When you are talking about a no-slipping condition, the static coefficient of friction, 0.23 in this case, is what matters.
There is a gravity weight component of 70.7 sin 21.6 = 26.0 N parallel to the incline. The maximum static friction force than can resist motion, in the same direction, is
70.7 cos 21.6 * 0.23 = 65.73 N
That is more than enough to prevent the block from sliding. No additional force is needed.
F = 0
A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?
Can someone please tell me what steps to do? I don't understand this one.
2 answers
See, I thought that was right, too. But I plug that into my computer and it says it's not right. So I'm not sure what I'm doing wrong...
3 answers