更新1:
The Clockwise motion = anti , menas there is no torque , means there is no rotation? So how can it against the step.
Torque and force
2010-11-13 9:07 am
A wheel of mass M has radius R. It is standing vertically on the floor , and we wantto exert a horizontal force F at its axle so that it will climb a step against which it rests. The step has height h where h<R What minimum force F is needed?
回答 (3)
2010-11-14 8:03 am
✔ 最佳答案
Taking moment about the point of contact of the wheel at the step.Perpendicular distance of F from the point of contant = (R - h)
Perpendicular distance of the weight of the wheel from the point of contact
= square-root[R^2 - (R-h)^2]
= square-root[h(2R-h)]
Hence, F x (R-h) = Mg.(square-root[h(2R-h)])
where g is the acceleration due to gravity
i.e. F = Mg.(square-root[h(2R-h)]/(R-h))
2010-11-14 5:31 am
2 forces actually
2010-11-14 4:30 am
This question is quite difficult, three forces acting at different positions which can not be drawn on one line.
2010-11-14 20:51:38 補充：
I guess there should have net torque for the wheel to rotate and ''climb'' the step.
3 forces : weight, applied force, normal reaction from the contact between the step and the wheel, but if the question does not tell if there is also the forth force which is friction.
2010-11-14 20:51:38 補充：
I guess there should have net torque for the wheel to rotate and ''climb'' the step.
3 forces : weight, applied force, normal reaction from the contact between the step and the wheel, but if the question does not tell if there is also the forth force which is friction.
收錄日期: 2021-04-27 13:52:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101113000051KK00093