Finding the work done by Spiderman

[Lnewqban]

Would the response be different if Spiderman just climbs a 6 meters high wall and then walks horizontally on it other 10.4 meters?

 

[ChiralSuperfields]

Ok, so what is the angle between the force mg and the displacement $\vec r$?

Thank you for your reply @haruspex! The angle is 120 degrees

 

[ChiralSuperfields]

"Equilateral" means all sides are the same size (equal sides). You don't need to split it and to form right angle triangles. You seem to have a tendency to pick the most complicated ways to solve things.,

Thank you for your reply @nasu!

Yeah, I do have that tendency, I will lose it hopefully when I solve more problems :)

 

[ChiralSuperfields]

Hi @Callumnc1. I’d like to mention an alternative method.

Do you know how to find (changes in) gravitational potential energy? If so, you can use:

Work done by gravity = - (change in gravitational potential energy)

It’s well worth thinking about why the above method is equivalent to the ‘dot product’ method.

Thank you for your reply @Steve4Physics!

Thanks for mentioning that method, I forgot you could solve that way. I will try it out!

I think it is equivalent because the dot product finds the component of two physical quantities in the same direction then multiplies them (weight x vertical displacement). I guess from the energy method, you are also x the weight by the vertical displacement (mgh).

Is my understanding correct?

Many thanks!

 

[ChiralSuperfields]

How are those two forces different?
Acting only vertically, weight can only resist vertical displacement between lowest and highest points.
The muscular energy from Spiderman is the only cause of the horizontal displacement (r→), about which the problem is not asking.
Re-visit post #3 and focus only on pure vertical work.

Thank you for reply @Lnewqban!

Do you mean tension and the weight of spider man? Sorry, what do mean muscular energy is the only cause of the displacement in the i hat direction?

Many thanks!

 

[ChiralSuperfields]

Would the response be different if Spiderman just climbs a 60 meters high wall and then walks horizontally on it other 10.4 meters?

View attachment 322488

Thank you for your reply @Lnewqban!

I don't think so since when spider man is walking horizontally, his COM is moving perpendicular to his weight so the dot product is zero

 

[Lnewqban]

Thank you for reply @Lnewqban!

Do you mean tension and the weight of spider man? Sorry, what do mean muscular energy is the only cause of the displacement in the i hat direction?

Many thanks!

"By repeatedly bending at the waist, ..."

Is there any displacement in the direction of tension?

 

[Lnewqban]

Thank you for your reply @Lnewqban!

I don't think so since when spider man is walking horizontally, his COM is moving perpendicular to his weight so the dot product is zero

What if you follow the approach proposed in post #34?

 

[ChiralSuperfields]

"By repeatedly bending at the waist, ..."

Is there any displacement in the direction of tension?

Thank you for your reply @Lnewqban !

I don't think so since spider man can be modelled as a particle in circular motion when he bends his waist (assuming COM at waistline). I think the displacement is tangent to the tension.

Many thanks!

 

[haruspex]

Would the response be different if Spiderman just climbs a 60 meters high wall and then walks horizontally on it other 10.4 meters?

You mean 6m, not 60m, right?

 

[ChiralSuperfields]

What if you follow the approach proposed in post #34?

Thank you for reminding me! Here it is also @Steve4Physics:

$W = -\Delta U$
$W = U_i - U_f$
$W = -(80)(9.81)(6)$
$W = -4.7 kJ$

Thanks!

 

[haruspex]

"By repeatedly bending at the waist, ..."

Is there any displacement in the direction of tension?

Not sure what your point is. The supposition is that he works it up like someone on a swing.

 

[Lnewqban]

You mean 6m, not 60m, right?

Just corrected in post #36.
Thank you, @haruspex

 

[ChiralSuperfields]

Not sure what your point is. The supposition is that he works it up like someone on a swing.

Thank you for your replies @haruspex and @Lnewqban !

How hard would it be to calculate the work done by spider man since we do don't how many times he would have to swing to be eventually 6m higher than his initial height?

Many thanks!

 

[nasu]

Forget about it. How much work do you when you walk 6 m horizontally?

 

[jbriggs444]

Thank you for your reply @Lnewqban !

