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

  t

 

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