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?

4.A vertical spring stretches 20 cm when a 5.0 kg block is hung from its end. Calculate the spring constant. This block is then displaced an additional 10 cm downward and released. Find the period, frequency, and amplitude of the block's motion.

5.A certain guitar string of length 1.0 m has a mass of 7.0 g. When properly tuned, the string has a third harmonic of 770 Hz. What is the tension for properly tuning this string?

6.A solid hoop (I = MR2) starts from rest at the top of a ramp that is 3.0 m long and is inclined at an angle of 300 relative to the horizontal. The mass of the solid hoop is 12 kg and its radius is 10 cm. What is the linear velocity of the hoop when it reaches the bottom of the ramp?

Answer 1

1. Let F1 be the force applied by the left beam and F2 applied by the right beam to the track. Let us calculate the torque about the point of intersection of left beam and the track.

Steps for solving the problem are as follows:

1. Write out the torque due to F1:
2. Write out the torque due to F2:	τ2 = (3 - 0.7)F2 = 2.3F2
3. Write out the torque due to gravity:	τg = (3/2 - 0.2)mg = 1.3mg
4. Apply the equilibrium condition for torque:
	2.3F2 - 1.3 mg = 0 F2 = 1.3mg/2.3 = 1.3*40/2.3 = 22.6 N
5. Apply the equilibrium condition for force:
	F1 = mg - F2 = 40 - 22.6 = 17.4N