What is the distance from the pivot when the seesaw is balanced?

[Tepictoc]

Homework Statement

It's a balanced seesaw but the pivot isn't in the exact middle and two weights are on it. One at the very end of each side.
The weight on the left is 4.8 kg.[m1]
The weight on the right is 5.5 kg[m2]
The length of the entire seesaw is 5m
L1 is the distance from the pivot on the left side to the end of the board where m1 is.
Calculate the distance L1 from the pivot when the board is balanced.

Homework Equations

Distance = Torque/force
Sum of all moments = 0

The Attempt at a Solution

m1*g=47.088 N This force is acting counterclockwise
m2*g=53.955 N This force is acting clockwise
The normal force from the pivot is m1g+m2g = 101,043 N

I'll use L2 to identify the length from the pivot to the right end since L1 is the length from pivot to the left end.
53.955*L2 = 47.088*L1

This is where I get stuck when trying to calculate the lengths of L1 and L2.
I could calculate the length of L1 and L2 if I knew the torque by dividing it by the force but it's unknown isn't it?

 

[gneill]

How are L1 and L2 related via the total length of the seesaw arm?

 

[abrewmaster]

you don't need to know the torque because the torque of the one end of the seesaw should be equal and opposite to the torque of the other just like you have in your one equation. You know that your total length of the seesaw is 5m so set up a second equation and then you have two equations and two unknowns.

 

[Tepictoc]

How are L1 and L2 related via the total length of the seesaw arm?

L1 < L2
The entire length of the seesaw is 5m. In the picture the pivot isn't at 2.5m, it's closer to L1.

you don't need to know the torque because the torque of the one end of the seesaw should be equal and opposite to the torque of the other just like you have in your one equation. You know that your total length of the seesaw is 5m so set up a second equation and then you have two equations and two unknowns.

I don't understand the last part.
Also sorry if I am confusing, English is not my first language.

 

[gneill]

L1 < L2
The entire length of the seesaw is 5m. In the picture the pivot isn't at 2.5m, it's closer to L1.

Also sorry if I am confusing, English is not my first language.

Write an equation that relates L1, L2, and L, where L is the total length of the seesaw arm.

Hint: If you happened to know what L1 is, how would you determine what L2 is?

 

[Tepictoc]

Write an equation that relates L1, L2, and L, where L is the total length of the seesaw arm.

Hint: If you happened to know what L1 is, how would you determine what L2 is?

L = L1+L2
5 = L1 +L2

If I knew L1 then I would know what L2 is but how do I get L1?
If L1 was 2m then L2 would obviously be 3m.

 

[haruspex]

L = L1+L2
5 = L1 +L2

If I knew L1 then I would know what L2 is but how do I get L1?
If L1 was 2m then L2 would obviously be 3m.

You have two equations relating L1 and L2, the one in the OP and the one for L1+L2 above. So it's a straightforward problem in simultaneous equations: use one equation, in the form L2= something, to substitute for L2 in the other.
Btw, you've made the arithmetic unnecessarily complicated. Given m1 g L1 = m2 g L2 you can cancel out g.

 

[Tepictoc]

You have two equations relating L1 and L2, the one in the OP and the one for L1+L2 above. So it's a straightforward problem in simultaneous equations: use one equation, in the form L2= something, to substitute for L2 in the other.
Btw, you've made the arithmetic unnecessarily complicated. Given m1 g L1 = m2 g L2 you can cancel out g.

L1+L2 = 5m
4.8L1 = 5.5L2

L2 = 4.8L1
L1+4.8L1 = 5
Am i on the right path? I am not sure what to do next.

Or is this another way to do it?
L1*4.8 = L2*5.5
L1 = 0.872L2
0.872L2+0.872L2 = mg
1.744L2 = mg
L2 = 5.5/1.744 = 3.15
L1 = 0.872*3.15 = 2.74
Actually then the total becomes too high(5.8m)

 

[gneill]

Rather than plugging in numbers so soon, why not manipulate the variables. You've got two equations:

L1 + L2 = L
m1*L1 = m2*L2

Solve for L1. Then plug the given numbers into the resulting expression.

