Question

Two men carry a weight of 2 KN by means of two ropes fixed to the weight one rope is inclined to 45 and othrer at 30 with their vertices. Find the tension in each type​

Answer 1

Answer:

$T_1\ =\ 1794.11\ N$

$T_2\ =\ 1470\ N$

Explanation:

Given,

• Weight of the block = W = 2 KN
• Angle of inclination of the first rope = $\theta_1\ =\ 45^o$
• Angle of inclination of the second rope = $\theta_2\ =\ 30^o$

Let $T_1\ and\ T_2$ be the tensions in the first and second rope respectively.

From the f.b.d. of the block,

$\sum\ F_x\ =\ 0\\\Rightarrow T_1cos45^o\ =\ T_2cos30^o\\\Rightarrow T_1\ =\ \dfrac{T_2cos30^2}{cos45^o}\\\Rightarrow T_1\ =\ 1.22T_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

$\sum F_y\ =\ 0\\\Rightarrow T_1sin45^0\ +\ T_2sin30^0\ -\ W\ =\ 0\\\Rightarrow 1.22T_2sin45^o\ +\ T_2sin30^o\ =\ 2000\ N\Rghtarrow 1.36T_2\ =\ 2000\ N\\\Rightarrow T_2\ =\ \dfrac{2000}{1.36}\\\Rightarrow T_2\ =\ 1470\ N$

From the eqn (1), we get,

$T_1\ =\ 1.36 T_2\ =\ 1.36\times 1470\ =\ 1794.11\ N$