# Man jumping onto a see-saw to launch bricks vertically

#### lioric

Homework Statement
A man of mass 100kg wants to launch a brick into the air using a see saw.
He wants to launch the bricks of mass 5kg, 10 meters vertically in to the air.
The see saw is placed parallel to the ground using a support.
The bricks are placed on one end of the see saw.
The see saw is made of rigid material with a fulcrum in the center.
The see saw beam is 5m long
If the man jumps onto the far end of the seesaw (5 meters from the bricks) from a height of h meters from where he would land on the see saw.
Assume no air resistance, total energy convertion.

a) Find the value of h
b) What would be the value of h if the fulcrum was placed 1m from the bricks
c) what would be the value of h if the fulcrum was placed 1 m from where the man was to land.
Homework Equations
W=mg
v=u+at
v²=u²+2as
s=ut+1/2at²
F=ma
GPE=mgh
KE=1/2mv²
Moment = force x perpendicular distance
Work done = force x distance moved

I drew a diagram for the a) part
The person is h meters high
So GPE= 100 x 9.8x h
GPE= 980h j
KE = 980h when the person hits the see saw
KE=1/2mv²
980h=0.5 x 5 x v²

Now it v²=u²+2as
For the brick going up to 10m
v = 0
u=?
a=-9.8ms-²
s=10m

u²=2 x 9.8 x 10
u=14m/s

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 5 x 14²
980h = 490
h=490/980
h=0.5m

Oh crap
I used the mass of bricks for the person

980h = 0.5 x 100 x 14²
980h = 9800
h=9800/980
h=10m

Is this correct?
I assumed that the v for both sides of the see saw is same cause the fulcrum is in the middle

But don't know how to attempt it when the fulcrum is not in the middle as in part b) and c)

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#### PeroK

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Gold Member
2018 Award
Problem Statement: A man of mass 100kg wants to launch a brick into the air using a see saw.
He wants to launch the bricks of mass 5kg, 10 meters vertically in to the air.
The see saw is placed parallel to the ground using a support.
The bricks are placed on one end of the see saw.
The see saw is made of rigid material with a fulcrum in the center.
The see saw beam is 5m long
If the man jumps onto the far end of the seesaw (5 meters from the bricks) from a height of h meters from where he would land on the see saw.
Assume no air resistance, total energy convertion.

a) Find the value of h
b) What would be the value of h if the fulcrum was placed 1m from the bricks
c) what would be the value of h if the fulcrum was placed 1 m from where the man was to land.
Relevant Equations: W=mg
v=u+at
v²=u²+2as
s=ut+1/2at²
F=ma
GPE=mgh
KE=1/2mv²
Moment = force x perpendicular distance
Work done = force x distance moved

View attachment 245060

I drew a diagram for the a) part
The person is h meters high
So GPE= 100 x 9.8x h
GPE= 980h j
KE = 980h when the person hits the see saw
KE=1/2mv²
980h=0.5 x 5 x v²

Now it v²=u²+2as
For the brick going up to 10m
v = 0
u=?
a=-9.8ms-²
s=10m

u²=2 x 9.8 x 10
u=14m/s

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 5 x 14²
980h = 490
h=490/980
h=0.5m

Is this correct?
I assumed that the v for both sides of the see saw is same cause the fulcrum is in the middle

But don't know how to attempt it when the fulcrum is not in the middle as in part b) and c)

Are you saying that the man jumps from only $0.5m$ and reaches $14m/s$ when he hits the see-saw?

What principles are you applying here to decide what happens when the man lands on the see-saw?

#### lioric

Are you saying that the man jumps from only $0.5m$ and reaches $14m/s$ when he hits the see-saw?

What principles are you applying here to decide what happens when the man lands on the see-saw?
Oh crap
I used the mass of bricks for the person

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 100 x 14²
980h = 9800
h=9800/980
h=10m

#### PeroK

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Gold Member
2018 Award
Oh crap
I used the mass of bricks for the person

We can assume that u=14m/s is the velocity that the man is when he hits the see saw

980h = 0.5 x 100 x 14²
980h = 9800
h=9800/980
h=10m
You are missing something fundamental. If the brick were tied to the see-saw, then the whole system would rotate as a rigid body. But, the see-saw is not a fully rigid body. The brick will be projected into the air faster than the man lands on the other end. Why?

The physical reason is that a shockwave travels along the see-saw. This is similar to hitting a tennis ball with a racket. The ball leaves the racket much faster than you have moved your arm.

In any case, you need to find the physical laws that govern the transfer of energy from the man to the brick in this case. Any ideas? Hint: what two things are conserved?

#### haruspex

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Gold Member
2018 Award
We can assume that u=14m/s is the velocity that the man is when he hits the see saw
Why?

#### lioric

Using equations of motion it takes initial 14m/s to lift an object up to 10m

#### lioric

You are missing something fundamental. If the brick were tied to the see-saw, then the whole system would rotate as a rigid body. But, the see-saw is not a fully rigid body. The brick will be projected into the air faster than the man lands on the other end. Why?

The physical reason is that a shockwave travels along the see-saw. This is similar to hitting a tennis ball with a racket. The ball leaves the racket much faster than you have moved your arm.

In any case, you need to find the physical laws that govern the transfer of energy from the man to the brick in this case. Any ideas? Hint: what two things are conserved?
Conservation of energy and conservation of momentum

#### haruspex

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2018 Award
Using equations of motion it takes initial 14m/s to lift an object up to 10m
But that's the launch velocity needed for the brick. Why is that the speed of the man on hitting the seesaw? (Not saying it isn't, just querying your assumption.)

#### PeroK

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2018 Award
Conservation of energy and conservation of momentum
What momentum, specifically?

