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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

	$T_{1} = m_{1} g \sin 30 ° + m_{1} a_{1}$ $T_{2} = m_{2} g \sin 45 ° + m_{2} a_{2}$ $m_{3} (a_{1} \cos 30 ° + a_{2} \cos 45 °) = m_{3} g - T_{1} \cos 30 ° - T_{2} \cos 45 °$ $m_{3} (a_{1} \cos 30 ° + a_{2} \cos 45 °) = m_{3} g - (m_{1} g \sin 30 ° + m_{1} a_{1}) \cos 30 ° - (m_{2} g \sin 45 ° + m_{2} a_{2}) \cos 45 °$ $\Rightarrow (m_{1} + m_{3}) a_{1} \cos 30 ° + (m_{2} + m_{3}) a_{2} \cos 45 ° = (m_{3} - m_{1} \sin 30 ° \cos 30 ° - m_{2} \sin 45 ° \cos 45 °) g$ $\Rightarrow 32 \sqrt{3} a_{1} + 22 \sqrt{2} a_{2} = (32 - 7 \sqrt{3}) g . . . (i)$ $m_{3} (a_{1} \sin 30 ° - a_{2} \sin 45 °) = T_{2} \sin 45 ° - T_{1} \sin 30 °$ $m_{3} (a_{1} \sin 30 ° - a_{2} \sin 45 °) = (m_{2} g \sin 45 ° + m_{2} a_{2}) \sin 45 ° - (m_{1} g \sin 30 ° + m_{1} a_{1}) \sin 30 °$ $\Rightarrow (m_{1} + m_{3}) a_{1} \sin 30 ° - (m_{2} + m_{3}) a_{2} \sin 45 ° = (m_{2} \sin^{2} 45 ° - m_{1} \sin^{2} 30 °) g$ $\Rightarrow 32 a_{1} - 22 \sqrt{2} a_{2} = - 3 g . . . (i i)$ $(i) + (i i) :$ $32 (1 + \sqrt{3}) a_{1} = (29 - 7 \sqrt{3}) g$ $\Rightarrow a_{1} = \frac{(18 \sqrt{3} - 25) g}{32} \approx 0.193 g$ $\Rightarrow a_{2} = \frac{(9 \sqrt{3} - 11) \sqrt{2} g}{22} \approx 0.295 g$ $a_{3} = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos 75 °} \approx 0.392 g$
	Commented by Tawa11 last updated on 17/Aug/22

	$Great sir$
	Commented by peter frank last updated on 18/Aug/22

	$thank you$
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