Find T 1 in the shown figure- Assume g -10 m - s 2 B- 240 N- 260 ND- 280 N

The correct option is B 240 NUsing FBD For Equilibrium ∑ F_x=0 T_1 cos 37=T_2 cos 53 T_1×4/5=T_2 nbsp;×3/5 ⇒ 4T_1=3T_2 ∑ F_y=0 T_1 sin 37+ T_2 sin 53=40 g 3 ...

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Byju's Answer

Standard XII

Physics

Tension in a String

Find T 1 in t...

Question

Find $T_{1}$ in the shown figure. (Assume $g = 10 {m/s}^{2}$ )

220 N

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240 N

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260 N

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280 N

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Solution

The correct option is B $240 N$
Using FBD

For Equilibrium
$\sum F_{x} = 0$
$T_{1} cos 37 = T_{2} cos 53$
$T_{1} \times \frac{4}{5} = T_{2} \times \frac{3}{5}$
$\Rightarrow 4 T_{1} = 3 T_{2}$
$\sum F_{y} = 0$
$T_{1} sin 37 + T_{2} sin 53 = 40 g$
$\frac{3 T_{1} + 4 T_{2}}{5} = 40 g$
putting $T_{2} = 4 T_{1} / 3$
$\Rightarrow 3 T_{1} + \frac{16 T_{1}}{3} = 5 \times 40 g$
$T_{1} = \frac{(5 \times 40 \times 10) 3}{25} = 240 N$

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Find T1 in the shown figure. (Assume g=10 m/s2)

The correct option is B 240 NUsing FBD For Equilibrium ∑Fx=0 T1cos37=T2cos53 T1×45=T2×35 ⇒4T1=3T2 ∑Fy=0 T1sin37+T2sin53=40 g 3T1+4T25=40 g putting T2=4T1/3 ⇒3T1+16T13=5×40 g T1=(5×40×10)325=240 N

Find $T_{1}$ in the shown figure. (Assume $g = 10 {m/s}^{2}$ )