 

[Tepictoc]

Rather than plugging in numbers so soon, why not manipulate the variables. You've got two equations:

L1 + L2 = L
m1*L1 = m2*L2

Solve for L1. Then plug the given numbers into the resulting expression.

I am pretty sure I understand what you mean but I get stuck after a while again.

L1 = L-L2
m1*L-L2 = m2*L2
4.4*5-L2 = 5.5*L2
22-L2 = 5.5*L2

 

[SteamKing]

I am pretty sure I understand what you mean but I get stuck after a while again.

L1 = L-L2
m1*L-L2 = m2*L2
4.4*5-L2 = 5.5*L2
22-L2 = 5.5*L2

And how do we solve an equation in one variable? Collecting like terms on one side of the equal sign?

Hint: algebra

 

[Tepictoc]

And how do we solve an equation in one variable? Collecting like terms on one side of the equal sign?

Hint: algebra

Is this correct?

22-L2 = 5.5*L2
22-L2-5.5L2 = 5.5L2-5.5L2
22-6.5L2 = 0
-6.5L2 = 22
6.5L2/6.5 = 22/6.5
L2 = 3.384 m

which means 5-3.384 = 1.616
L1 = 1.6 m
L2 = 3.3 m

 

[haruspex]

L1+L2 = 5m
4.8L1 = 5.5L2

Yes

L2 = 4.8L1

How did you get that ?!

L1*4.8 = L2*5.5
L1 = 0.872L2

i think you have that backwards

0.872L2+0.872L2 = mg

Where did that come from?

 

[Tepictoc]

Yes
How did you get that ?!

i think you have that backwards

Where did that come from?

I tried using the method to calculate a hanging weight held up by two strings but I realize that's completely wrong here.

I tried solving it with algebra here as suggested:
L1 = L-L2
m1*L-L2 = m2*L2
4.4*5-L2 = 5.5*L2
22-L2 = 5.5*L2

Is this correct?

22-L2 = 5.5*L2
22-L2-5.5L2 = 5.5L2-5.5L2
22-6.5L2 = 0
-6.5L2 = 22
6.5L2/6.5 = 22/6.5
L2 = 3.384 m

which means 5-3.384 = 1.616
L1 = 1.6 m
L2 = 3.3 m

 

[SteamKing]

Is this correct?

22-L2 = 5.5*L2
22-L2-5.5L2 = 5.5L2-5.5L2
22-6.5L2 = 0
-6.5L2 = 22
6.5L2/6.5 = 22/6.5
L2 = 3.384 m

which means 5-3.384 = 1.616
L1 = 1.6 m
L2 = 3.3 m

22-6.5L2 = 0

should simplify to

6.5L2 = 22

 

[Tepictoc]

22-6.5L2 = 0

should simplify to

6.5L2 = 22

But the answer is correct?

Also Thanks a lot everyone for your help!

 

[gneill]

I am pretty sure I understand what you mean but I get stuck after a while again.

L1 = L-L2
m1*L-L2 = m2*L2 <--- Why replace L1 when that's what you're looking for? Replace L2 instead.
4.4*5-L2 = 5.5*L2 <--- Uh Oh. Caught out by lack of parentheses. a*(b-c) ≠ a*b - c
22-L2 = 5.5*L2 <--- !

Watch your algebra! If you replace variable "L1" with "L - L2", then if L1 was multiplied by something in the equation then that multiplication distributes over both L and L2. Use parentheses around things that you substitute in order to avoid inadvertently "dropping" operations. So: m1*L1 = m2*L2 ----> m1*(L - L2) = m2*L2.

But as I indicated above, since you're looking for L1, don't substitute it away! Replace L2 instead, leaving you with an equation in just L1.

