How Far Can a Person Walk on an Overhanging Plank Before It Tips?

[wchvball13]

Homework Statement

A uniform plank of length 4.6 m and weight 210 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support. To what distance x can a person who weighs 448 N walk on the overhanging part of the plank before it just begins to tip?

Homework Equations

W1(x1)=W2(x2) = Xcm
W1+W2

The Attempt at a Solution

210(3.5) + 658(1.1)
210 + 448

Xcm=1.87m

Not even sure if this is right, and I'm stuck after this. The way I understand it, once the center of gravity passes the second support, the plank will start to tip. But I don't know how to figure out how to calculate how far the person can walk before the center of gravity gets to that point.

 

[PhanthomJay]

Homework Statement

A uniform plank of length 4.6 m and weight 210 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support. To what distance x can a person who weighs 448 N walk on the overhanging part of the plank before it just begins to tip?

Homework Equations

W1(x1)=W2(x2) = Xcm
W1+W2

The Attempt at a Solution

210(3.5) + 658(1.1)
210 + 448

Xcm=1.87m

Not even sure if this is right, and I'm stuck after this. The way I understand it, once the center of gravity passes the second support, the plank will start to tip. But I don't know how to figure out how to calculate how far the person can walk before the center of gravity gets to that point.

your moment calculations are way off. Hint: The plank will start to tip when the reaction at the left support goes to zero...does that help?

 

[wchvball13]

I don't understand...

W1(x1)=W2(x2) = Xcg
W1+W2

Is this the wrong equation?

 

[PhanthomJay]

I don't understand...

W1(x1)=W2(x2) = Xcg
W1+W2

Is this the wrong equation?

I don't know what you're trying to do with this equation which has some arbitrary numbers in it. When the left support reaction goes to zero, then the moment about the right support, from the cg of the planks weight, must balance the moment about the right support from the person's weight. Equate the moments and solve for the distance from the person to the right support. As he walks beyond that point, the plank tips because the left support cannot withstand an upward load (assuming its not bolted down, in which case she'd never tip).

 

[wchvball13]

ok I'm still completely lost...could you maybe baby step me through this?

 

[PhanthomJay]

ok I'm still completely lost...could you maybe baby step me through this?

The first step is to recognize that the planks weight of 210N acts at its cg of 2.3m from the left support..at its midpoint. So it acts at 1.2m to the left of the right support. So now equate moments about the right support, with the knowledge that the left support provides no support at the tipping point:
210(1.2) = 448(x). Solve for x, the distance to the right of the right support that the person must be for the beam to just start to tip. I get x= .56m. Aternatively, if you like your equation better and you understand where it's coming from, then with the cg of the system at the right support, you can sum moments about the left support and come up with
(210 +448)(3.50) = 210(2.3) + 448(3.5 + x), which yields the same result in a rather cumbersome manner. So why sum about the left support when it is much easier to sum about the right support
(210 + 448)(0) = 210(1.2) - 448(x) = 0; x = .56m.
Does this help or only serve to confuse?