Question

Find the tensions in the strings T, and T2 in kgwt. please ans qo​

Answer 1

Explanation:

T2 cos 53° = 10 kg wt

so T2 = 10x 5/3= 50/3

T2 cos 37° = T1 = 50/3x4/5 = 200/15=40/3

so T1 = 40/3

T2 = 50/3

hence second option is right

Answer 2

The tension $\sf{T_1}$ is acting horizontally rightwards.

• $\vec{\sf{T_1}}=\sf{T_1\ \hat i}$

The tension $\sf{T_2}$ makes an angle 37° with the horizontal and so 53° with the vertical.

• $\vec{\sf{T_2}}=\sf{T_2\cos37^{\circ}\left(-\hat i\right)+T_2\cos53^{\circ}\left(\hat j\right)}$

• $\vec{\sf{T_2}}=\sf{-\dfrac{4T_2}{5}\ \hat i+\dfrac{3T_2}{5}\ \hat j}$

The block has weight 10 kgwt acting vertically downwards.

• $\vec{\sf{W}}=\sf{-10\ \hat j}$

For the system being in equilibrium, the net force acting at the point should be zero.

$\longrightarrow\vec{\sf{T_1}}+\vec{\sf{T_2}}+\vec{\sf{W}}=\sf{0}$

$\longrightarrow\sf{T_1\ \hat i-\dfrac{4T_2}{5}\ \hat i+\dfrac{3T_2}{5}\ \hat j-10\ \hat j=0}$

$\longrightarrow\sf{\left(T_1-\dfrac{4T_2}{5}\right) \hat i+\left(\dfrac{3T_2}{5}-10\right) \hat j=0\ \hat i+0\ \hat j}$

Equating corresponding components, we get two equations,

$\longrightarrow\sf{\dfrac{3T_2}{5}-10=0}$

$\longrightarrow\underline{\underline{\sf{T_2=\dfrac{50}{3}\ kgwt}}}$

And,

$\longrightarrow\sf{T_1-\dfrac{4T_2}{5}=0}$

$\longrightarrow\sf{T_1-\dfrac{4\times50}{5\times3}=0}$

$\longrightarrow\underline{\underline{\sf{T_1=\dfrac{40}{3}\ kgwt}}}$

Hence (2) is the answer.