Question

1.A horizontally suspended 3.0 m track for fluorescent lights is held in place by two vertical beams that are connected to the ceiling. The track and lights jointly weigh 40 N. One beam is attached 20 cm from one end of the track and the other beam is attached 50 cm from the opposite and of track. What upward force does each beam apply to the track?

2.Three 5.0 kg spheres are distributed as follows. Sphere A is located at the origin, sphere B is located at x = 0.30 m y = 0 m, and sphere C is located at x = 0 m and y = 0.40 m. Find the net gravitational force vector that sphere A experiences.

3.A 33 kg child runs with a speed of 2.5 m/s tangential to the rim of a stationary merry-go-round. The merry-go-round has a moment of inertia of 630 kg.m2 and a radius of 2.7 m. When the child jumps on the merry-go-round, the entire system begins to rotate. With what angular velocity does the merry-go-round rotate after the child jumps onto it?

Answer 1

1.A horizontally suspended 3.0 m track for fluorescent lights is held in place by two vertical beams that are connected to the ceiling. The track and lights jointly weigh 40 N. One beam is attached 20 cm from one end of the track and the other beam is attached 50 cm from the opposite and of track. What upward force does each beam apply to the track?

Let the Force applied by Left beam = FL
Let the Force applied by Right beam = FR

FL + FR = 40 ------1

Taking the moment along the Left end,
40 * (1.5 - 0.2) = FR * (3 - 0.5 - 0.2)
FR = 22.6 N

FL = 40 - 22.6
FL = 17.4 N

Force by the Left & Right beam are, 17.4 N & 22.6 N Respectively !!

2.Three 5.0 kg spheres are distributed as follows. Sphere A is located at the origin, sphere B is located at x = 0.30 m y = 0 m, and sphere C is located at x = 0 m and y = 0.40 m. Find the net gravitational force vector that sphere A experiences.

m = 5.0 kg
(o,o) , ( 0.3 , o) , (0 , 0.4)

Gravitational Force on A due to sphere B,
Fx = G*m^2/r^2
Fx = (6.67*10^-11 * 5^2)/(0.3^2)
Fx = 1.85*10^-8 N   

Gravitational Force on A due to sphere C,
Fy = G*m^2/r^2
Fy = (6.67*10^-11 * 5^2)/(0.4^2)
Fy = 1.04*10^-8 N

Fnet = sqrt(Fx^2 + Fy^2)
Fnet = sqrt((1.85*10^-8)^2 + (1.04*10^-8)^2)
Fnet = 2.12 * 10^-8 N

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