A10 m beam of mass 280 kg extends over a ledge. the beam is not attached, but simply rests on the surface. a 50 kg student intends to position the beam so that he can walk to the end of it. how far from the edge of the ledge can the beam extend?

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  

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y