I don't think so since spider man can be modelled as a particle in circular motion when he bends his waist (assuming COM at waistline).

You cannot model an object which bends at the waist as a point-like object. Nor can a point-like object pump up the amplitude of its swing. It will just swing back and forth in an arc with a fixed amplitude. It has no degrees of freedom to influence anything.

With an extended body, one can pump a rope or swing in either of two modes.

Mode 1: You utilize the angular momentum of your body to shift the ropes or chains fore and aft as you swing. You shift aft during the forward swing and fore during the aft swing. This works for low amplitudes, though it is rather difficult to get started. From a seated posture, the main things you are doing is leaning back with legs extended forward and pulling back on the chain or leaning forward with legs folded beneath the seat and pushing forward on the chain. From a standing posture, the main thing you are doing is leaning forward and pushing forward on the chain or leaning back and pulling back.

Mode 2: You shift your body up and down. Suddenly up at the bottom of the stroke. Slowly down anywhere near the end points. The more work you feel like you are doing, the more energy you are pumping into the system. This works exceptionally well at high amplitudes and is best done from a standing posture.

As a child, I had a chance to practice extensively since we had a longer-than-usual swing on one of the elm trees in our back yard. Maybe 15 feet of chain. It had a flat wooden seat and could be pumped from either a standing or a seated posture. From a standing posture, it was possible to get the chains a bit above the horizontal. I do not know how our parents failed to have heart attacks watching us every day.

 

[haruspex]

How hard would it be to calculate the work done by spider man since we do don't how many times he would have to swing to be eventually 6m higher than his initial height?

Many thanks!

The question makes the assumption that there is little wasted energy.

As an analogy, consider a pendulum length R with a square bob side 2r, mass M, hanging at rest. But I'll treat the bob mass as a point mass at its centre.
An ant mass m at the top of the bob runs across to one corner. By conservation of angular momentum, it succeeds in displacing the mass centre of the ant+bob system horizontally by about $\frac{mr^2}{MR}$. This exerts a torque about the pendulum axis.
As the pendulum swings, the ant finds itself at the lower end of the top edge of the bob. When at the limit of the swing, the ant runs up to the top corner. The process repeats, with the ant's work in running uphill each time gradually increasing the amplitude of the swing.

 

[ChiralSuperfields]

Forget about it. How much work do you when you walk 6 m horizontally?

Thank you for your reply @nasu!

No work is done since the COM moves parallel to the direction of spider mans weight

 

[ChiralSuperfields]

You cannot model an object which bends at the waist as a point-like object. Nor can a point-like object pump up the amplitude of its swing. It will just swing back and forth in an arc with a fixed amplitude. It has no degrees of freedom to influence anything.

With an extended body, one can pump a rope or swing in either of two modes.

Mode 1: You utilize the angular momentum of your body to shift the ropes or chains fore and aft as you swing. You shift aft during the forward swing and fore during the aft swing. This works for low amplitudes, though it is rather difficult to get started. From a seated posture, the main things you are doing is leaning back with legs extended forward and pulling back on the chain or leaning forward with legs folded beneath the seat and pushing forward on the chain. From a standing posture, the main thing you are doing is leaning forward and pushing forward on the chain or leaning back and pulling back.

Mode 2: You shift your body up and down. Suddenly up at the bottom of the stroke. Slowly down anywhere near the end points. The more work you feel like you are doing, the more energy you are pumping into the system. This works exceptionally well at high amplitudes and is best done from a standing posture.

As a child, I had a chance to practice extensively since we had a longer-than-usual swing on one of the elm trees in our back yard. Maybe 15 feet of chain. It had a flat wooden seat and could be pumped from either a standing or a seated posture. From a standing posture, it was possible to get the chains a bit above the horizontal. I do not know how our parents failed to have heart attacks watching us every day.

Thank you for your reply @jbriggs444 !

Do you mean it was possible to get the chains a bit above the vertical?

Many thanks!

 

[haruspex]

Do you mean it was possible to get the chains a bit above the vertical?

No, a bit above horizontal. A swing amplitude of over 90°.
(How do you get something above vertical?)

 

[ChiralSuperfields]

The question makes the assumption that there is little wasted energy.

