Draw a diagram showing the forces acting on a plank and find x

[kuruman]

yx-z((l/2)-x)=0
x-z((l/2)-x)=0

Why did y disappear from the first line to the second?
Please be more careful with your algebra.

 

[Richie Smash]

I divided the left side by y, so therefore i would have gotten 0 divided by y on the next side, thus eliminating y

Or perhaps this equation is harder than it looks

 

[kuruman]

divided the left side by y, so therefore i would have gotten 0 divided by y on the next side, thus eliminating y

Starting with $$yx-z(l/2-x)=0$$
Dividing both sides by y gives $$x-\frac{z(l/2-x)}{y}=0$$
You have to divide both terms on the left by y. Just remove the parentheses by multiplying the two terms inside them by -z.

 

[kuruman]

Ok I've managed to simplify to x² - (zl/2) +z=0

Still incorrect because
(a) y disappeared from the equation. Since y stands for W_m, not having y in the expression means that x does not depend on how much mass you put at the end of the plank. This does not make sense.
(b) You have a term in $x^2$. The equation you started from does not have two $x$ terms multiplied together so the simplified version should not have them either.

Please try again,

 

[Richie Smash]

Ok here is another attempt, after simplifying I have gotten (2xy)/z+2x-l=0

 

[kuruman]

That is correct. Can you solve for x?

 

[Richie Smash]

yes it would be x= zl/2y

 

[kuruman]

Let's see if that works. I will put in the value $x=zl/2y$ in the left side and see if I get zero.
$L.S. = yzl/2y-z(l/2-zl/2y)$
simplify the $y$ in the first term and remove the parentheses
$L.S. = zl/2-zl/2+z^2l/2y=z^2l/2y$
It is not zero, so your solution is incorrect. Please try again. You may wish to verify that your solution works by plugging it back in the equation like I did here.

 

[Richie Smash]

Alright, I'll have to tackle this tmrw, but thank you for continuing to help

 

[Richie Smash]

here is my latest attempt

x= lz/(2y+2z)

However I am not sure how I would solve for x given the only variable I have value for is L

 

[kuruman]

However I am not sure how I would solve for x given the only variable I have value for is L

Read the statement of the problem that you posted in #1 carefully. The mass of the plank and that of the added mass are given. It looks like when you renamed W_plank and W_mass z and y in post #35, you lost track of what's what. There is a lesson to be learned here.

 

[Richie Smash]

So you're saying I don't need Force to find length, but simply the mass, or perhaps the centre of gravity? so it would be (l/2-x)/2 *10 would be the moment of the centre of the plank, and 15 *x/2 for the other one

 

[kuruman]

So you're saying I don't need Force to find length, but simply the mass, or perhaps the centre of gravity? so it would be (l/2-x)/2 *10 would be the moment of the centre of the plank, and 15 *x/2 for the other one

That's not what I'm saying. You have reached the expression

x= lz/(2y+2z)

This gives you a value for x if you know l, z and y. I am saying that if you replace symbols l, z and y with numbers as per post #1, you will get a number for x.

 

[Richie Smash]

Ok, so x = 0.4m, the question is solved.

But the new questions arise
The first being, Why does mass by length work? I know moments to be Force by Distance.

The second was your point, the centre of gravity of the mass relative to the right end of the top plank would be x/2 = 0.2m

Was there an easier way to do this?

 

[kuruman]

Your answer 0.4 m is correct.

The first being, Why does mass by length work? I know moments to be Force by Distance.

That's because if you put the forces (in this case weights) in, you get
$m_{mass}~g~x-m_{plank}~g~(L/2-x)=0$
Since $g$ appears in both terms on the left and the right side is zero, we can divide by it to get
$m_{mass}~x-m_{plank}~(L/2-x)=0$

The second was your point, the centre of gravity of the mass relative to the right end of the top plank would be x/2 = 0.2m

That's not how it works. The center of gravity from the right end is given by
$$X_{cg}=\frac{W_{mass}~x_{mass}+W_{plank}~x_{plank}}{W_{mass}+W_{plank}}$$
where $x_{mass}$ and $x_{plank}$ are distances measured from the right end. Here $x_{mass}=0$ and $x_{plank}=L/2$
so
$$X_{cg}=\frac{0+W_{plank}~L/2}{W_{mass}+W_{plank}}=\frac{W_{plank}~L}{2(W_{mass}+W_{plank})}=x=0.4 m$$
So the cg is at the same position as the edge of the lower plank. That's no coincidence and the point to be made is this.

For the net torque to be zero, the cg must be vertically above or below the pivot point. If you stack the planks with their edges flush, there will be no tipping when you put some mass on the right end because the cg of the assembly {top plank + mass} is to the left of the edge so a point can be found on the bottom plank to provide zero torque. Now if you pull the top plank with the mass to the right, the top assembly will tip if its cg goes past the pivot point at the edge because you run out of points on the bottom plank that can provide zero torque. Therefore to find $x$ all you have to do is find where the cg of the top assembly is with respect to the right edge and that's your answer. That's where I was leading you at first and that's the easier way to do this. Now you know.

A simple convincing experiment about the cg and tipping thresholds is this. Go to the middle of the room and stand on your toes. Keep the position long enough to convince yourself that you can do it. Then stand facing a bare wall with your toes right up against the baseboard. Try standing on your toes. You will fail. Why? Answer: When you stand on your feet, your cg is somewhere above your heels, not above your toes and the torque due to gravity is zero. When you stand on your toes, you have to lean a bit forward to bring your cg above the pivot point at your toes which again makes the torque due to gravity zero. Now when you try to stand on your toes facing the wall, you are prevented from leaning forward as the torque due to gravity keeps you standing on your heels.

 

[Richie Smash]

Hey thanks so much Kuruman, for all the help, I truly have a better understanding now, and I bookmarked this thread in case I need some reminding, that makes a lot of sense thanks for having so much patience when I first saw this question I was clueless.