✔ 最佳答案
The centripetal force in this case must horizontally pointing towards the centre of circular path.
Let N be the normal reaction and f be the friction.
First, you should resolve both x-components and y-components of the Normal reaction and the friction.
Therefore, you should have :
For normal reaction==>one vector pointing upwards (Ncos30) and one vector pointing left side(Nsin30).
For friction==>one vector pointing upwards (fsin30)and one vector pointing right side (fcos30).
Two equations can be set up then.
1. fsin30+Ncos30=mg
2. fcos30-Nsin30=(mv^2)/r
So, by substitution of N,
fcos30-mgtan30+fsin30tan30=(mv^2)/r
You can get the ans. by substituting the value of m,v and r.
Please note that,
v can be found by (rpi)/t, where pi=3.141592654, r=5m and t=10s
After finding f, N can also be found by substitution.
If still have problems, feel free to ask.
2010-11-09 23:15:23 補充：
It is the x-component of friction provides the centripetal force.
2010-11-09 23:22:03 補充：
There is no a specific force called centripetal force, centripetal force must come from certain existing force. For example, the centripetal force of the circular motion of planets is the gravitioal force between them. In this case, the centripetal force is the x-componet of friction.
2010-11-09 23:23:04 補充：
Actually, you can call that it is the friction being the centripetal force, but it is not precise.
2010-11-09 23:28:11 補充：
The forces produce circular motion we call them a centripetal force.
Within a circular motion, you should note that a centripetal force can come from more than one force. You will learn more later.
