Answer to Question #81913 in Physics for cyril earl jay remedio

Question

The center of gravity of a non-uniform bamboo pole of length three meter is 1 frpm its lerger end and its weight is 50 newtons. If it is to be supportes at its midpoint, what vertical force must be applied at smaller end of the pole to keep it a horizontal position.

Answer 1

Taking moments about the midpoint:

F(3/2)=W(3/2-1)

The vertical force must be applied at smaller end of the pole to keep it a horizontal position:

F=W(3/2-1)/((3/2) )=W/3=50/3=16.7 N.