- #1
QaH
- 6
- 0
Homework Statement
A uniform plank 6.1 m long rests on two supports, 2.5 m apart (see figure below). The gravitational force on the plank is 130 N. The left end of the plank is 1.5 m to the left of the left support, so the plank is not centered on the supports. A person is standing on the plank one tenth of a meter to the right of the right support. The gravitational force on this person is 814 N. How far to the right can the person walk before the plank begins to tip?
Homework Equations
The Attempt at a Solution
The center of mass, since the plank is uniform, is 6.1m/2=3.05m
This means that the center of mass is 4m-3.05m=0.95m from the right support.
Our person is standing 0.1 meters to the right of the right support.
0.95m(130N)=(0.1m+x)814N
solving for x I got
[0.95m(130N)/814N]-0.1m=x
Last edited: