Answer to Question #260279 in Mechanics | Relativity for feifie

Question

A 1750 N

N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A

A and B

B), each 1.20 m

mlong and weighing 0.260 N

N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A

A is attached. Ignore the effect of the weight of the wires on the tension in the wires.

Answer 1

A 1750 N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.20 m long and weighing 0.260 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached. Ignore the effect of the weight of the wires on the tension in the wires. If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling?

Beam A support $= \frac{2}{3} \times 1750 \;N = 1166.6 \;N$

Beam B support $= \frac{1}{3} \times 1750 \;N = 583.3 \;N$

Mass of each wire $= \frac{0.260}{9.8} = 0.02653 \;kg$

Linear density $= \frac{0.02653}{1.20}=0.0221 \;kg/m$

Speed:

$V_A = \sqrt{\frac{T}{U}} = \sqrt{\frac{1166.6}{0.0221}} = 229.75 \;m/s \\ V_B = \sqrt{\frac{583.3}{0.0221}} = 162.44 \;m/s \\ T_A = \frac{1.20}{229.75} = 0.00522306 \\ T_B = \frac{1.2}{162.44} = 0.00738734 \\ Time \; delay = 0.00738734 -0.00522306 = 0.00216034 \;s$