# Can you solve this problem on Mechanics?

Nov 13, 2016

Let the acceleration of the system be $a$ and tension on the connecting string be $T$

Considering the forces on ${W}_{B}$ we can write

$25 \times g \times \sin 30 - 0.2 \times 25 \times g \times \cos 30 - T = 25 \times a \ldots . . \left(1\right)$

Considering the forces on ${W}_{A}$ we can write

$15 \times g \times \sin 30 - 0.4 \times 15 \times g \times \cos 30 + T = 15 \times a \ldots . . \left(2\right)$

Adding (2) and (1) we get

$\left(25 + 15\right) \times g \times \sin 30 - \left(0.2 \times 25 + 0.4 \times 15\right) \times g \times \cos 30 = \left(25 + 15\right) \times a$

=>(40xxgxxsin30-(5+6)xxgxxcos30=40xxa

$\implies 40 \times g \times \sin 30 - 11 \times g \times \cos 30 = 40 \times a$

$\implies a = \frac{40 \times 32 \times \frac{1}{2} - 11 \times 32 \times \frac{\sqrt{3}}{2}}{40}$

$= 16 - 4.4 \sqrt{3} = 8.38 f t {s}^{-} 2$

Taking $\text{acceleration due to gravity} g = 32 f t {s}^{-} 2$

Inserting the value of $a$ in (1)

$25 \times 32 \times \sin 30 - 0.2 \times 25 \times 32 \times \cos 30 - T = 25 \times 8.38$

=>25xx32xx1/2-0.2xx25xx32xxsqrt3/2-T=25xx8 38

$\implies 400 - 80 \times \sqrt{3} - T = 209.5$

$\implies 261.44 - T = 209.5$

$\implies T = 261.44 - 209.5 = 51.94 p o u n \mathrm{da} l$