Two-masses-5-kg-and-M-are-hanging-with-the-help-of-light-rope-and-pulley-as-shown-below-If-the-system-is-in-equilibrium-then-M-

Question Number 17530 by Tinkutara last updated on 07/Jul/17

Two masses 5 kg and M are hanging

with the help of light rope and pulley

as shown below . If the system is in

equilibrium then M =

Commented by Tinkutara last updated on 07/Jul/17

Commented by mrW1 last updated on 07/Jul/17

Tcos 53 = 5 \times \cos 37

T = 5 \times \frac{\cos 37}{\cos 53} = 5 \times \frac{\cos 37}{\sin 37}

M = 5 \times \sin 37 + T \times \sin 53

= 5 \times \sin 37 + 5 \times \frac{\cos^{2} 37}{\sin 37}

= \frac{5}{\sin 37} = 8.31 kg

Commented by 1234Hello last updated on 07/Jul/17

Thanks Sir! I forgot the options . But

can you explain why M = 5 \sin 37 °

+ T \sin 53 ° ?

Commented by mrW1 last updated on 07/Jul/17

I {don}^{'} t understand what you mean with

it is not in the options . You can solve

it in different ways . Since 37 + 53 = 90,

the three forces build a right angeled

triangle, so you can directly get

M = 5 / \sin 37 ° . If you have no calculator,

you can use \sin 37 \approx \frac{\sin 30 + \sin 45}{2} \approx \frac{0.5 + 0.7}{2} = 0.6

\Rightarrow M \approx 5 / 0.6 \approx 8.3 .

Commented by ajfour last updated on 07/Jul/17

5 mgcos 37 = Tcos 53

\Rightarrow 4 mg = 3 T

mgsin 37 + Tsin 53 = Mg

3 mg + 4 T = 5 Mg

\Rightarrow 3 mg + \frac{16 mg}{3} = 5 Mg

M = \frac{5 m}{3} = \frac{5 \times 5 kg}{3} = 8 \frac{1}{3} kg .

Commented by mrW1 last updated on 07/Jul/17

equilibrium in x - direction :

Tcos 53 = 5 \times \cos 37

equilibrium in y - direction :

M = 5 \times \sin 37 + T \times \sin 53

Commented by Tinkutara last updated on 08/Jul/17

Thanks ajfour and mrW 1 Sir!