Question-175066

Question Number 175066 by peter frank last updated on 17/Aug/22

Answered by mr W last updated on 17/Aug/22

Commented by mr W last updated on 17/Aug/22

T_{1} = m_{1} g \sin 30 ° + m_{1} a_{1}

T_{2} = m_{2} g \sin 45 ° + m_{2} a_{2}

m_{3} (a_{1} \cos 30 ° + a_{2} \cos 45 °) = m_{3} g - T_{1} \cos 30 ° - T_{2} \cos 45 °

m_{3} (a_{1} \cos 30 ° + a_{2} \cos 45 °) = m_{3} g - (m_{1} g \sin 30 ° + m_{1} a_{1}) \cos 30 ° - (m_{2} g \sin 45 ° + m_{2} a_{2}) \cos 45 °

\Rightarrow (m_{1} + m_{3}) a_{1} \cos 30 ° + (m_{2} + m_{3}) a_{2} \cos 45 ° = (m_{3} - m_{1} \sin 30 ° \cos 30 ° - m_{2} \sin 45 ° \cos 45 °) g

\Rightarrow 32 \sqrt{3} a_{1} + 22 \sqrt{2} a_{2} = (32 - 7 \sqrt{3}) g \dots (i)

m_{3} (a_{1} \sin 30 ° - a_{2} \sin 45 °) = T_{2} \sin 45 ° - T_{1} \sin 30 °

m_{3} (a_{1} \sin 30 ° - a_{2} \sin 45 °) = (m_{2} g \sin 45 ° + m_{2} a_{2}) \sin 45 ° - (m_{1} g \sin 30 ° + m_{1} a_{1}) \sin 30 °

\Rightarrow (m_{1} + m_{3}) a_{1} \sin 30 ° - (m_{2} + m_{3}) a_{2} \sin 45 ° = (m_{2} \sin^{2} 45 ° - m_{1} \sin^{2} 30 °) g

\Rightarrow 32 a_{1} - 22 \sqrt{2} a_{2} = - 3 g \dots (i i)

(i) + (i i) :

32 (1 + \sqrt{3}) a_{1} = (29 - 7 \sqrt{3}) g

\Rightarrow a_{1} = \frac{(18 \sqrt{3} - 25) g}{32} \approx 0.193 g

\Rightarrow a_{2} = \frac{(9 \sqrt{3} - 11) \sqrt{2} g}{22} \approx 0.295 g

a_{3} = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos 75 °} \approx 0.392 g

Commented by Tawa11 last updated on 17/Aug/22

Great sir

Commented by peter frank last updated on 18/Aug/22

thank you