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

`F_1= Mg- (2Mv^2L)/(Rd)`

and

`F_2=(2Mv^2L)/(Rd)`

Step-by-step 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

⇒
`(Mv^2L)/(r) -(Mgd)/(2)+(F_1 d)/(2) = 0`\

solving the above equation we get

`F_1= Mg- (2Mv^2L)/(Rd)`

therefore,

`F_2=(2Mv^2L)/(Rd)`