Be careful: are there any external forces on the system? Apart from gravity on the man and the brick? Is momentum in the up/down direction conserved during the impact of the man and the launching of the brick?

#### haruspex

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2018 Award
The question is is quite hard to answer if you take the physics of it seriously. Instead, it seems you are expected to rely on this utterly unrealistic assumption: total energy conversion. I interpret that as meaning all the man's lost PE is converted into man's KE, thence into brick KE, and so into brick PE. That's all you need - forget the seesaw.

#### PeroK

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2018 Award
The question is is quite hard to answer if you take the physics of it seriously. Instead, it seems you are expected to rely on this utterly unrealistic assumption: total energy conversion. I interpret that as meaning all the man's lost PE is converted into man's KE, thence into brick KE, and so into brick PE. That's all you need - forget the seesaw.
I see it differently. The man collides with the see-saw loses some KE, which is transferred to the brick.

I assume we ignore any rotational KE the see-saw gains durng the short-duration collision that launches the brick.

#### haruspex

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2018 Award
loses some KE
But how much? Seems to me they are saying "all".
Edit: @PeroK has just pointed out to me that what I am suggesting would violate a physical law. So the next option is to interpret full conversion of energy as no loss of mechanical energy. The man gets to keep some.
Combined with that other law that provides enough info for a solution.

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#### lioric

What momentum, specifically?

Be careful: are there any external forces on the system? Apart from gravity on the man and the brick? Is momentum in the up/down direction conserved during the impact of the man and the launching of the brick?
This is question meant for a lower like grade 10
We can say that the momentum is conserved
Otherwise we'll have to take account a lot of ways energy gets wasted. So ya momentum is conserved

Nope no external forces
But to makes things interesting we can say the force exerted by the brick on the could oppose the person's force

#### lioric

But that's the launch velocity needed for the brick. Why is that the speed of the man on hitting the seesaw? (Not saying it isn't, just querying your assumption.)
Well I saw a similar question about a contractor launching a brick to his friend

I used the idea from that
But I guess since the fulcrum is in the middle and we are treating the see saw as a rigid body no energy is wasted in to bending the seesaw while over coming the inertia
But mainly the rigid body rotating from the center can have equal speeds equidistant from the pivot

#### PeroK

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This is question meant for a lower like grade 10
We can say that the momentum is conserved
Otherwise we'll have to take account a lot of ways energy gets wasted. So ya momentum is conserved

Nope no external forces
But to makes things interesting we can say the force exerted by the brick on the could oppose the person's force
There is an external force at the fulcrum. You can see this because the man cannot speed up, so his momentum is reduced. And, the brick is launched upwards, which equates to more loss of momentum in the downward direction. The upward force at the fulcrum, therefore, is an external force which gives an impluse to the system (of man, brick and see-saw) in the upwards direction.

This problem may be too advanced for you. You'll have to decide. The key concept you are missing is conservation of angular momentum: in this case about the fulcrum. We choose to look at angular momentum about the fulcrum because the external force at the fulcrum has no effect on this quantity.

#### lioric

The question is is quite hard to answer if you take the physics of it seriously. Instead, it seems you are expected to rely on this utterly unrealistic assumption: total energy conversion. I interpret that as meaning all the man's lost PE is converted into man's KE, thence into brick KE, and so into brick PE. That's all you need - forget the seesaw.
Ya as I said this is meant for grade 10s so they haven't gone through the whole complications. They can use momentum, moment, energy and equations of motion.
And we take everything as an isolated system. No air resistance, total energy and momentum conservation.

#### haruspex

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2018 Award
We can say that the momentum is conserved
As @PeroK has pointed out, there is an impulse from the fulcrum, so linear momentum is not conserved. What is a third conservation law?
we are treating the see saw as a rigid body no energy is wasted in to bending the seesaw
Energy loss is not a simple question of rigid or otherwise. Nothing is completely rigid. The seesaw will behave like a spring, though a very stiff one. Energy loss in it depends on its elasticity, not its rigidity.

#### lioric

As @PeroK has pointed out, there is an impulse from the fulcrum, so linear momentum is not conserved. What is a third conservation law?

Energy loss is not a simple question of rigid or otherwise. Nothing is completely rigid. The seesaw will behave like a spring, though a very stiff one. Energy loss in it depends on its elasticity, not its rigidity.
Sorry
I was referring to the deforming of the see saw
But I can understand

#### lioric

What is a third conservation law?
Are you referring to Newton's third law which related to conservation of angular momentum?

#### haruspex

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Are you referring to Newton's third law which related to conservation of angular momentum?
Yes. What axis should you choose?

#### jbriggs444

Homework Helper
In the context of angular momentum, you should understand "axis" in the sense of "axis of rotation", not in the sense of a coordinate axis.

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#### lioric

Personally, I would attack the problem by considering the see saw with a fulcrum in the middle as an ideal massless energy-conserving force reverser. Except for the reversal of direction, it is exactly as if the falling man had collided with the brick through an ideal massless energy-conserving spring. Both rebound from what amounts to an elastic collision.
Yes that is exactly how I want this question to be attempt
By the children

So what your saying is I need to put momentum in to this and treat it like a collision.
How should I try out the part with the fulcrum not in the middle?

#### lioric

In the context of angular momentum, you should understand "axis" in the sense of "axis of rotation", not in the sense of a coordinate axis.
So I guess the action of the man jumping on the seesaw would cause a clockwise moment about the fulcrum

#### haruspex

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So I guess the action of the man jumping on the seesaw would cause a clockwise moment about the fulcrum
Right, but think about angular momentum about that axis. Would it be conserved and why or why not?

"Man jumping onto a see-saw to launch bricks vertically"

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