Finding the work done by Spiderman

[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.

 

[member 731016]

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

 

[member 731016]

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?)

 

[member 731016]

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!

 

[member 731016]

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 $\frac{mr-M\frac{mrR}{M(R+r)}}{M+m}=\frac{mr(1-\frac R{R+r})}{M+m}=\frac{mr^2}{(M+m)(R+r)}\approx \frac{mr^2}{MR}$.

 

[member 731016]

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.

 

[member 731016]

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]

 

[member 731016]

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]

 

[member 731016]

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!

 

[member 731016]

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 !

 

[member 731016]

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.

 

[member 731016]

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!

 

[haruspex]

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!

So r is a constant, but you integrate wrt it?

 

[member 731016]

So r is a constant, but you integrate wrt it?

Thank you for your reply @haruspex!

I think take it out of the work integral that will wrt to theta

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

I think take it out of the work integral that will wrt to theta

Many thanks!

Right.
Don't just throw integrals together that seem to have the right ingredients. Think what the integral is saying.
In the present case, you are considering spiderman moving as the rope swings, so the independent variable is the angle. As it swings through a small angle $d\theta$, Spiderman moves a distance (what?) at an angle (what?) to the vertical force (what?) thereby doing work (what?).

 

[member 731016]

Right.
Don't just throw integrals together that seem to have the right ingredients. Think what the integral is saying.
In the present case, you are considering spiderman moving as the rope swings, so the independent variable is the angle. As it swings through a small angle $d\theta$, Spiderman moves a distance (what?) at an angle (what?) to the vertical force (what?) thereby doing work (what?).

Thank you for your reply @haruspex ! That is good advice! I have tried to include my reasoning below.

Since we know that the when the spider man moves an infinitesimal displacement $\vec {ds} = dx\hat i + dy\hat y$ when a $F_g$ is applied by the earth

$dW = F_g \cdot \vec {ds}$

We know the path spider man travels though is a circular arc of length $S = r\theta$ (The arc length is dependent of the theta)

Given that the arc length of the path is $S = r\theta$, we take the derivative of arc length with respect to theta giving $ds = r\theta d\theta$

$dW = Fr\cos\theta d\theta$ Where $d\theta$ is the angle between the weight and displacement vector $\vec ds$ tangent to the circular path at each point

Ok so know we need to get $\cos\theta$ in terms of the angle the rope makes with the vertical $\phi$ (I just defined that variable)

However, do you please know how do you do that?

I think maybe from geometry (if I proved it correctly) $\theta = \phi$

Many thanks!

 

[haruspex]

the angle the rope makes with the vertical ϕ (I just defined that variable)

When you wrote S=rθ, weren’t you taking θ to be that variable?

 

[member 731016]

When you wrote S=rθ, weren’t you taking θ to be that variable?

Thank you for your reply @haruspex !

I think I changed the variables. I now take $\phi$ as the angle between the vertical and the rope and $d\theta$ as the angle between the force of gravity and differential displacement $\vec {ds}$

Many thanks!

 

[haruspex]

I think I changed the variables. I now take $\phi$ as the angle between the vertical and the rope and $d\theta$ as the angle between the force of gravity and differential displacement $\vec {ds}$

Then this is wrong:

Given that the arc length of the path is $S = r\theta$, we take the derivative of arc length with respect to theta giving $ds = r\theta d\theta$

$dW = Fr\cos\theta d\theta$ Where $d\theta$ is the angle between the weight and displacement vector $\vec ds$ tangent to the circular path at each point

Please update those equations using your new definitions.

Edit: just noticed your step from $S = r\theta$ to $ds = r\theta d\theta$ is also wrong.

 

[member 731016]

Then this is wrong:

Please update those equations using your new definitions.

Edit: just noticed your step from $S = r\theta$ to $ds = r\theta d\theta$ is also wrong.

Thank you for your reply @haruspex!

I see my mistake, it should be $ds = rd\theta$

I will post the update equations soon.

Many thanks!

 

[member 731016]

Here are the new equations @haruspex ,

We start with $dW = Fr\cos\theta d\theta$

And we must integrate from $\phi_1$ to $\phi_2$ so $dW = \int_{\phi_1}^{\phi_2} Fr\cos\theta d\theta$

Therefore, since $\theta$ depends on $\phi$ we must get $\theta$ in terms of $\phi$. I did this geometrically and proved that $\theta = \phi$.

Am I correct?

Many thanks!

 

[haruspex]

Here are the new equations @haruspex ,

We start with $dW = Fr\cos\theta d\theta$

And we must integrate from $\phi_1$ to $\phi_2$ so $dW = \int_{\phi_1}^{\phi_2} Fr\cos\theta d\theta$

Therefore, since $\theta$ depends on $\phi$ we must get $\theta$ in terms of $\phi$. I did this geometrically and proved that $\theta = \phi$.

Am I correct?

Many thanks!

If ϕ is the angle between the vertical and the rope then S=rφ.
When ϕ=0, what is the angle between $\vec ds$ and the gravitational force?

 

[member 731016]

If ϕ is the angle between the vertical and the rope then S=rφ.
When ϕ=0, what is the angle between $\vec ds$ and the gravitational force?

