- #1

issacnewton

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- 31

## Homework Statement

The masses of blocks ##A## and ##B## are 20kg and 10 kg, respectively. The blocks are initially at rest on the floor and are connected by a massless and frictionless pulley. An upward force ##\vec{F}## is applied to the pulley. Find the accelerations ## \vec{a_A}## of block ##A##, and ## \vec{a_B}## of block ##B## when F is (a) 124 N ;(b) 294 N ; (c) 424 N.

## Homework Equations

Newton's second law of motion. Relative acceleration formula

## The Attempt at a Solution

I think it would be convenient to work in the pulley's frame. Since that is a non-inertial frame, I will need to introduce fictitious force in the analysis, so that I can use Newton's equations of motion as usual. Now let ##x_1## and ##x_2## be the coordinate of the blocks ##A## and ##B## as measured downward from the pulley. So, I am taking downward direction as positive. Since the system is accelerating upwards, I will need to introduce a fictitious force ##F## acting downward on each block. Let ##L## be the length of the string, then we have ##x_1+x_2=L##, which means ##\ddot{x_1} + \ddot{x_2} = 0##. Now equations of motion in the pulley's frame would be $$m_1g +F - T = m_1\ddot{x_1}$$ and $$m_2g +F - T = m_2\ddot{x_2}$$ where ##T## is the tension in the string and ##m_1## and ##m_2## are the masses of blocks ##A## and ##B## respectively. Subtracting two equations we get $$(m_1-m_2)g = m_1 \ddot{x_1} -m_2 \ddot{x_2} = m_1 \ddot{x_1} + m_2\ddot{x_1}$$ Solving, we get $$\ddot{x_1} = \frac{(m_1-m_2)g}{(m_1+m_2)}$$ Does it look alright ? There is no ##F## involved here. So I think I made mistakes somewhere.