Answer:
[tex]F_1= Mg- \frac{2Mv^2L}{Rd}[/tex]
and
[tex]F_2=\frac{2Mv^2L}{Rd}[/tex]
Explanation:
distance of Center of mass from his feet = L
Mass of man = M
Radius of turn = R
Speed = v
distance between the feet =d
Centrifugal force = MĂ—vĂ—v/R = F
Moment due to this force = FL
now as per the situation in the question
force on feet = Force on feet 1 + force on feet 2 = F1 + F2 = Mg
Now, balancing the moment we get by taking moment about feet 2
moment about feet 2 ( outward in the turn) = moment due to centrifugal force ( + ve ) + moment due to weight ( -ve ) + moment due to normal on foot 1 ( + ve) = 0
⇒[tex]\frac{Mv^2L}{r} -\frac{Mgd}{2}+\frac{F_1 d}{2} = 0[/tex]\
solving the above equation we get
[tex]F_1= Mg- \frac{2Mv^2L}{Rd}[/tex]
therefore,
[tex]F_2=\frac{2Mv^2L}{Rd}[/tex]
