A 1750-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (?A and ?B?), each 1.25 m long and weighing 0.360 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. 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? Which pulse arrives first? (Neglect the effect of the weight of the wires on the tension in the wires.)

Solution 54P Step 1 of 6: The total force F= 1750 N is due to the sum of tension in wires A and B. F=T + T =1750 N a b Length of wire, L= 1.25 Weight, W=0.360 N Using W= mg mass m=(W/g) m=(0.360/9.8) m=0.0367 kg Step 2 of 6: In order to calculate the time delay between two pulses of reaching ceiling , we need to calculate the tension and speed on two wires A and B. To calculate the tension, If we sum moments about one end, we can find the tension at the other. For A, 1750 (3)=T Lb T = 1750 b 3 T = 583.33 N b Similarly for B, Using =T + T =1750 N a b T =1750 N-T a b T =1750 N- 583.33 N a T =1166.67 N a Step 3 of 6: To calculate the speed, For a wire with tension T and mass per unit length , the speed of the wave is given by v = T m Using = L Substituting m=0.0367 kg and L= 1.25 m 0.0367 kg = 1.25 m =0.2936 kg/m Using in speed equation, v = 0.2936 kg/m...1