Static equilibrium question

In summary, a 6.00m plank with a mass of 30.0kg is placed horizontally across two horizontal bars of a scaffold, with 4.50m between the bars and 1.50m of the plank hanging over one side. A painter with a mass of 70.0kg walks on the overhanging part of the plank, causing it to tip. By considering the torques about the pivot point, it can be determined that the torque due to the painter's weight must equal the torque due to the plank's weight for equilibrium. The equation to solve this is 70g*d-30g*(4.5/2)=0, where "d" is the distance between the pivot point
  • #1
rlmurra2
19
0
A uniform plank of length 6.00m and mass 30.0 kg rests horizontally across two horizontal bars of a scaffold. the bars are 4.50 m apart, and 1.50m of the plank hangs over one side of the scaffold. How far can a painter of mass 70.0kg walk on the overhanging part of the plank before it tips?

This is really confusing me. I have the forces identified: mg going down, normal force up, and the weight of the man going down. I am guessing the equation to solve this would look something like n-mg-70kg(d)=0...I really don't know how to incorporate the lengths into it.
 
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  • #2
You need to consider torques about the pivot point. As the painter walks further, at some point the other end of the plank begins to rise off of its support. At that point the torque due to the painter's weight exactly balances the torque due to the plank's weight. That's the equation you want.
 
  • #3
I don't know how to set up the equation to solve it...and again, where do the lengths (the distance b/t the bars, 1.50m hanging off...) fit in?
 
  • #4
You need only worry about forces that create a torque about the pivot point. There are only two: the weight of the painter and the weight of the plank. Since the plank is in equilibrium, the torques must be equal and opposite.

To find the torque due to the weight of the painter: Multiply his weight by the distance he is from the pivot point. Since that distance is the unknown you asked to find, just call it "d".

To find the torque due to the weight of the plank: First realize that the weight of the plank acts at the center of the plank (since it's uniform). Find the distance from that center to the pivot point. Multiply the weight of the plank by that distance to get the torque.

Hint: Since weight = mg, don't actually calculate the weight. When you set up your equation, the "g" will cancel from both sides.
 
  • #5
Okay I'm getting closer...do I need to worry about any upward forces, such as the normal force? So far I tried (weight of painter)*d-(weight of plank)(2.25m)=0 but it didnt work, also tried (weight of plank)(3m) (since 3 is half of 6, the length of the plank), but still didnt work. I think I'm leaving something out.
 
  • #6
rlmurra2 said:
...do I need to worry about any upward forces, such as the normal force?
Does the normal force exert a torque? (No.)
So far I tried (weight of painter)*d-(weight of plank)(2.25m)=0
Where did you get the 2.25m from? What's the distance of the middle of the plank to the pivot point? (Draw yourself a picture showing the plank, the supports, and the center of mass of the plank.)
 
  • #7
Solution : If “a” is the pivot point, then at the moment of equilibrium the “reaction” force in “b” is zero. Then , Sum of Forces in Y = Na + Nb – 70g – 30g = 0
But, Nb = 0, => Na = 100g
Sum of torques in “b” = 0
70g * d + Na * 0 - 30g * (4.5/2) + Nb * 4.5 = 0
=> 70g * d + Na * 0 - 30g * (4.5/2) + 0 * 4.5 = 0
=> 70g * d - 30g * (4.5/2) = 0
=> d = [ 30g * (4.5/2) ] / 70g
 
  • #8
gvicencio said:
Solution : If “a” is the pivot point, then at the moment of equilibrium the “reaction” force in “b” is zero. Then , Sum of Forces in Y = Na + Nb – 70g – 30g = 0
But, Nb = 0, => Na = 100g
This correct. The normal force, at the moment when the plank just starts to tip, is totally at point a (the pivot point). But since we are taking torques about point a, the normal force there will exert no torque since it is zero distance from the pivot.


Sum of torques in “b” = 0
70g * d + Na * 0 - 30g * (4.5/2) + Nb * 4.5 = 0
=> 70g * d + Na * 0 - 30g * (4.5/2) + 0 * 4.5 = 0
=> 70g * d - 30g * (4.5/2) = 0
=> d = [ 30g * (4.5/2) ] / 70g
This is incorrect. You've made the same error as rlmurra2. See my last post for a hint.
 
  • #9
Sorry, a typo error! (can happen..) However, we need to differentiate form and substance
You detected the typo error (good!) but didn’t “see” the whole exposition (?) … think about ..
Donald Knuth pay $2.56 to the first person who finds and reports anything that remains typographically, or politically incorrect in his books….
In this case I will not pay since this typo error is too basic. …

Just change [ Sum of torques in “b” = 0 ]
with : [ Sum of torques in “a” = 0 ] ,which is the right equation and too obvious [*]

All the other exposition is perfect.

70g * d + Na * 0 - 30g * (4.5/2) + Nb * 4.5 = 0
=> 70g * d + Na * 0 - 30g * (4.5/2) + 0 * 4.5 = 0
=> 70g * d - 30g * (4.5/2) = 0
=> d = [ 30g * (4.5/2) ] / 70g


[*] This is like chess… Hope now you could ‘see’ better the whole game...

