# Plank problem/Center of Gravity/Torque/Level Arm(p.26)

1. Oct 31, 2014

### gcombina

1. The problem statement, all variables and given/known data
A horizontal, 10-m plank weighs 100 N. It rests on two supports that are placed 1.0 m from each end as shown in the figure. How close to one end can an 800-N person stand without causing the plank to tip?
(a) 0 m (d) 0.7 m

(b) 0.3 m (e) 0.9 m

(c) 0.5 m

2. Relevant equations
T = Force x level arm

3. The attempt at a solution

T = F x L

mg x (center of gravity of the plank) = mg x level arm
[ (90N)(9.8) ] ( 4.5 ) = [ (800N)(9.8) ] ( l )

level arm = .5

My question is why do "center of mass" needs to be used?

I originally did 90N but didn't work so I had to do 4.5 for the center of mass

Thanks!

#### Attached Files:

• ###### q26.png
File size:
3.5 KB
Views:
448
2. Oct 31, 2014

### haruspex

You've forgotten the rest of the plank.
As opposed to using what? Or do you mean why is the centre of mass at 4.5?
I've no idea what you mean. How exactly did you "do" 90N?
If there's some approach you took that didn't work, please post all the working, and explain how you know it did not work.

3. Oct 31, 2014

### gcombina

because i did the torque from the left side which is 9.0 meters

I am doing the torque on the left = torque on the right

the torque on the left starts from the second fulcrum to the left

see below

I did mg x length
[ (90N)(9.8) ] ( 9.0 m ) = FL

And this way did NOT work

I had to change it to

[ (90N)(9.8) ] ( 4.5 m ) = Fl

4. Oct 31, 2014

### BvU

Hello and welcome back! non-Shouting contest misunderstanding ironed out, I hope :)

I completely agree with the anti-clockwise torque (torque on the left).
The 4.5 m is the average distance of the wood to the left of the pivot point to that pivont point
Now the right-hand side.

And if you want do do some checking of the way you obtain your answer: have the support in the center and check that the person has to be 5m from the right end of the plank.

5. Oct 31, 2014

### gcombina

:)

6. Oct 31, 2014

### haruspex

Right, so you want to know why the second is correct?
If all of the mass of the 9m of plank were at 9m from the fulcrum then of course the torque would be 90*9*9.8. But it is spread evenly from zero distance to 9m. If you think of the plank as made of mass elements $\rho dx$ at distance x, the torque of an element is $\rho x dx$. On average, an element is at distance 4.5, so treat it as though the whole mass is at that distance.

Now, as BvU said, let's move on the the other side of the equation. You had that wrong in the OP, as I mentioned in my first comment of post #2.