A gymnast stands on a balance beam of length L that is supported above the gym f

Question

A gymnast stands on a balance beam of length L that is supported above the gym floor by two supports at its ends. Lets place the origin O at the point where the gymnasts foot touches the beam. Let's say that the gymnast is 1/5 L from the right end, the mass of the gymnast is m, and the mass of the beam is M. Let the contact forces exerted on the beam by the right and left supports be Fcr and Fcl ,respectively.

(a) In terms of these quantities, what are the magnitude and direction of the torque exerted by the right support?

(b) What are the magnitude and direction of the torque exerted by the left support?

(c) What are the magnitude and direction of the torque exerted by the beam's weight?

Explanation / Answer

net torque acting on the beam is Zero

Torque due to (Fcr + Fcl + weight of the gymnast + weight of the beam ) =

let the axis of rotation is about the origin

Fcr*(L/5) - Fcl*(4L/5) + 0 + (M*g*(3L/10)) = 0

Torque exerted by right support is Fcr*(L/5) = [Fcl*(4L/5)] - [(3*M*g*L)/10]

direction is along perpendicular to the rod i.e +Z-axis(say horizontal is x-axis and vertical is Y-axis)

B) torque exerted by the left support is Fcl*(4L/5) = Fcr*(L/5) - (M*g*(3L/10))

direction is along -Z -axis

C) torque exerted by the beam's weight is (M*g*(3L/10)) = Fcl*(4L/5) - Fcr*(L/5)

direction is alonf +Z-axis

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.