Answer:
the extension of the beam over the ledge is
x = 4.24 kg
Explanation:
The length of the beam is L= 10 m
The mass of the beam is M = 280 kg
The mass of the student is m = 50 kg
For the equilibrium the net torque on the beam in
counterclockwise direction about the point of
pivot should be zero.
Στ = Mg(L/2 - x) - m×g×x = 0
Mg(L/2 - x) - m×g×x = 0
Mg(L/2 - x) = m×g×x
M(L/2 - x) = m×x
(280 kg)((10 m/2) - x) = (50 kg)(x)
calculating the value of x from the above equation we get
x= 4.24 Kg
the extension of the beam over the ledge is
x = 4.24 kg