The diagram shows two masses connected via a string across two pulleys. The masses are initially held in place, then allowed to move freely. If $m_1$ = 4.7 kg and $m_2$ = 1.8 kg. a) Which direction will the masses accelerate? b) Draw the FBD for each mass. c) Determine the magnitude of acceleration of the masses. d) Determine the force of tension in the string. ![[Unbalanced_System.png]] ### ☀️ ==Answer== a) $m_1$ is greater in mass than $m_2$, therefore, $m_1$ will accelerate downwards and $m_2$ will accelerate upwards. b) ![[Free_Body_Diagram_two_masses.png]] ==A good practice is to base the positive direction for each mass on the direction of acceleration.== c) Write down the equations for both masses. Note that both have $\vec{F}_T$. $m_1$ $\vec{F}_{net} = \vec{F}_{g1} +\vec{F}_T$ $m_1a = m_{1}g -F_T$ $m_2$ $\vec{F}_{net} = \vec{F}_T + \vec{F}_{g2}$ $m_2a = F_T - m_{2}g$ $m_2a + m_{2}g = F_T$ $m_1a = m_{1}g -(m_2a + m_2g)$ $m_1a + m_2a = m_{1}g -m_2g)$ $a(m_1 + m_2) = g(m_{1} -m_2)$ $a = \frac{g(m_{1} -m_2)}{(m_1 + m_2)}$ $a = \frac{9.8 \frac{m}{s^2}(4.7 kg -1.8 kg)}{(4.7 kg + 1.8 kg)}$ $a = 4.372 \frac{m}{s^2}$ d) Start with the $m_2$ statement $m_2a + m_{2}g = F_T$ $1.8 kg(4.372 \frac{m}{s^2} + 9.8 \frac{m}{s^2}) = F_T$ ==The masses accelerate with magnitude 4.4== ==$\frac{m}{s^2}$== ==and the force of tension in the string is 26 N.== $25.51 N = F_T$