Question

Two masses M1=2kg and M2 are attached by a massless cord over a solid pulley wheel of mass M=4kg, and radius R=5cm. Static Friction between the cord and the pulley makes the pulley rotate counter-clockwise when the system is released from rest, M1 accelerates with a magnitude of 3.92 m/s2.

a) Draw and label the forces acting on the two blocks, and the pulley. (6 points)

b) Find the tension in the cord between the pulley and M1 (6 points)

c) Find the tension between the pulley and M2 (6 points)

d) Find M2 (6 points)

e) if someone uses their finger to impose a frictional torque which reduces the acceleration to 0 m/s2, rewrite the torque equation, and use it to evaluate the coefficient of kinetic friction, between the finger and the wheel, if the radial force exerted by the finger is 56N (6 points)

Answer 1

(a) The free body diagram, of a pulley and two blocks system is represented as follows: ма M a AT. Mg M& M Figure: Free body diagram of two blocks And a pulley. Here, M, is the mass of left block, M, is the mass of right block, T, is the tension in the left string, T, is the tension in the right string, r is the radius of the pulley, and a is the acceleration. (b) The expression in the right string is represented as follows: T1 - Mg = Ma T = M(a+g) = (2 kg)(9.8 m/s? +3.92 m/s²) = 27.44 N Therefore, the tension between the pulley and mass M is 27.44 N. The torque in the pulley is represented as follows: T=la Here, I is the moment of inertia and a is the angular acceleration.

The inertia is represented as follows: I = 2M2 The angular acceleration is represented as follows: Therefore, t=la = }(M)(a)(v) = }(4 ks)(392 m(3)s em [10 m] Therefore, = (4 kB)(3.92 m/s) s cm 10 come) = 0.392 N.m The torque is represented as follows: T=r(T2-T) (10 m ) (0.392 N·m) = 5 cm - |(T2-T;) [ 1 cm =(0.303)+(27.44 N) (0.05) = 35.28 N

Therefore, the tension in the pulley and M, is 35.28 N. (d) The expression on the left string is represented as follows: M,g-T, = M.a T, MT-a (35.28 N) (9.8 m/s? –3.92 m/s?) = 6 kg Therefore, the mass M, is 6 kg The expression for the frictional force due to the finger is represented as follows: f = uN Here, N is the normal force and u is the coefficient of friction. The moment of force expression for the given situation is represented as follows: r=f(r) I=UNR UN(R) (0.392 N·m) (56 N)(0.05 m) Therefore, (0.392 N·m) (56 N)(0.05 m) = 0.14 Therefore, the coefficient of friction is 0.14 N.