Quote :"Imagination is more important than knowledge" , Albert Einstein
 
  • #10
gvicencio said:
All the other exposition is perfect.

70g * d + Na * 0 - 30g * (4.5/2) + Nb * 4.5 = 0
=> 70g * d + Na * 0 - 30g * (4.5/2) + 0 * 4.5 = 0
=> 70g * d - 30g * (4.5/2) = 0
=> d = [ 30g * (4.5/2) ] / 70g
Still wrong for the same reason. :wink:
 
  • #11
“same reason” ( a typo error ?). Maybe you need more time to understand…
(Doesn’t means knowledge, intelligence…. some times lack of concentration…)...
So take your time to process with "more quality"
Math, Physics… are “Sciences” (with perfect rules and demonstrations) not just “subjective statements”.
The solution indicated is “the right solution”. Now, if you don’t understand there is nothing that I could do for you. (Maybe study more……)
note : This is a basic problem explained in any "introductory physics course" (Mechanic Static ) go to any basic book to check your concepts…..
Same problem is well-explained and documented in hundred of places…
 
  • #12
gvicencio said:
“same reason” ( a typo error ?). Maybe you need more time to understand…
You are making the same mistake that the original poster made. (No, I'm not talking about a typo!) See my comment in post #6.
(Doesn’t means knowledge, intelligence…. some times lack of concentration…)...
So take your time to process with "more quality"
Math, Physics… are “Sciences” (with perfect rules and demonstrations) not just “subjective statements”. The solution indicated is “the right solution”. Now, if you don’t understand there is nothing that I could do for you. (Maybe study more……)
note : This is a basic problem explained in any "introductory physics course" (Mechanic Static ) go to any basic book to check your concepts…..
Same problem is well-explained and documented in hundred of places…
:zzz:

Thanks for the lecture, but all you are doing is further embarrassing yourself. Your solution is still "the wrong solution".
 
  • #13
For rlmurra2

Please don't forget that the weight of the plank is acting at its mid point which is at 3m from each end and thus at 1.5 m from the pivot point. So the solution is simpelly
(weight of painter)*d-(weight of plank)(1.5m)=0 [1.5m of plank is hanging]

Sorry Mr.Doc Al:smile:
 
  • #14
That is correct, of course. But I wanted rlmurra2 to draw a diagram and discover that for himself. :wink:
 
  • #15
....../\....../\......
...[70]...[Na]...[30]...[Nb]......
...\/.......\/.........
...<---d------..A.......B.....
=============================================
......^......^......
<------1.5------><-------- 4.5 ---------><------1.5----->

<------------------------ 7.5 --------------------------->

=> d = [ 30g * (4.5/2) ] / 70g

My solution was for the above problem ( I made a mistake when I read the statement of the problem,
thinking that scaffold was used as a typical configuration symetrical...)
Now I realize that the right section of my draw don't exist (the 1.5 in the right in B
... (and also the numbers match)
So the diagram is :

....../\....../\
...[70]...[Na]...[30]......[Nb]
...\/.....\/......
...<---d------..A.......B
=================================
......^......^
<------1.5------><-------- 4.5 ---------->

<--------------------6.0 ---------------->

Hence :
70g * d + Na * 0 - 30g * ([6/2]-1.5) + Nb * 4.5 = 0

=> 70g * d + Na * 0 - 30g * ([6/2]-1.5) + 0 * 4.5 = 0
=> 70g * d - 30g * ([6/2]-1.5) = 0

=> d = [ 30g * ([6/2]-1.5) ] / 70g

General solution (in this particular case) :
d max = [(L/2)-C]*(m1/m2)

C : distance of point "A"

( "C" mts of the plank hangs over one side of the scaffold. )
 

1. What is static equilibrium?

Static equilibrium is a state in which an object is at rest and has no net force acting on it. This means that the object is not moving and is in a stable position.

2. How is static equilibrium different from dynamic equilibrium?

Static equilibrium refers to an object at rest, while dynamic equilibrium refers to an object that is moving at a constant velocity. In both cases, the net force on the object is zero, but in static equilibrium, the object is not moving at all.

3. What are the conditions for static equilibrium?

The conditions for an object to be in static equilibrium are that the net force acting on the object is zero and the net torque (rotational force) acting on the object is also zero. This means that the forces and torques acting on the object must be balanced.

4. Can an object be in static equilibrium if it is accelerating?

No, an object cannot be in static equilibrium if it is accelerating. In order for an object to be in static equilibrium, it must be at rest or have a constant velocity. If an object is accelerating, it is not in a state of balance and is therefore not in static equilibrium.

5. How is static equilibrium used in real-world applications?

Static equilibrium is used in many engineering and design applications, such as building structures and bridges. It is also important in physics and mechanics, as it helps us understand the forces and torques acting on an object and how they affect its stability. Static equilibrium is also used in sports, such as balancing on a balance beam or tightrope.

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