Uniform plank problem: Center of gravity, equilibrium, and torque

[pintsize131]

Homework Statement

A uniform plank of length 4.7 m and weight 207 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 454 N walk on the overhanging part of the plank before it just begins to tip?
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Homework Equations

t_net=0
F_net=0
t=(radius)(F sin \theta)

I'm not exactly sure if there are any other equations, or even if these are the right equations- I'm just so confused. I assumed that "radius" in the torque equation was equal to "x" in the drawing, and that the force was the weight of the man.

The Attempt at a Solution

t₁=(r)(F sin \theta₁)
r=t₁/454 sin 90

t₂=(r)(F sin \theta₂)
r=t₂/454 sin 91
(I did this trying to figure out the torque that would cause the board to start to tip.)
Set the equations equal to one another:

t₁/ 454 (sin 90) = t₂/ 454 (sin 91)
t₁/ 454 = t₂/ 454 (sin 91)
454*(sin 91)*t₁ = 454*t₂

I hit a dead end here. I'm so lost. Please help.

 

[tiny-tim]

welcome to pf!

hi pintsize131! happy new year, and welcome to pf!

why are you using 91°?

hint: the plank will tip if the net torque about the right-hand support is clockwise

 

[pintsize131]

Happy new year to you too!

I honestly have no idea what I am doing. My teacher assigned this homework over break, and told us to teach ourselves, so I need a crash course in torque.

I have no idea what that means. Literally, I'm clueless.

 

[tiny-tim]

ok, torque simply means what you already know … Fdsinθ

(except in this case sinθ is so obviously zero that you needn't even mention it, just use Fd)

so just add the Fd for all the forces (measuring d from the pivot point, in this case the right support) …

what do you get? ​

 

[pintsize131]

I get stuck almost immediately.

The torque for the weight of the board:
t=Fd
t=(Weight force?) (1.1 m)
I don't know how to find the weight for just that section of the board.
Am I even supposed to find that?

 

[tiny-tim]

I don't know how to find the weight for just that section of the board.

you're allowed to assume that the weight of the whole board is at its centre of mass

 

[pintsize131]

thanks! That helped! I got the question right. :)