Question

A man of mass M stands on a railroad car that is rounding an unbanked turn of radius R at speed v. His center of mass is height L above the car, and his feet are distance d apart. The man is facing the direction of motion. How much weight is on each of his feet

Answer 1

Answer:

$F_1= Mg- \frac{2Mv^2L}{Rd}$

and

$F_2=\frac{2Mv^2L}{Rd}$

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 = M×v×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

⇒$\frac{Mv^2L}{r} -\frac{Mgd}{2}+\frac{F_1 d}{2} = 0$\

solving the above equation we get

$F_1= Mg- \frac{2Mv^2L}{Rd}$

therefore,

$F_2=\frac{2Mv^2L}{Rd}$