Q.
The adjacent figure is the part of a horizontally stretched net. Section AB is stretched with a force of 10 N. The tensions in the sections BC and BF are
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Solution:
As shown in figure. (ii)
$ {{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 $ and $ 2T\sin {{30}^{o}}=10 $ $ 2T.\frac{1}{2}=10\Rightarrow T=10N $ Thus, the tension in section BC and BF are 10 N and 10 N respectively.
