1) A 200 g rod hangs from a pin located at 20 cm from the left edgeof the rod. T

Question

1) A 200 g rod hangs from a pin located at 20 cm from the left edgeof the rod. There is 300g mass hanging from the rod at a distanceof 50 cm from the left edge of the rod. If the mass of the rod isnot neglected, what is the net torque about the point at which thepin is located?

2) It takes 2N to open the door if it is pushed at a distance of80cm from its axis of rotation, and if the force is applied in thedirection perpendicular to the plane of the door. What will be theforce required to open the same door if the door is now pushed at adistance of 50 cm axis of rotation and the force is applied at anangle of 30 degrees from the plane of the door?

Explanation / Answer

1)the mass concentrated at 20 cm is (500/70) * 20 = 142.8 g (m) the net torque about the point at which the pin islocated = F * r = mg * r g = 9.8 m/s^2 and r = 20 cm = 20 * 10^-2 m 2)the torque is constant therefore F * r * sin = constant or F1 * r1 * sin1 = F2 * r2 * sin2 or F2 = (F1 * r1 * sin1/r2 * sin2) F1 = 2 N,r1 = 80 cm,1 = 90o,r2 = 50 cm and2 = 30o

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