Find the forces exerted by the two supports of a 4.0m, 50-kg diving board when a 60-kg woman stands at its end. The two supports are 0.8m apart. This time consider the mass of the diving board.

Answer 1

x1= 0.8m -the distance between supports.
x2= 1.2 m - the distance between the right-hand support anf the center of mass of the board,
x3 =3.2 m –the distance between the right-hand support and the diver.

m1= 50 kg, m1•g = 490 N,
m2= 60 kg, m2•g = 588 N.
F1 is the downward force on the left-hand support,
F2 is the upward force on the right-hand support.
Three torques act about the right support
τ1=+F1•x1,
τ2 = - m1•g•x2,
τ3 = - m2•g•x3,
Σ τ= τ1 +τ2 +τ3 = 0,
F1•0.8 - 490•1.2 - 588•3.2 = 0,
F1=(490•1.2 + 588•3.2)/0.8 =3087 N
ΣF = - F1 + F2 - m1•g - m2•g = 0,
-3087+F2-490-588 =0
F2 = 3087+ 490+588=4165 N