Chapter 5B Rotational Equilibrium

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T30 0Example 4(Cont.)F yF x5 m200 N200 N 800 NT30 03 m 2 m800 NChoose axis of rotation at wall (least information)(ccw(ccw):(cw(cw):rTr = T (8 m)sin 30 0 = (4 m)T): (200 N)(5 m) + (800 N)(10 m) = 9000 Nm(4 m)T = 9000NmT = 2250 N
T30 0Example 4(Cont.)F yTF x x5 mT30 0 T y30 03 m 2 m200 N800 N200 N800 NF(up) = F(down):T y + F y = 200 N + 800 NF y = 200 N + 800 N - T y ;F y = 1000 N - (2250 N)sin 30 0F y = 1000 N - T sin 30 0F y = -125 NF(right) = F(left):F x = T y = (2250 N) cos 30 0F x = 1950 N or F = 1954 N, 356.3 0
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Chapter 5B Rotational Equilibrium

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