Question

Consider two masses: Mi=2.0 kg and M2 =1.5 kg. Mass Mı moves on a horizontal surface where the coefficient of kinetic friction 4k = 0.40. Mass M2 is hanging freely. Two masses are connected by a strong cord of negligible mass that extends over a pulley. M = 2.0 kg 2 MK = 0.40 My = 1.5 kg a) Draw Free Body Diagrams for the two objects b) Write the equations for the two masses in the direction of motion (both x and y direction ) c) Find acceleration of the system. d) Find the tension in the string connecting the two blocks.

Answer 1

T ta T Given, My = 20 kg ng mig M2 ja M = 1.5 Мк = 0.40 Mzg where, tx friction force N Normal force (b) For the mass M motion along direction: Ef = ma - » T - fx = M, a For the mass M, , o motion along - direction; {fy » N- Mig=0 Mig N = Mi lo For the Mass M2 of motion equation along x-anis Fx - and along t direction equation of motion = M₂ a * Mz g - T = M2 a { Fy

from equation is and 1, we get T = M, at tk = M, at man = M, a t Mi from equation (111) (iv) M, Mxg we yer T = M2g - Mza . (۷) from equation (iv) and (w), we get M, at Me Mx g = M2 g - M₂ A (M + My) a M2 g - Mx Mig M₂ g - Mx Mig Mit", (15 - 0.4x2) x 9.8 (5+2) m/s 1.96 The acceleration of the system is 26 m 1.96 mpya in the string connecting (a) The tension on the two blocks is. T = M₂ g - My a M,13-) = 1.5/98-1999 = 11.76 w
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