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Question

Mathematics Question on Vector Algebra

A body weighing 13kg13\, kg is suspended by two strings 5 m and 12m12 \,m long, their other ends being fastened to the extremities of a rod 13m13\, m long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are

A

12kg12\, kg and 13kg13\, kg

B

5kg5\, kg and 5kg5\, kg

C

5kg5 \,kg and 12kg12 \,kg

D

5kg5 \,kg and 13kg13\, kg

Answer

5kg5 \,kg and 12kg12 \,kg

Explanation

Solution

T2cos(π2θ)=T1cosθT1cosθ=T2sinθT_{2}\,cos\left(\frac{\pi}{2}-\theta\right)=T_{1}\,cos\,\theta \Rightarrow T_{1}\,cos\,\theta =T_{2}\,sin\,\theta T1sinθ+T2cosθ=13T_{1}\,sin\,\theta+T_{2}\,cos\,\theta =13. OC=CA=CB\because OC=CA=CB ?AOC=?OAC\Rightarrow ?AOC=?OAC and?COB=?OBC?COB=?OBC sinθ=sinA=513\therefore sin\theta=sinA=\frac{5}{13} and cosθ=1213cos\theta=\frac{12}{13} T1T2=512T1=512T2\Rightarrow \frac{T_{1}}{T_{2}}=\frac{5}{12} \Rightarrow T_{1}=\frac{5}{12}T_{2} T2(512513+1213)=13T_{2}\left(\frac{5}{12}\cdot\frac{5}{13}+\frac{12}{13}\right)=13 T2(16912.13)=13T_{2}\left(\frac{169}{12.13}\right)=13 T2=12kgs.T1=5kgs.T_{2}=12\,kgs. \Rightarrow T_{1}=5\,kgs.