A 4-m long plank with a weight of 80 N is placed on a dock with 1 m of its lengt
ID: 1696901 • Letter: A
Question
A 4-m long plank with a weight of 80 N is placed on a dock with 1 m of its length extended over the water, and 3 m on the dock. The plank is uniform in density so that the center of gravity of the plank is located at the center of the plank. A boy with a weight of 150 N is standing on the plank and moving out slowly from the edge of the doctor onto the 1 m portion of the plank.a) What is the torque exerted by the weight of the plank about the pivot point at the edge of the dock? (Treat all the weight as acting through the center of gravity.)
b) How far from the edge of the dock can the boy move until the plank is just on the verge of tipping?
c) How can the boy test this conclusion without falling in the water? Explain.
Explanation / Answer
a). Torque exerted by the weight of the plank about the edge. T=80*1=80(Nm) b). Let x be distance from boy to the edge of the dock. T'=x*150. total torque=T-T'=80-x*150=0 so x=8/15=0.533(m). c) He should not go too fast, just slowly to that point.