A) 10 N, 11 N
B) 10 N, 6 N
C) 10 N, 10 N
D) cant be calculated due to insufficient data
Correct Answer: C
Solution :
From the figure. \[{{T}_{1}}\cos {{30}^{o}}={{T}_{2}}\cos {{30}^{o}}\] \[\therefore \] \[{{T}_{1}}={{T}_{2}}=T\](Let) Again\[{{T}_{1}}\sin {{30}^{o}}+{{T}_{2}}\sin {{30}^{o}}=10\] \[2T\sin {{30}^{o}}=10\] \[2T.\frac{1}{2}=10\Rightarrow T=10N\] \[\therefore \] Tension in section BC and BF are 10 N and IONYou need to login to perform this action.
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