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

[linuxux]

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.

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

torques must be equal so,

\tau_{clockwise}=\tau_{counterclockwise}
\left(L_{1}\right)\left(784N\right)=\left(4.9\cdot10^{3}N\right)\left(.5m\right)
\left(L_{1}\right)=\frac{\left(4.9\cdot10^{3}N\right)\left(.5m\right)}{784N}=3.1mso, the man can stand 3.1m from the fulcrum on the side over the cliff?

 

[linuxux]

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.

 

[Doc Al]

(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.

 

[linuxux]

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?

 

[Doc Al]

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.

 

[linuxux]

okay, it seems much smaller now, but I am really not sure since in each case it seems right, but is it right now? if it is i got one more question for you: https://www.physicsforums.com/showthread.php?p=1275102#post1275102 , thanks for the help btw.

 

[Doc Al]

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:

torques must be equal so,

\tau_{clockwise}=\tau_{counterclockwise}
\left(F_1\right)\left(0.5m\right)=\left(4.9\cdot10^{3}N\right)\left(6.0m\right) ---> 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.

F_1 D_1 = F_2 D_2

 

[linuxux]

okay, the number i got now is much more reasonable, but is it right?

 

[Doc Al]

torques must be equal so,

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


so, the man can stand 3.1m from the fulcrum on the side over the cliff?

Looks good to me.