Physics Help Forum D'Alembert, two blocks and an inclined plane

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 Aug 21st 2017, 08:46 PM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 D'Alembert, two blocks and an inclined plane Ok, I am still confused about this. This time I considered all the forces and all the particles. However this is how I broke it down. $\displaystyle F_{applied} = (\frac{m_1}{\sin \alpha}+m_2)g$ $\displaystyle F_{inertial} = (m_2 +\frac{m_1}{\sin \alpha})\ddot{s}$ However the solution misses out the sin(alpha) in the applied force. Are constraints always ignored when writing down applied forces? Attached Thumbnails
 Aug 22nd 2017, 09:10 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,287 The F_applied is $\displaystyle m_1 g \sin \alpha - m_2 g$. You have m_1 divided by sin alpha rather than multiplied by sin alpha. And for F_inertial it's simply $\displaystyle \sum m a = (m_1 + m_2) a$.

 Tags blocks, dalembert, inclined, plane

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