Net torque and net force in equilibrium Help

[renaldocoetz]

Net torque and net force in equilibrium.. Help please

Homework Statement

The diagram shows a uniform plank of length 5.0m and weight 225N that rests horizontally on 2 supports, with 1,1 m of the plank hanging over the right support.

http://www.postyourimage.com/view_image.php?img_id=Oc9PYdcDMUhBrGB1196709689

To what distance x can a person who weighs 450N walk on the overhanging part of the plank before it just begins to tip?

Homework Equations

Net force on Y-axis = 0

Net torque on right support = 0

Net torque on left support = 0

The Attempt at a Solution

I used these 3 formulas to end up with 3 unknowns in order to work out x but ended up with 0x = 234 :(

Net torque right post = -F_Nl(3,9) + W_p(1,4) - W_mx = 0

Net torque left post = -W_p(2,5) + F_Nr(3,9) - W_m(3,9 + x)

F_Nr + F_Nl -225 -450 = 0

i used counter-clockwise rotation as positive

I combined these 3 formulas in an attempt to get x but it didnt work quite so well. Is this a wrong approach or might my mistake just be in my algebra?

Please help any1
thxPS
F_Nl = normal force of left support on plank upwards
F_Nr = normal force of right support on plank upwards
W_p = weight of plank
W_m = weight of man

 

[Doc Al]

All you need is one equation: Find torques about the right support. Hint: When the plank is just about to tip, what's the normal force from the left support?

 

[renaldocoetz]

hmm.. k so when the plank is just about to tip the normal force on left support is -225?

then torque = -(W of man)(x) -(-225)(3,9) ?

 

[Doc Al]

hmm.. k so when the plank is just about to tip the normal force on left support is -225?

No. Realize that when it starts to tilt, the plank will lose contact with the left support. So what must the normal force be at that point?

Also: Where does the weight of the plank act?

 

[renaldocoetz]

oh yeah so the normal force is 0 on left suport then? and the weight of the plank acts in the middle i believe

 

[Doc Al]

oh yeah so the normal force is 0 on left suport then? and the weight of the plank acts in the middle i believe

Exactly. So set up the equation for the net torque about the right support.

 

[renaldocoetz]

net torque = -(W of man)(x) + (W of plank)(1,4) ?Just wondering.. if the net force is zero on left suport, and the plank has already started to tip is the equation still right considering the question is how far can the person walk BEFORE the plank starts to tip? so obv he has already walkd to far?

 

[Doc Al]

net torque = -(W of man)(x) + (W of plank)(1,4) ?

Good.

Just wondering.. if the net force is zero on left suport, and the plank has already started to tip is the equation still right considering the question is how far can the person walk BEFORE the plank starts to tip? so obv he has already walkd to far?

We're finding the point where the normal force from the left support just becomes zero, so the man's torque still balances the torque due to the plank's weight. If he keeps walking, his torque will be greater than the torque due to the plank's weight and he'll surely fall.

 

[renaldocoetz]

that makes sense :) so can i set before-mentioned equation equal to 0? by doing this i get that the man can walk 0,36 m before it begins to tip