A merry go around makes one complete revolution in?

A merry go around makes one complete revolution in 7.74s. A 53.7kg child sits on the horizontal floor of t he merry-go-round 2.82m from the center. Gravity=9.8





a. find the horizontal force of friction that acts on the child. Answer in units of N





b. Find the horizontal force of friction that acts on the child. Answer in units of N





c. What minimum coefficient of static friction is neccessary to keep the child from slipping?

A merry go around makes one complete revolution in?
a and b are the same question.





Since Friction is the Fnet, and the Fnet is the centripetal force. Friction is the centripetal force





Ffriction=Fc


Fc= 4*pi^2*m*r / T^2


Fc= 4*pi^2*53.7kg*2.82m / (7.74s)2


Fc= 100N


since Fc is 100N then Friction is also 100N





Fnormal=Fweigh


Fweigh=mg


Fweigh= (53.7)(9.8)


Fweigh= 526.26N


Fnormal is also 526.26N





Friction=mu(Fnormal)


100N=mu(526.26N)


mu= .19
Reply:way too long!!!!lol
Reply:do your own homework..

wreath