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CH12.Problems JH.131. Take the lower center as rotation center then: M2 (L/4) = M3 (3L/4)  M2= 3 M3 Now the upper center: M1 (L/4) = (M2+M3)

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Presentation on theme: "CH12.Problems JH.131. Take the lower center as rotation center then: M2 (L/4) = M3 (3L/4)  M2= 3 M3 Now the upper center: M1 (L/4) = (M2+M3)"— Presentation transcript:

1 CH12.Problems JH.131

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7 Take the lower center as rotation center then: M2 (L/4) = M3 (3L/4)  M2= 3 M3 Now the upper center: M1 (L/4) = (M2+M3) (3 L/4)  M1 = 3(M2+M3) = 3* 4 = 12kg  Only Torques, no forces!

8 -Ta + Tc cos30 = 0 Tc Sin = 0 Only forces no torques

9 Project! Mg (L/2 sin alpha) is cw torque 2Mg (L) Sin(90-alpha) is CCW torque Then 2 Cos(alpha) = Sin(alpha) /2 Or Tan (alpha) = 4  alpha = 76 deg

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13 Any triangle =180; so, beam makes 60 with vertical

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