Question

1. A street light is supported by two wires and each makes a different angle from each other. The one on the left is 32 degrees from the vertical and the one on the right is 43 degrees from the vertical. Find the tension in each wire if the street light has a mass of 60.0 kg. See notes in week 5 announcement on this problem.

2. There are three masses m₁ = 2.0 kg, m₂ = 5.0 kg and m₃ = 7.0 kg. The mass are all affixed to a massless rod of length 15cm. m₁ and m₃ are at the ends and m₂ is 10 cm from left where m₁ is. Calculate the center of mass of the system and specify from which end it is from.

3. A pair of hedge trimmers are 45 cm long from the pivot to the end of the handles. If the person applies of a force of 15 N on each handle and the blades are 3.5 cm long, how much force is applied to the branch?

4. A 245 kg person lays on board that is supported only on the ends and that has a mass of 12.0 kg. If the person and the board are both 205 cm long and the person

Answer 1

1)

the key is that cables only support force vectors that lie along their length.

and the net horizontal force at the lamp = 0

for the 32 degree cable vertical component = sin32 =0.530 horiz comp = cos 32=0.848

for the 43 degree cable V = sin43= 0.682 , H= cos43 = 0.731

we have to multiply the h&v components of the 43 deg cable by .848/.731 to make the horizontal components cancel.

43 deg cable becomes v = .791 and h = .848

the verical load is shared by the cables in proportion to the V components

32 deg = 0.530 43deg = .791 sum = 1.321

(.530/1.321)*60 = 24.07kg for the 32 degree cable.

60-24.07 = 35.93 kg on the 43 degree cable

24.07cosec 32 = 24,07*1.8887 = 45.42kg tension in 32 degree cable
35.93cosec43 = 35.93*1.07 = 38.53 kg tension in the 43 deg cable

3)

Torque = Force * distance from pivot point in meters
For the person, Torque = 15 * 0.45 = 6.75
For the branch, Torque = F * 0.035

F * 0.035 = 6.75
F = 6.75