As an analogy, consider a pendulum length R with a square bob side 2r, mass M, hanging at rest. But I'll treat the bob mass as a point mass at its centre.
An ant mass m at the top of the bob runs across to one corner. By conservation of angular momentum, it succeeds in displacing the mass centre of the ant+bob system horizontally by about $\frac{mr^2}{MR}$. This exerts a torque about the pendulum axis.
As the pendulum swings, the ant finds itself at the lower end of the top edge of the bob. When at the limit of the swing, the ant runs up to the top corner. The process repeats, with the ant's work in running uphill each time gradually increasing the amplitude of the swing.

Thank you for your analogy @haruspex!

Sorry, could you tell me how you got $\frac{mr^2}{MR}$?

Many thanks!

 

[ChiralSuperfields]

No, a bit above horizontal. A swing amplitude of over 90°.
(How do you get something above vertical?)

Thank you for your reply @haruspex !

Oh whoops sorry, I understand now. I was thinking of the horizontal being at the bottom of the swing. Why are you allowed to have the horizontal up there (where the swing has an amplitude greater than 90 degrees)? Are both places where you define the horizontal valid?

Many thanks!

 

[haruspex]

Thank you for your analogy @haruspex!

Sorry, could you tell me how you got $\frac{mr^2}{MR}$?

Many thanks!

If the ant accelerates at a and the bob at A the other way, the torque balance at the pendulum pivot gives $MA(R+r)=maR$. So the ratio of speeds is $M(R+r):mR$, and the ratio of distances travelled is the same.
When the ant has gone distance r (roughly - the bob moves the other way so it will be a bit less), the bob has moved $\frac{mrR}{M(R+r)}$. So the net movement of the mass centre is $\frac{mr-M\frac{mrR}{M(R+r)}}{M+m}=\frac{mr(1-\frac R{R+r})}{M+m}=\frac{mr^2}{(M+m)(R+r)}\approx \frac{mr^2}{MR}$.

 

[ChiralSuperfields]

If the ant accelerates at a and the bob at A the other way, the torque balance at the pendulum pivot gives $MA(R+r)=maR$. So the ratio of speeds is $M(R+r):mR$, and the ratio of distances travelled is the same.
When the ant has gone distance r (roughly - the bob moves the other way so it will be a bit less), the bob has moved $\frac{mrR}{M(R+r)}$. So the net movement of the mass centre is $\frac{mr-M\frac{mrR}{M(R+r)}}{M+m}=\frac{mr(1-\frac R{R+r})}{M+m}=\frac{mr^2}{(M+m)(R+r)}\approx \frac{mr^2}{MR}$.

Thank you for very much your reply @haruspex !

Sorry, I am having trouble visualizing this situation, could you kindly draw a diagram? I think this will also help me understand why angular momentum is not conserved in this situation

Many thanks!

 

[haruspex]

I was thinking of the horizontal being at the bottom of the swing

Horizontal is the orientation of a line or a plane, not a particular height. In an apartment block, all the floors are horizontal, preferably.

 

[ChiralSuperfields]

Horizontal is the orientation of a line or a plane, not a particular height. In an apartment block, all the floors are horizontal, preferably.

Thank you for your reply @haruspex!

 

[jbriggs444]

Do you mean it was possible to get the chains a bit above the vertical?

A swing normally hangs straight down. The chains are vertical. "Vertical" is in the direction of gravity. "Horizontal" is perpendicular to this.

As you swing back and forth the angle (measured from the vertical center line) of the chains reaches larger and larger peaks depending on how hard you pump.

If you pump hard enough, the peaks can reach 90 degrees. That is, the chains at the top of the arc are horizontal.

For a child swinger, problems arise when you pass the horizontal. When you reach the top of the arc and swing back down, the chains will have gone slack. You will fall downward on slack chains until they come taut again in an inelastic collision. The collision is jarring and, hence, dangerous. This collision also absorbs energy. Once you reach this situation, you can no longer pump gradually, adding a little energy with each oscillation. Instead, you need to inject all of the excess energy required to surpass the horizontal anew on every half cycle. It becomes hard work. [Plus, you are spending the next quarter cycle trying to get back to a stable orientation when you should have been prepping for the pump action at the bottom -- it keeps you busy dancing on the edge of control]

 

[ChiralSuperfields]

A swing normally hangs straight down. The chains are vertical. "Vertical" is in the direction of gravity. "Horizontal" is perpendicular to this.

