Rotational Equilibrium Ex) Mass Safe = 15,000 kg Mass Beam = 1500 kg Find FA (force of left leg on table) FB (force of right leg on table)
Using Left leg as pivot point, St = 0 Draw all the torques
St = 0 Torque CW = Torque CCW
10.0m(15,000N) + 15.0m(150,000N)= 20mFB
FB = 120,000 N SFy = 0
0 = -15,000N - 150,000N + 120,000N + FA
FA = 45,000 N
Giancoli - p. 247) 2, 4, 6
ŠTony Mangiacapre.,
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