Question

Two masses are arranged as shown in the picture below with m1 = 5.98 kg and m2 = 2.29 kg. a) What minimum coefficient of static friction would be necessary to keep the system stationary? b) When the blocks are released, m2 is observed to accelerate downward with an acceleration of 1.73 m/s2. What is the tension in the rope? c) What is the coefficient of kinetic friction between m1 and the surface? d)

Frictional m2

Answer 1

m_1 = 5, 98 kg m_2 = 2, 29 kg m_1 { Sigma fx = m_1 a = T - fr Sigma fy = 0 = N_1 - P_1 { Sigma fy = m_2 a = P_2 - T the Sertem Stationary implies Sigma fx, Sigma fy = 0 implies { T = fr N_1 = P_1 {P_2 = T { T = fr = T = P_2 = m_2 g = 22, 442 Nw implies fy = me e N_1 = T = P_2 = 22, 442 Nw implies mu e = 22, 442/N_1 = 22, 42Nw/58, 604 N mu_e = 0, 38 minimum coefficient of static friction (2) (b) m_2 Has a_m2 = 1, 73 m/s^2 Take the sames diagram but f_r = mu_c, N because there are movement Sigma fx = m_1 a = T - fr sigma fy = m_2 a = P_2 - T sigma fy = (1) = N_1 - P_1 {m_1 a = T - fr m_2 a = P_2 - TN_1 = P_1 = m_1 g, am_2 = am_1 = a = 1, 73 m/s^2 the system have the same acceleration, both masses, have the sames velocity (speed) implies a(m_1 + m_2) = P_2 - fr implies fr = P_2 - a(m_1 + m_2) fr = m_2 g - a(m_1 + m_2) = [2, 29 times 9, 8 - 1, 73 (2, 29 + 5, 98)]N fr = (22, 442 - 14, 307) N = 8, 135Nw