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

pintsize131
Messages
5
Reaction score
0

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?
https://mail.google.com/mail/?ui=2&ik=bff73eefc2&view=att&th=12d42f81bf015ff9&attid=0.1&disp=inline&realattid=f_gievtrao0&zw

Homework Equations


tnet=0
Fnet=0
t=(radius)(F sin [tex]\theta[/tex])

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 [tex]\theta[/tex]1)
r=t1/454 sin 90

t2=(r)(F sin [tex]\theta[/tex]2)
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.
 
Last edited:
on Phys.org
welcome to pf!

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

why are you using 91°? :confused:

hint: the plank will tip if the net torque about the right-hand support is clockwise :wink:
 
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.
 
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? :smile:
 
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?
 
Last edited:
pintsize131 said:
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 :wink:
 
thanks! That helped! I got the question right. :)
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
17
Views
4K