Consider a rod of total length 4 m that is free to pivot above its center. The linear mass density of the rod is given by λ(x) = 6 x4 (kg/m), where x is the distance from the center of the rod. The rod is in outer space, so you don't have to worry about any gravitational torques. There is a 168 N force that acts perpendicularly to the rod at its right end, and there is a 512 N force that acts halfway between the left end of the rod and its center. This force acts at an angle of 33 degrees to the vertical. This scenario is shown below:
Consider a rod of total length 4 m that is free to pivot above its center. The linear mass density of the rod is given by λ(x) = 6 x4 (kg/m), where x is the distance from the center of the rod. The rod is in outer space, so you don't have to worry about any gravitational torques. There is a 168 N force that acts perpendicularly to the rod at its right end, and there is a 512 N force that acts halfway between the left end of the rod and its center. This force acts at an angle of 33 degrees to the vertical. This scenario is shown below:
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Consider a rod of total length 4 m that is free to pivot above its center. The linear mass density of the rod is given by λ(x) = 6 x4 (kg/m), where x is the distance from the center of the rod. The rod is in outer space, so you don't have to worry about any gravitational torques. There is a 168 N force that acts perpendicularly to the rod at its right end, and there is a 512 N force that acts halfway between the left end of the rod and its center. This force acts at an angle of 33 degrees to the vertical. This scenario is shown below:
Calculate the
(Please answer to the fourth decimal place - i.e 14.3225)
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