Question

Refer to Fig. 4 The inclined plane is fixed to the ground and its inclined surface is smooth. Determine the angular accelerat
m=0.5kg, r=0.2m Fig. 4 m=3kg m=1.5kg ground 30°
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Answer #1

n=0.5kg, 8= 0.2m T2 a T m= 15kg al Im=3rg 11.5gsino 30 39 Step=1 39 39- Ti 73 - 1.57 Simo da ---12) Id Tix0.2 -T2 x0.2 T2 Id134 1.59 Simo)- 17,- 12) = 4.50 -3 From Equation I 2 = 4.5x0.22 (39 - 15g sino) 0.2 g (3- 1.5 Sim 3o) = (0.9+ I 1 / 2 Jd I =

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