4. By means of a rope whose mass is negligible, two blocks are suspended over a

Question

4. By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be treated as a uniform solid cylindrical disk, (moment of inertia I = MR^2/2). The downward acceleration of the 44.0-kg block is observed to be exactly one-half the acceleration due to gravity. Noting that tension in the rope is not the same on each side of the pulley, find: the mass of the pulley. The tension exerted on the mass of 44 Kg. 5. A 3.0-kg ball and a 1.0-kg ball are placed at opposite ends of a 1 meter beam with mass M=0.5 Kg, so that the system in the equilibrium as shown. a. What is the value of a? b. What is the normal force applied on the support point?

Explanation / Answer

4a. Let T1 and T2 be tensions in the string on both sides. T1 being on 44 kg

From FBD on each bodies;

44g - T1 = 44(g/2) -----------(1)

T2 - 11g = 11(g/2) ------------(2)

(T1-T2) r = (1/2)(Mr^2) (a/r) -----------(3)

Adding 1,2 gives

33g - 55(g/2) = M/2

M = 11g

Tension T1 = 22g

5.A

Uniform beam mass =0.5kg/m

Torque equilibrium about support gives

3g(a) + 0.5a(a/2) = 1g(1-a)+(0.5)(1-a)(1-a)/2

Solving gives 1/4(2a-1) + 4ga = g taking g = 10

a = 10.25/40.5 = 0.253;

Normal force is sum of all forces down ward = 3+1 +0.5 = 4.5 kg force

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