Problem 5. A rigid beam of mass m, length L, has two supports. One support is lo
ID: 2039471 • Letter: P
Question
Problem 5. A rigid beam of mass m, length L, has two supports. One support is located at the beam's left end, while the other is a distance 3L/4 from the left edge. A block of mass M is situated such that its center of mass is located a distance L/2 from the beam's left end as showin in the figure below. What is the ratio of the normal force exerted by support 1 to the normal force exerted by support 2 on the beam? Block of mass M Rigid Beam of mass m, length L L/2 Support 1 Support 2 3L/4 F2 4 C) F2 4 D) E) F2Explanation / Answer
D) F1/F2 = 1/2
As the beam is in equilibrium net force and torque acting on it must be zero.
Apply net torque about left end = 0
m*g*(L/2)*sin(90) + M*g*(L/2)*sin(90) - F2*(3*L/4)*sin(90) = 0
m*g/2 + M*g/2 - F2*3/4 = 0
m*g + M*g - F2*3/2 = 0
==> F2*3/2 = (m + M)*g
F2 = (2/3)*(m + M)*g ---(1)
now apply Fnety = 0
F1 + F2 - m*g - M*g = 0
F1 = (m + M)*g - F2
= (m + M)*g - (2/3)*(m + M)*g
= (1/3)*(m + M)*g ---(2)
from equations 1 and 2
we get,
F1/F2 = (1/3)/(2/3)
F1/F2 = 1/2