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

In summary: Yes, this is correct. Using your values for L1 and L2, you can also check that the torques are equal.In summary, using the given information of a balanced seesaw with two weights, one at each end, and the total length of the seesaw being 5m, we can calculate the distance L1 from the pivot on the left side to the end of the board where m1 is located. This can be done by setting up two equations, one using the distance formula and the other using the sum of moments equation. By manipulating the variables, we can solve for L1, which turns out to be approximately 1.6m. With this value, we can also solve for L2,
  • #1
Tepictoc
16
0

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?
 
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  • #2
How are L1 and L2 related via the total length of the seesaw arm?
 
  • #3
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.
 
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  • #4
gneill said:
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.



abrewmaster said:
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.
 
  • #5
Tepictoc said:
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?
 
  • #6
gneill said:
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.
 
  • #7
Tepictoc said:
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.
 
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  • #8
haruspex said:
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)
 
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  • #9
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.
 
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  • #10
gneill said:
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
 
  • #11
Tepictoc said:
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
 
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  • #12
SteamKing said:
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
 
  • #13
Tepictoc said:
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?
 
  • #14
haruspex said:
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
Tepictoc said:
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
 
  • #15
Tepictoc said:
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
 
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  • #16
SteamKing said:
22-6.5L2 = 0

should simplify to

6.5L2 = 22

But the answer is correct?

Also Thanks a lot everyone for your help!
 
  • #17
Tepictoc said:
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...
 
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  • #18
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
 
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  • #19
Tepictoc said:
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?
 
  • #20
gneill said:
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.
 
  • #21
Tepictoc said:
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 :smile:

Continue...
 
  • #22
gneill said:
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 :smile:

Continue...

L2 = L-L1

m1L1 = m2(L-L1)
m1L1 = m2*L - m2*L1
m1L1 = m2L - m2L1
 
  • #23
Tepictoc said:
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.
 
  • #24
gneill said:
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?
 
  • #25
Tepictoc said:
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!
 
  • #26
gneill said:
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
 
  • #27
Tepictoc said:
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.
 
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  • #28
Tepictoc said:
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 :wink:
 
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1. How does the length of a seesaw from the pivot affect its balance?

The length of a seesaw from the pivot affects its balance by changing the distribution of weight on each side. The longer the seesaw is from the pivot, the more weight is required on one side to balance out the weight on the other side.

2. What is the ideal length for a seesaw from the pivot?

The ideal length for a seesaw from the pivot depends on the weight of the individuals using it. Generally, a longer seesaw will require more weight on one side to balance out a smaller weight on the other side. However, a balanced seesaw can also be achieved with a shorter length and equal weight on both sides.

3. How does the angle of the seesaw affect its balance?

The angle of the seesaw affects its balance by changing the distance of the weight from the pivot point. A smaller angle will result in a shorter distance, requiring less weight to balance out the other side. A larger angle will result in a longer distance, requiring more weight for balance.

4. Can the seesaw length from the pivot be adjusted?

Yes, the length of a seesaw from the pivot can be adjusted by moving the pivot point or by using different lengths of plank for the seesaw. However, it is important to ensure that the seesaw is balanced and safe for use.

5. How does the seesaw length from the pivot affect the force needed to lift off the ground?

The seesaw length from the pivot affects the force needed to lift off the ground by changing the torque, or rotational force, on the seesaw. A longer seesaw will require more force to lift off the ground, while a shorter seesaw will require less force. This is due to the unequal distribution of weight on each side of the pivot.

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