Question

007 (part 1 of 2) 10.0 points A 0.091 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A 0.568 kg object hangs vertically from the 5.75 cm mark. A second mass is attached at another mark to keep it horizontal and in rotational and translational equilibrium. If the tension in the string attached to the ceiling is 16.9 N, find the value of the sec- ond mass. The acceleration due to gravity is 9.8 m/s. Answer in units of kg.

Answer 1

m₂ g 5.75cm 40cm 50cm Xu 100 cm

Gravitational acceleration = g = 9.8 m/s²

Mass of the meter stick = M = 0.091 kg

Mass of the first object = m₁ = 0.568 kg

Mass of the second object = m₂

Tension in the string = T = 16.9 N

Balancing the forces on the meter stick,

T = m₁g + m₂g + Mg

16.9 = (0.568)(9.8) + m₂(9.8) + (0.091)(9.8)

m₂ = 1.065 kg

Position of the first object = X₁ = 5.75 cm = 0.0575 m

Position of the string = X₂ = 40 cm = 0.4 m

Position of the center of mass of the meter stick = X₃ = 50 cm = 0.5 m

Position of the second object = X₄

Distance of the first mass from the string = D₁

D₁ = X₂ - X₁ = 0.4 - 0.0575 = 0.3425 m

Distance of the center of mass of the meter stick from the string = D₃

D₃ = X₃ - X₂ = 0.5 - 0.4 = 0.1 m

Distance of the second mass from the string = D₄

Taking moment balance about the string,

m₁gD₁ = MgD₃ + m₂gD₄

m₁D₁ = MD₃ + m₂D₄

(0.568)(0.3425) = (0.091)(0.1) + (1.065)D₄

D₄ = 0.1741 m

D₄ = X₄ - X₂

0.1741 = X₄ - 0.4

X₄ = 0.5741 m

X₄ = 57.41 cm

1) Value of mass of the second object = 1.065 kg

2) The mark at which the second object is attached = 57.41 cm