Question

m,= 3 kg and m;= 15 kg are connected by strings over pulleys to m;= 2 kg in Figure. There isn't friction force on surface. The direction of acceleration a is given in Figure. (a) Draw the free-body diagram.

(b) Find the acceleration

(c) Find the tensions in the strings.(g=9.8 m/s2 sin37=0.6 cos 37=0.8)

Answer 1

(a)

$(b) \; T_1 – m_1g = m_1a \\ T – T_1 = m_2a \\ m_3gsinθ – m_1g = (m_1+m_2+m_3)a \\ a = \frac{g(m_3sinθ – m_1)}{m_1+m_2+m_3} \\ a = \frac{9.8(15sin37º – 3)}{15+2+3} = 0.3 \;m/s^2 \\ a=0.3\;m/s^2 \\ (c) \; T_1 = m_1(g+a) \\ T_1 = 3(9.8 +0.3) = 30.3\;N \\ T = m_3(gsinθ- a) \\ T = 15(9.8 \times 0.6-0.3) = 83.79 \;N \\ T = 83.79\;N$