# Balance Problem

## Homework Statement

A plank is supported from two points. How far can a gril that weights 48 kg walk on the plank before the plank starts to rotate?. Middle of plank is its gravity point.

Picture: I have added a picture that might tell a bit more(not so good at English :/)

https://www.physicsforums.com/attachment.php?attachmentid=24296&stc=1&d=1268318353

m1= mass of the plank = 80kg
m2= mass of gril = 48kg
L plank = 5m
Support point A= 2.5m from center of plank mass/gravity
Support point B= 0.3m from center of plank mass/gravity

Summ F=0
Summ M=0
G=mg
M=Fr

## The Attempt at a Solution

I have tryed a few equitions but have not had any luck.

Na= support point A
Nb= support point B
Gp= Plank gravity center
Gg= Girl Gravity
Na+Nb-Gp-Gg=0

Thx for Help

#### Attachments

• Picture Balance.JPG
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tiny-tim
Homework Helper
Welcome to PF!

Hi procol! Welcome to PF! (btw, she's a"girl" … but i like her hair! )

Hint: when the plank starts to turn, Na will be zero.

So take moments about B. Hi procol! Welcome to PF! (btw, she's a"girl" … but i like her hair! )

Hint: when the plank starts to turn, Na will be zero.

So take moments about B. Ok thx alot I will try now :D

I counted the force which point B applys on the plank if the gril is not standing on it and i got to 700N.
I used that in this equition: Nb x Nr = Gg x r -------> rg = Nb x Nr / Gg

(700N x 0.3m) / (9.81m/s^2 x 48kg) = 0.44597m....

Is this Right :D?

tiny-tim
Homework Helper
Hi procol! I counted the force which point B applys on the plank if the gril is not standing on it and i got to 700N.
I used that in this equition: Nb x Nr = Gg x r -------> rg = Nb x Nr / Gg

(700N x 0.3m) / (9.81m/s^2 x 48kg) = 0.44597m....

Is this Right :D?

Yes, except for the 700N … I don't understand where you got that from. Hi procol! Yes, except for the 700N … I don't understand where you got that from. The 700N I got from counting what support forces the two points A and B have to support the plank without it rotateing or moving when the gril is not standing on the plank.

F=0

Na + Nb - Gp = 0

M=0

-Na x ra + Nb x rb = 0

Na x ra = Nb x rb

Nax 2.5m = Nb x 0.3m

2.5Na - 0.3Nb = 0

Making equition par of both equitions with force and force momment:

2.5Na - 0.3Nb = 0
Na + Nb - Gp = 0

2.5Na - 0.3Nb = 0
Na + Nb - Gp = 0 x 0.3

2.5Na - 0.3Nb = 0
0.3Na + 0.3Nb - 0.3Gp = 0

2.8Na - 0.3Gp = 0

2.8Na = 0.3Gp

Na = (3/28)Gp

G = 80kg x 9.81m/s^2 = 784.8N

Na = 784.8N x (3/28) = 84.085N

Gp - Na = Nb

784.8N - 84.085N = 700,714N = 700N there I got it.

tiny-tim
Homework Helper
The 700N I got from counting what support forces the two points A and B have to support the plank without it rotateing or moving when the gril is not standing on the plank.

No, the forces when the girl isn't there are irrelevant.

Just use the forces on your diagram. (and remember, when the plank starts to turn, Na will be zero)

No, the forces when the girl isn't there are irrelevant.

Just use the forces on your diagram. (and remember, when the plank starts to turn, Na will be zero)

ok :D ty