Balance a 12m Board with 500kg and 80kg Man

In summary, the weight of the board is 4900 N and the weight of the man is 784 N. The weight of the board acts 0.5 m from the fulcrum. By setting the torque created by the weight of the board equal to the torque created by the weight of the man, it can be determined that the man can stand 3.1 m from the fulcrum on the side over the cliff before the board tips.
  • #1
linuxux
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Homework Statement



a uniform 12.0m long board has mass 500kg. it rest 5.5m over a cliff. How far can a 80kg man walk on the board before it tips?

Homework Equations



equilibrium and torque

The Attempt at a Solution



im not quite sure about any portion of my work, because i resorted to canceling units to produce the one that made sense.

[tex]W_{board}=500kg\cdot9.8=4.9\cdot10^{3}N[/tex]
[tex]W_{man}=80kg\cdot9.8=784N[/tex]
the fulcrum is at at position: 12.0m-5.5m = 6.5m

torques must be equal so,

[tex]\tau_{clockwise}=\tau_{counterclockwise}[/tex]
[tex]\left(L_{1}\right)\left(784N\right)=\left(4.9\cdot10^{3}N\right)\left(.5m\right)[/tex]
[tex]\left(L_{1}\right)=\frac{\left(4.9\cdot10^{3}N\right)\left(.5m\right)}{784N}=3.1m[/tex]so, the man can stand 3.1m from the fulcrum on the side over the cliff?
 
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  • #2
actually i just realized that if any addition weight cases it to tip, that means the fulcrum is in the center, which it is not.
 
  • #3
(1) Recalculate the weight of the board--you are off by a decimal.
(2) Where does the weight of the board act? How far is that from the fulcrum?

In calculating torques, always measure distances from the fulcrum.
 
  • #4
okay, i recalculated the weight of the board, but i got the same answer, i did discover the weight of the man was wrong however. so the weight of the board acts at a point .5m from the fulcrum, so the ground has an equal force in the opposite direction, but its at that same point, so the first torque equation, i used should equal the weight of the board for F1. hows is it now?
 
  • #5
linuxux said:
okay, i recalculated the weight of the board, but i got the same answer, i did discover the weight of the man was wrong however.
You have the weight of the board equal to 500*9.8 = 490; that's not right. (Exponent problem.) The weight of the man is fine.

so the weight of the board acts at a point .5m from the fulcrum, so the ground has an equal force in the opposite direction, but its at that same point, so the first torque equation, i used should equal the weight of the board for F1. hows is it now?
Yes, the weight of the board acts 0.5m from the fulcrum. Set the torque created by the weight of the board equal to the torque created by the weight of the man.
 
  • #7
linuxux said:
okay, it seems much smaller now, but I am really not sure since in each case it seems right, but is it right now?
Your torque equation doesn't make sense:

linuxux said:
torques must be equal so,

[tex]\tau_{clockwise}=\tau_{counterclockwise}[/tex]
[tex]\left(F_1\right)\left(0.5m\right)=\left(4.9\cdot10^{3}N\right)\left(6.0m\right)[/tex] ---> the only reason i did this is because the "m" units would cancel
You know both forces, so why are you treating F_1 as an unknown? The unknown is the distance from the fulcrum to the man. And why are you multiplying the weight of the board by 6.0 m?? You just told me that that force acts 0.5 m from the fulcrum.

[tex]F_1 D_1 = F_2 D_2[/tex]
 
  • #8
okay, the number i got now is much more reasonable, but is it right?
 
  • #9
linuxux said:
torques must be equal so,

[tex]\tau_{clockwise}=\tau_{counterclockwise}[/tex]
[tex]\left(L_{1}\right)\left(784N\right)=\left(4.9\cdot10^{3}N\right)\left(.5m\right)[/tex]
[tex]\left(L_{1}\right)=\frac{\left(4.9\cdot10^{3}N\right)\left(.5m\right)}{784N}=3.1m[/tex]


so, the man can stand 3.1m from the fulcrum on the side over the cliff?
Looks good to me.
 

1. How can a 12m board be balanced with 500kg and 80kg man?

The key to balancing a 12m board with a 500kg and 80kg man lies in understanding the principles of center of mass and weight distribution. By placing the board on a fulcrum and adjusting the position of the man and weights, it is possible to achieve a state of equilibrium where the board remains balanced.

2. What is the significance of the 12m length in balancing the board?

The length of the board plays a crucial role in balancing it with the man and weights. A longer board provides a larger area for weight distribution, making it easier to achieve balance. Additionally, the longer length also allows for a greater distance between the fulcrum and the weights, which can aid in balancing the board.

3. Can any type of board be used for this experiment?

While it is possible to balance a 12m board with a variety of materials, it is important to consider the strength and stability of the board. The board must be able to support the weight of the man and the weights without breaking or bending. It is recommended to use a sturdy and durable material such as wood or metal for the board.

4. What is the role of the man's weight in balancing the board?

The man's weight acts as a counterbalance to the weights on the board. By adjusting the position of the man, it is possible to change the center of mass and achieve balance. The man's weight should be distributed evenly on the board and should be positioned in a way that creates an equal amount of tension on either side of the fulcrum.

5. Is it possible to balance the board with different weight combinations?

Yes, it is possible to balance the board with different weight combinations. The key is to ensure that the total weight on each side of the fulcrum is equal. This can be achieved by adjusting the position of the man and the weights on the board. However, it is important to note that the man's weight should always be greater than the weights on the board in order to maintain stability.

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