A uniform plank of lent light 5m and weight 225N Rests horizontally on two supports with 1.1m of the plank hanging over the right support. To what distance,X, can a person weighing 450 walk On the overhanging part of the plank before it just begins to tip?
6 answers
Sorry for the typos. It should say, "of length"
The sum of torques has to be zero. So, the sum respectively the reference point (the right support) is
225•1.4 = 450•x
x=225•1.4/450 = 0.7 m
225•1.4 = 450•x
x=225•1.4/450 = 0.7 m
Why does it have to be zero?
The two conditions of equilibrium are:
1. Concurrent Equilibrium : the sum of vector forces through a point is zero.
2. Coplanar equilibrium: the sum of forces in a plane is zero and the sum of the torques around the axis of the plane is zero.
Here we've used the 2nd rule
1. Concurrent Equilibrium : the sum of vector forces through a point is zero.
2. Coplanar equilibrium: the sum of forces in a plane is zero and the sum of the torques around the axis of the plane is zero.
Here we've used the 2nd rule
Should the 1.4 that you used be 1.1?
1.4 is the distance between the center of the plank (its center of mass) and right support.
1.4 = (5/2)-1.1
1.4 = (5/2)-1.1
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