Ques 5
Asked by rbhatt16
21st February 2018,
10:23 PM
Answered by Expert
Answer:
Tensions acting on the steel wires are shown in figure.
case(1)
M1 = 10 kg, M2 = 20 kg;
Stress for the wire between M1 and M2 is given by T1 = [ (10+M) g ] / ( 0.003×10-4 ) N/m2
Maximum Load for breaking stress will be obtained from [ (10+M) g ] / ( 0.003×10-4 ) = 8×108 N/m2
Solving for M we will get M = 14.5 kg
Stress for the wire above M2 is given by T2 = [ (30+M) g ] / ( 0.006×10-4 ) N/m2
Maximum Load for breaking stress will be obtained from [ (30+M) g ] / ( 0.006×10-4 ) = 8×108 N/m2
Solving for M we will get M = 19 kg
Since the maximum laod for breaking of wire between M1 and M2 is 14.5 kg, we can take the load limit as 14.5 kg. if the load exceeds this limit, then lower wire, i.e. wire between M1 and M2 will break first.
case(2)
M1 = 10 kg, M2 = 36 kg;
Stress for the wire between M1 and M2 is given by T1 = [ (10+M) g ] / ( 0.003×10-4 ) N/m2
Maximum Load for breaking stress will be obtained from [ (10+M) g ] / ( 0.003×10-4 ) = 8×108 N/m2
Solving for M we will get M = 14.5 kg
Stress for the wire above M2 is given by T2 = [ (46+M) g ] / ( 0.006×10-4 ) N/m2
Maximum Load for breaking stress will be obtained from [ (46+M) g ] / ( 0.006×10-4 ) = 8×108 N/m2
Solving for M we will get M = 3 kg
Since the maximum laod for breaking of wire above M2 is 3 kg, we can take the load limit as 3 kg. if the load exceeds this limit, then upper wire, i.e. wire above M2 will break first.
Answered by Expert
22nd February 2018,
3:43 PM
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