A beam with a weight of 100 N nd a length of 12.0 meters is supported at each end. It is supporting a machine with a weight of 1800 N which is located 5.0 m from the left end of the beam. A.) draw a free body diagram for the beam, clearly labeling lol forces, write down the equations which are obtained from Newton’s second law. B.) what is the force of support at the right end? C.) what is the force of support at the left end?

Respuesta :

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

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