Also, your 4.8kg seems to have become 4.4kg...

 

[Tepictoc]

Simplifying
4.8L1 = 5.5(5-L2)
4.8L1 = (5 * 5.5-L2 * 5.5)
4.8L1 = (27.5-5.5L2)

Solving
4.8L1 = 27.5 -5.5L2

Divide each side by 4.8.
L = 5.729166667 -1.145833333L2

L1+L2 = 5
5.72-1.14L2 + L2 = 5

 

[gneill]

Simplifying
4.8L1 = 5.5(5-L2)

Nooooo. You've replaced L2 with (5 - L2). That can't be right.

It really would be better to work with just the symbols until the very end. Use the symbols.

So, what's L2 in terms of L and L2?

 

[Tepictoc]

nooooo. You've replaced l2 with (5 - l2). That can't be right.

It really would be better to work with just the symbols until the very end. Use the symbols.

So, what's l2 in terms of l and l2?

L2 = L-L1

4.8L1=5.5(L-L1)
and then I'm not sure.

 

[gneill]

l2 = l-l1

4.8l1=5.5(l-l1)

Lower case L's are confusing with the fonts used here. Best stick to upper case L's.

No numbers! Use symbols. That 4.8 is m1. That 5.5 is m2. Otherwise, looks better

Continue...

 

[Tepictoc]

lower case l's are confusing with the fonts used here. Best stick to upper case l's.

No numbers! Use symbols. That 4.8 is m1. That 5.5 is m2. Otherwise, looks better

Continue...

L2 = L-L1

m1L1 = m2(L-L1)
m1L1 = m2*L - m2*L1
m1L1 = m2L - m2L1

 

[gneill]

L2 = L-L1

m1L1 = m2(L-L1)
m1L1 = m2*L - m2*L1
m1L1 = m2L - m2L1

Good so far. Can you isolate L1? Hint: first move all the terms containing L1 to the left hand side.

 

[Tepictoc]

Good so far. Can you isolate L1? Hint: first move all the terms containing L1 to the left hand side.

m1L1 = m2L - m2L1
Moving all L1s to the left
m1L1+m2L1 = m2L

I don't know how to isolate L1.
Do I divide both sides by m2?

 

[gneill]

m1L1 = m2L - m2L1
Moving all L1s to the left
m1L1+m2L1 = m2L

I don't know how to isolate L1.
Do I divide both sides by m2?

No, factor L1 out of the terms on the left. Basic algebra!

 

[Tepictoc]

No, factor L1 out of the terms on the left. Basic algebra!

L1(m1+m2) = m2L
10.3L1 = m2L
10.3L1/10.3 = 5.5*5/10.3
L1 = 2.6699 m
L2 = 5-2.6699
L2 = 2.3301 m

 

[haruspex]

L1(m1+m2) = m2L
10.3L1 = m2L
10.3L1/10.3 = 5.5*5/10.3
L1 = 2.6699 m
L2 = 5-2.6699
L2 = 2.3301 m

Looks right.

 

[gneill]

L1(m1+m2) = m2L
10.3L1 = m2L
10.3L1/10.3 = 5.5*5/10.3
L1 = 2.6699 m
L2 = 5-2.6699
L2 = 2.3301 m

Yup. Those results are good.

I'll just point out that, because you've once again plugged in numbers before finishing off the algebra, you've perhaps missed out on the elegant (and easily memorized) form of the final arrangement:

From your L1(m1+m2) = m2L, dividing both sides by m1+m2 yields:
$$L1 = \frac{m2}{m1 + m2} L$$
similarly, L2 will be found to be:
$$L2 = \frac{m1}{m1 + m2} L$$
So a simple fraction of the total mass yields the distance of each mass from the center of mass of the system, the place where it where it will balance. This form of ratio of masses will come up several times in different types of problems and is worth remembering