The system as represented in this figure is in equilibrium around the axis of ro

Question

The system as represented in this figure is in equilibrium around the axis of rotation A (pivot). The force F is applied at the point C and makes an angle of 30 with the horizontal. In this problem we neglect the mass of the rod AD. Choose the CCW direction as positive. Don't forget to add the weight w1 and w2 of the objects of mass 10kg and 20 kg respectively. Using the concepts of the net Force = 0 and the Net torque = 0 to a) Determine the value of the Force F. b) Determine the value of AC AD-1.5m, AC = ? , AB .80 m, F-7 g=9.80 m/s2 (p: AC is not the lever arm of the force F) A B 30 F D 10kg 20kg

Explanation / Answer

By Fnet = 0 in vertical,

F sin 30 degree = w1+w2

F = [10*9.8+20*9.8]/sin 30 degree = 588 N answer

Now torque = 0

588 sin 30 degree* AC = w1*AB + w2*AD

294 AC = 98*0.8 +196*1.5

AC = [98*0.8 +196*1.5]/294 =  1.267 m answer

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