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

In summary, a person who weighs 454 N can walk on the overhanging part of the plank before it just begins to tip if the net torque about the right-hand support is clockwise.
  • #1
pintsize131
5
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?
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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.
 
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  • #2
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:
 
  • #3
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.
 
  • #4
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:
 
  • #5
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?
 
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  • #6
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:
 
  • #7
thanks! That helped! I got the question right. :)
 
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FAQ: Uniform plank problem: Center of gravity, equilibrium, and torque

What is the Uniform Plank Problem?

The Uniform Plank Problem is a physics problem that involves finding the center of gravity, equilibrium, and torque of a uniform plank resting on a fulcrum or pivot point.

What is the Center of Gravity?

The center of gravity is the point at which the weight of an object can be considered to be concentrated. In the case of the uniform plank, it is the point where the weight of the plank is evenly distributed on both sides.

How do you find the Center of Gravity of a Uniform Plank?

The center of gravity of a uniform plank can be found by balancing the plank on a pivot point and marking the point where it is perfectly balanced. This point is the center of gravity.

What is Equilibrium in the Uniform Plank Problem?

Equilibrium refers to a state in which all forces acting on an object are balanced, resulting in no net force or movement. In the uniform plank problem, equilibrium is achieved when the plank is perfectly balanced on the pivot point.

What is Torque in the Uniform Plank Problem?

Torque is a measure of the rotational force acting on an object. In the uniform plank problem, torque is calculated by multiplying the weight of the plank by the distance between the center of gravity and the pivot point.

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