In the inclined plane arrangement below, the mass m? slides down the incline wit

Question

In the inclined plane arrangement below, the mass m? slides down the incline with an acceleration a as shown. The masses are m?-10.0 kg, mz-1.0 kg and 0-30°. The radius and mass of the pulley are 0.11 m and 1.1 kg, respectively. If the pulley has a constant frictional torque of 0.52 N.m when the system is in motion, and the kinetic friction coefficient between mi and the incline is 0.27 find the magnitude of the acceleration of the system (in m/sA2). 2. a. 7.3 b. 3.6 c. 2.1 d. 1.0 e.r .52 T1 T2 m2 no

Explanation / Answer

le the acceleration be a m/s ^2

let tension in left part of pulley be T1 and T2 in right part

for incline part

m1*g*sin(30)-T1-0.27*m1*gcos(30)=m1*a.............(1)

for m2

T2-m2*g=m2*a...........(2)

for pulley

(T1-T2)*0.11=0.52=>T1-T2=4.73............(3)

adding 1,2,3 we got

a=1.05 m/s^2

so option (d) is correct

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