Question #121378
A mass of 1019Kg is supported by two wires A and B each made of a different material. Before loading the wires are of equal length. After loading they are still of equal length. Find the load carried by each wire.
(WA = 3066N) (WB = 6930N)
Wire
Diameter mm
E Nmm-2
A

1

206,000
B
2

116,000
1
Expert's answer
2020-06-10T18:24:40-0400

Solution.

m=1019kg;g=9.81N/kg;m=1019kg; g=9.81N/kg;

dA=1mm;d_A=1mm;

dB=2mm;d_B=2mm;

EA=206000N/mm2;E_A=206000N/mm^2;

EB=116000N/mm2;E_B=116000N/mm^2;

Fg=FA+FB;F_g=F_A+F_B;

Fg=mg;F_g=mg;

Fg=1019kg9.81N/kg=9996N;F_g=1019kg\sdot9.81N/kg=9996N;

σ=FS;\sigma=\dfrac{F}{S}; σ=Eϵ    F=ESϵ;\sigma=E\epsilon \implies F=ES\epsilon;

S=πd24;S=\pi \dfrac{d^2}{4};

SA=3.14(1mm)24=0.785mm2;S_A=3.14\sdot\dfrac{(1mm)^2}{4}=0.785mm^2;

SB=3.14(2mm)24=3.14mm2;S_B=3.14\sdot\dfrac{(2mm)^2}{4}=3.14mm^2;

Fg=EASAϵ+EBSBϵ;F_g=E_AS_A\epsilon+E_BS_B\epsilon;

ϵ=FgEASA+EbSB;\epsilon=\dfrac{F_g}{E_AS_A+E_bS_B};

ϵ=9996N206000N/mm20.785mm2+116000N/mm23.14mm2=0.019;\epsilon=\dfrac{9996N}{206000N/mm^2\sdot0.785mm^2+116000N/mm^2\sdot3.14mm^2}=0.019;

FA=206000N/mm20.785mm20.019=3072N;F_A=206000N/mm^2\sdot0.785mm^2\sdot0.019=3072N;

FB=116000N/mm23.14mm20.019=6924N;F_B=116000N/mm^2\sdot3.14mm^2\sdot0.019=6924N;

Answer: FA=3072N;F_A=3072N;

FB=6924N.F_B=6924N.




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