counterclockwise acceleration is a positive torque or negative.
1 Answer | Add Yours
1. Force is any influence that can change an object's motion or geometric configuration. For instance, if a push a cart and it moves forward, I applied force to cart (hence, it accelerated). Force is usually derived from mass and acceleration (`F = ma` , one of Newton's laws of motion). Torque, on the other hand, is the tendency of a force to rotate an object (torque is sometimes referred to as the moment of force). Torque is given by the formula `tau = F x r` , the cross-product of Force applied and distance from the pivot (or fixed) point. A higher torque would mean higher tendency to rotate. For example, when you're opening a door, we usually push/pull away from the hinges, the pivot/fixed point (torque here is higher). If you try to push near the hinges, it would be really hard to open the door (torque is lower, you need to apply more force compared to when you simply push away from the hinges). In short, in a rotating body (such as a door), force is the influence you exert to make it rotate, but torque measures the tendency that the door will rotate given you applied that amount of force at that certain point. Hence, while some people will refer to torque as some kind of force involved in twisting/rotating, they are actually different. Another clue would be that force is newtons, while torque is newton-meters (though it is also not energy, as energy is scalar and torque is a vector). Another difference is direction - force is linear (acceleration is along a straight path), while torque is rotational (it has an angular acceleration: clockwise/counter clockwise).
2. This usually depends on the convention you are using, or on the assignment of your coordinates. Usually though, clockwise would be negative and counter-clockwise is positive (you may use the right hand rule convention; but it really depends on the sign convention you used to solve the problem).
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes