![[Perpendicular_Accelerating_objects.png]] Examine the diagram of two masses connected by a strong, taut string over a frictionless pulley where $m_1$ = 3.2 kg, $m_2$ = 4.1 kg, $\mu_s$ = 0.35, and $\mu_k$ = 0.28. a) Show that the two masses will accelerate. b) Determine the acceleration of the two objects. c) Determine the force of tension in the string. ### ==Answer== Start by drawing the FBDs for both $m_1$ and $m_2$. Label the forces. ![[Screen_Shot_2020-11-05_at_7.52.55_AM.png]] a) If $F_{g1}$ > $F_f$ then the masses will accelerate. The maximum friction to overcome is caused by static friction. $F_f$ = $\mu_s \cdot F_N$ $F_f$ = 0.35 (40.18 N) = 14.063 N Since 14.063 N < 31.36 N then the masses will accelerate. --- b) Use $\mu_k$ to determine $F_f$ because the objects are accelerating. $F_f$ = 0.28 (40.18 N) = 11.2504 N $m_1$ $\Sigma \vec{F}$ = $\vec{F}_{g1}$ + $\vec{F}_T$ $m_1 \vec{a}$ = $\vec{F}_{g1}$ + $\vec{F}_T$ $m_1a = m_1g - F_T$ $m_1a = m_1g - (m_2a + F_f)$ $m_2$ $\Sigma \vec{F}$ = $\vec{F}_T + \vec{F}_f$ $m_2\vec{a}$ = $\vec{F}_T + \vec{F}_f$ $m_2a = F_T - F_f$ ==$m_2a + F_f = F_T$== $m_1a + m_2a = m_1g - F_f$ $(m_1+m_2)a = m_1g-F_f$ $a = \frac{m_g - F_f}{m_1 + m_2}$ $a = \frac{31.36 N - 11.2504 N}{3.2 kg + 4.1 kg}$ a = 2.8 $\frac{m}{s^2}$ --- c) $m_2a + F_f = F_T$ (4.1 kg)(2.755 $\frac{m}{s^2}$) + 11.2504 N = $F_T$ 22.54 N = $F_T$ 23 N = $F_T$