a roller whose diameter is 20 in weight 72 lbs. what horizontal force is necessary to pull the roller over a brick 2 in high when (a) the force is applied to the center, (b) at the top

Answer 1

Interesting question.

the vertical height from the edge of the brick to the center is 18 inches. The horizontal distance from the edge of the brick to the center of gravity requires some calculation.

draw the figure. from the brick edge to the center is 20 inches. sketch a horizontal line frm the edge of the brick to the vertical downward radius. The You now have a triangle, 20 hypotenuse, 18 on the downward leg, and the horizontal leg will be d=sqrt(20^2-18^2)=sqrt(400-324)=sqrt 76 check that.

Now moments around the edge sum to zero.

F*18-weight*sqrt76=0
forceat center=72lbs*sqrt76/18

at the top, the process is similar.
Force*38-72sqrt76=0 solve for Force

Answer 2

the vertical height from the edge of the brick to the center is 8 inches. The horizontal distance from the edge of the brick to the center of gravity requires some calculation.

By illustration, from the brick corner to the center is 10 inches(since the given is diameter, divide it by 2 to get the radius which is equal to the distance from the corner of the brick to center of the roller). sketch a horizontal line from the corner of the brick to the vertical downward radius. Then now you have a triangle, 10 hypotenuse, 8 on the downward leg, and the horizontal leg will be d=sqrt(10^2-8^2)=sqrt(100-16)=6

Now moments around the edge sum to zero.

F*(10+8)-weight(6)=0
forceat center=72lbs

Force*(18)-72(6)=0 solve for Force

Answer is 24 lbs.