If the right support is 1.1m from the end, then 3.9m, or 78% hangs over the left. 78% of 225N is exerted downward from the 'center' of the left side of the board. That 'center' works out to be half of 3.9m, or 1.95m.
On the right side, you have 22% of the board. 22% of 225N acts downward 0.55m away (half of 1.1m) from the support. Then you have a person of 451N on the right side, as well, an unknown away from the support. You need to equal out all of the forces times distances.
(0.78 • 225) •1.95 = ((0.22 • 225) •0.55) + (451•X )
Solve for X.
Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 451 N walk on the overhanging part of the plank before it just begins to tip?
1 answer
1 answer