Question

A beam with negligible mass has masses m₁ and m₂ suspended at distances a and 2a from a pivot point, respectively. The beam is balanced and so not moving, and makes an angle θ with respect to the horizontal.

у 2a a Om₂ m,

Part 1

The force exerted along the y direction by the pivot on the beam is

a. ( m₁g − m₂g ) sin(θ)

b. m₁g − m₂g

c. m₁g + m₂g

d. ( m₁g + m₂g ) cos(θ)

e. ( m₁g + m₂g ) sin(θ)

f. ( m₁g − m₂g ) cos(θ)

Part 2

To balance the beam, the mass m₂ must be chosen to be

a. 2 m₁

b. 2 m₁ cos(θ)

c. ½ m₁ sin(θ)

d. ½ m₁

e. 2 m₁ sin(θ)

½ m₁ cos(θ)

Answer 1

Solution

Part1

Answer: c. m1g+m2g

Part2

Answer: d. (1/2)m1

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