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CH12.Problems

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) (3 L/4) M1 = 3(M2+M3) = 3* 4 = 12kg Only Torques, no forces!. -Ta + Tc cos30 = 0 Tc Sin30 -240 = 0 Only forces no torques. Project!

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CH12.Problems

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  1. CH12.Problems JH.131

  2. 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!

  3. -Ta + Tc cos30 = 0 Tc Sin30 -240 = 0 Only forces no torques

  4. 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

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

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