As you swing back and forth the angle (measured from the vertical center line) of the chains reaches larger and larger peaks depending on how hard you pump.

If you pump hard enough, the peaks can reach 90 degrees. That is, the chains at the top of the arc are horizontal.

For a child swinger, problems arise when you pass the horizontal. When you reach the top of the arc and swing back down, the chains will have gone slack. You will fall downward on slack chains until they come taut again in an inelastic collision. The collision is jarring and, hence, dangerous. This collision also absorbs energy. Once you reach this situation, you can no longer pump gradually, adding a little energy with each oscillation. Instead, you need to inject all of the excess energy required to surpass the horizontal anew on every half cycle. It becomes hard work. [Plus, you are spending the next quarter cycle trying to get back to a stable orientation when you should have been prepping for the pump action at the bottom -- it keeps you busy dancing on the edge of control]

Thank you for your help @jbriggs444 ! I think I understand about what you were meaning now :)

 

[haruspex]

Thank you for very much your reply @haruspex !

Sorry, I am having trouble visualizing this situation, could you kindly draw a diagram? I think this will also help me understand why angular momentum is not conserved in this situation

Many thanks!

During the part of the process I described, the ant running to one corner, I took angular momentum as conserved. But the result is that the mass centre of the ant+bob is displaced from being vertically below the pivot. So now gravity exerts a torque about the pivot.

 

[Jake357]

Homework Statement:: Please see below
Relevant Equations:: Please see below

For this problem,
View attachment 322398
The answer is $-4.70 kJ$. I am not sure what I am doing wrong.

My working is
View attachment 322399
$W = mgr\cos\theta$
$W = mgr\cos150$ (since angle between $\vec g$ and $\vec r$ is 150 degrees)
$W = -mgr\frac {\sqrt{3}}{2}$
$W = -mgr\frac {\sqrt{3}}{2}$
$W = (-80)(9.81)(12\sin60)(\frac {\sqrt{3}}{2})$
$W = -7063.2 J$

Would some please be to offer some guidance?

Many thanks!

The work done is proportional to the difference in the height from the ground, which means that
W=mg(h2-h1), which is also the difference in the potential energy of Spiderman.
The difference in height is: l-l cos 60=12-12 cos 60=6 m.
[Solution redacted by the Mentors]

 

[ChiralSuperfields]

During the part of the process I described, the ant running to one corner, I took angular momentum as conserved. But the result is that the mass centre of the ant+bob is displaced from being vertically below the pivot. So now gravity exerts a torque about the pivot.

Ok thanks you for your reply @haruspex!

That helps a bit more :) I will draw a diagram of the situation you describe then post it soon. I find diagrams quite helpful for understanding the setup.

Many thanks!

 

[ChiralSuperfields]

The work done is proportional to the difference in the height from the ground, which means that
W=mg(h2-h1), which is also the difference in the potential energy of Spiderman.
The difference in height is: l-l cos 60=12-12 cos 60=6 m.
W=mg(h2-h1)=6*9.8*80=4704 J=4.704 kJ

Thank you for your reply @Jake357 !

 

[ChiralSuperfields]

Third method to solve the work done by spider man:

I think we can solve this using the work integral with respect to theta and have the limits of integration as the initial angle spider man makes with the vertical and the finial angle spider man moves with the vertical.

I think we have to integrate with respect to theta, because for each differential displacement of spider man along his path the differential angle between spider man's weight and his differential displacement vector is not the same.

$W = \int_{\theta_i}^{\theta_f} mgr\cos\theta dr$
$W = \int_{\theta_i}^{\theta_f} mgr^2\cos\theta~d\theta$

In progress...

Many thanks!

 

[haruspex]

$W = \int_{\theta_i}^{\theta_f} mgr\cos\theta dr$

Please define r and θ in that equation.

 

[ChiralSuperfields]

Please define r and θ in that equation.

Thank you for your reply @haruspex!

r is the radius to where the rope hangs from so is the length of the rope
$\theta$ is the angle the rope makes with the vertical

Many thanks!