Answer:
A) [tex]F_{L}[/tex] + [tex]F_{R}[/tex] = W + [tex]W_{B}[/tex]
B) 800 N
C) 1100 N
Explanation:
A)
[tex]F_{L}[/tex] = Force on the left end
[tex]F_{R}[/tex] = Force on the right end
L = length of the beam = 12 m
[tex]W_{B}[/tex] = Weight of the beam = 100 N
W = weight of the machine = 1800 N
a = position of machine relative to left end = 5 m
Using equilibrium of force along the vertical direction
[tex]F_{L}[/tex] + [tex]F_{R}[/tex] = W + [tex]W_{B}[/tex]
B)
Using equilibrium of torque about the left end
[tex]F_{R}[/tex] (L) = W a + [tex]W_{B}[/tex] (0.5 L)
[tex]F_{R}[/tex] (12) = (1800) (5) + (100) (0.5 )(12)
[tex]F_{R}[/tex] = 800 N
C)
Using equilibrium of force along the vertical direction
[tex]F_{L}[/tex] + [tex]F_{R}[/tex] = W + [tex]W_{B}[/tex]
[tex]F_{L}[/tex] + 800= 1800 + 100
[tex]F_{L}[/tex] = 1100 N