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The masses are interconnected with the pulley system neglecting inertial and frictional effect of pulleys and cord, determine the acceleration of the mass $m_2. $

written 7.9 years ago by teamques10 ★ 68k

•   modified 4.5 years ago

Take $m_1 = 50$ kg and $m_2 = 40$ kg.

engineering mechanics

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written 7.9 years ago by teamques10 ★ 68k

$T_1 a_1=T_2.a_2 \\ 2T.a_1=T.a_2 \\ a_2=2a_1$

From FBD of block 1,

$∑f_y=0⟹2T+ma_1=50×9.81 \\ 2T+50a_1=490.5 ---(1) $

From FBD of block 2,

$∑f_y=0⟹T-ma_2=40×9.81 \\ T-40×2a_1=392.4 ---(2) \\ ∴T=280.28 N $

and $a_1= -1.40 m/s^2$

Acceleration of the mass $m_2=a_2=-2.80m/s^2$

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