A uniform plank of length 5.0 m and weight 225 N rests horizontally on two suppo

Question

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 475 N walk on the overhanging part of the plank before it just begins to tip?

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 475 N walk on the overhanging part of the plank before it just begins to tip?

Explanation / Answer

Draw the plank with the two supports with it overhanging 1.1 m

Now you apply the weight of the beam at the center so draw a dot at the center of the beam and put an arrow down and label it 225 N. This is 2.5 m to the left. Now Knowing its a 1.1 m overhang and the beam is a total of 5.0 m , from the center of the beam to the edge of the right support is 1.4 m , and finally the overhang of 1.1 m

2.5 + 1.4 + 1.1 = 5.0m

Now from the top of the beam go right from the right support and draw an arrow going down somewhere in the middle, it really doesnt matter where. After you do this draw a straight line up the edge of the right side of the right support and the distance between your 450 N person and the right support is delta x which means change of x. If you dont understand that just put an x. This is the unknow distance.

I imagine you have been taught moments so I wont go to far indepth about how to do the next step, email me if you need more help.

If you start from the right side in the center of the beam and go over 1.1 m to the left and stop this is yourreference point. This is because this is where the beam is going to pivot on when too much weight is to the right.

Now summing moments in the diagram means you are looking for all the weights with there distance to your reference point.

Looking left from your reference point we have the weight of the beam pointing downward 1.4 m away so this is ( 225 N x 1.4 m ) Now if the arrow was a rocket and you tied a string to it from your reference point you would notice that when it took off it went counter clockwise around your reference point, this means your last calculation is a positive ( 225N x 1.4 m ) If it went clockwise it would be negative.

Thats it to the left no more weight, now we look right and see the persons weight shooting downward at an unknown x variable. ( 450 N x ( x ) ). Using the rocket trick again this flies clockwise meaning its negative.

Now thats it, only two forces. You want to set these equal to 0 to solve for x because this is when the two weights will be perfectly balanced.

So once again when you sum moments of your reference point you get

( 225 N x 1.4m ) - 450 N ( x ) = 0
315 Nm = 450 N ( x )
0.7 m = x

So at 0.7 m it is perfectly balanced with no tipping at x > 0.7 m is when it starts tipping.

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