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

Question

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 (A) 10 N,11 N (B) 10 N, 6N (C) 10 N, 10 N (D) can't be calculated due to insufficient data

Tardigrade Admin · Accepted Answer

As shown in figure. (ii) T1 cos 30o=T2 cos 30o ∴ T1=T2=T (Let) Again T1 sin 30o+T2 sin 30o=10 and 2T sin 30o=10 2T.(1/2)=10⇒ T=10N Thus, the tension in section BC and BF are 10 N and 10 N respectively.

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

10 N,11 N

10 N, 6N

10 N, 10 N

can't be calculated due to insufficient data

As shown in figure. (ii)
$T_{1} cos 30^{o} = T_{2} cos 30^{o}$ $∴$ $T_{1} = T_{2} = T$ (Let) Again $T_{1} sin 30^{o} + T_{2} sin 30^{o} = 10$ and $2 T sin 30^{o} = 10$ $2 T . \frac{1}{2} = 10 \Rightarrow T = 10 N$ Thus, the tension in section BC and BF are 10 N and 10 N respectively.

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

10 N,11 N

10 N, 6N

10 N, 10 N

can't be calculated due to insufficient data

As shown in figure. (ii) T1​cos30o=T2​cos30o ∴ T1​=T2​=T (Let) Again T1​sin30o+T2​sin30o=10 and 2Tsin30o=10 2T.21​=10⇒T=10N Thus, the tension in section BC and BF are 10 N and 10 N respectively.

As shown in figure. (ii)
$T_{1} cos 30^{o} = T_{2} cos 30^{o}$ $∴$ $T_{1} = T_{2} = T$ (Let) Again $T_{1} sin 30^{o} + T_{2} sin 30^{o} = 10$ and $2 T sin 30^{o} = 10$ $2 T . \frac{1}{2} = 10 \Rightarrow T = 10 N$ Thus, the tension in section BC and BF are 10 N and 10 N respectively.