CH12.Problems JH.131. Take the lower center as rotation center then: M2 (L/4) = M3 (3L/4) M2= 3 M3...
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CH12.Problems JH.131
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Transcript of CH12.Problems JH.131. Take the lower center as rotation center then: M2 (L/4) = M3 (3L/4) M2= 3 M3...
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!
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
Any triangle =180; so, beam makes 60 with vertical