Thank you for your reply @haruspex !

True I did not realize that s =rφ.

I should then take the of the new arc length equation with respect to phi to get $\frac{ds}{d\phi} = r$

When $\phi$ is zero, then $\theta$ the angle between $\vec {ds}$ and $F_g$ is 90 degrees.

Are you trying to get me to find the relationship between $\phi$ and $\theta$ by considering a few physical situations?

If I consider the second case where $\phi$ is 90 degrees, then the angle between $\vec {ds}$ and $F_g$ is 180 degrees, correct?

Is there a way to find the relationship between $\phi$ and $\theta$ without considering intuitive cases along the circular path?

Many thanks!

 

[haruspex]

Is there a way to find the relationship between ϕ and θ without considering intuitive cases along the circular path?

Of course: draw the right diagram. The special cases are just an easy way to check your answer.
If you thought you had proved a different relationship, try to see where that proof went wrong.

 

[member 731016]

Of course: draw the right diagram. The special cases are just an easy way to check your answer.
If you thought you had proved a different relationship, try to see where that proof went wrong.

Thank you for your reply @haruspex !

Based on my special cases in post #81,

1. $\phi = 0$ then $\theta = 90$

2. $\phi = 90$ then $\theta = 180$

It looks to me like the relationship is $\theta = \phi+ 90$ so theta is always a phase shift of 90 degrees ahead of phi.

I will draw a diagram again to prove this and post it soon!

Many thanks!

 

[member 731016]

Here is are my diagrams @haruspex

Could you please give me some more guidance, I am still getting $\theta = \phi$

But I guess I'm meant to be looking at differential displacement so I can see here that $\theta = 180$

many thanks!

 

[haruspex]

Here is are my diagrams @haruspex
View attachment 322556
View attachment 322558

Could you please give me some more guidance, I am still getting $\theta = \phi$

But I guess I'm meant to be looking at differential displacement so I can see here that $\theta = 180$
View attachment 322560
many thanks!

You have drawn $\theta$ as the angle the final displacement makes to the vertical. You defined it as the angle the displacement element $\vec ds$ makes to the vertical at an intermediate point.
Draw a diagram with the rope at two angles to the vertical, $\phi$ and $\phi+d\phi$.
The displacement element is length $r.d\phi$. Note that it is at right angles to the rope.

 

[member 731016]

You have drawn $\theta$ as the angle the final displacement makes to the vertical. You defined it as the angle the displacement element $\vec ds$ makes to the vertical at an intermediate point.
Draw a diagram with the rope at two angles to the vertical, $\phi$ and $\phi+d\phi$.
The displacement element is length $r.d\phi$. Note that it is at right angles to the rope.

Thank you for your reply @haruspex !

I guess it sort of makes sense to consider incremental changes in phi if we are trying prove the change in theta. I will try that!

Many thanks!

 

[member 731016]

Here are the new diagrams @haruspex !

I'm assuming the displacement vector is only at a right angle to the rope where it extends from (the red line in this case). I'm not sure if the displacement vector makes at right angle with the dark orange line (finial rope position)

Many thanks!

 

[jbriggs444]

I'm not sure if the displacement vector makes at right angle with the dark orange line (finial rope position)

$ds$ is (or behaves like) an infinitesimal. The incremental displacement is infinitesimally close to being at right angles to both the initial and final angles. Its angle differs from the perpendicular by only $d \phi$ at most.

But that is window dressing. Surely you are after the incremental work done. This should be the vector dot product of the force of gravity, $\vec{mg}$ and the incremental displacement, $\vec{ds}$.

 

[member 731016]

$ds$ is (or behaves like) an infinitesimal. The incremental displacement is infinitesimally close to being at right angles to both the initial and final angles. Its angle differs from the perpendicular by only $d \phi$ at most.

But that is window dressing. Surely you are after the incremental work done. This should be the vector dot product of the force of gravity, $\vec{mg}$ and the incremental displacement, $\vec{ds}$.

Thank you for your reply @jbriggs444 !

Oh ok that makes sense, now. I have added that too my new diagram (which is very not too scale):

Yes I think I am after the incremental work ($dW = \int_{\phi_1}^{\phi_2} Fr\cos\theta~d\phi$) in terms of phi which I will integrate from $\phi_1 = 0$ to $\phi_2 = 60$ (Where $\theta$ is the angle between the differential displacement and spider man's weight).

Would you please know how to get theta in terms of phi from the diagram? I think (by considering two cases from post #83) that $\theta = \phi + 90$

Many thanks!

 

[member 731016]

I think I may have found the answer!!!

First method:

If we start with the big triangle,

Then,

$\phi + d\phi + 90 + \theta_2 = 180$
$\theta_2 = 90 - \phi - d\phi$

Now for the small triangle,

$90 + \theta_2 + \theta_1 = 180$
$\theta_1 = 90 - \theta_2$
$\theta_1 = 90 - (90 - \phi - d\phi)$
$\theta_1 = \phi + d\phi$

Therefore since $\theta = 90 + \theta_1$ (since $theta_1$ is vertically opposite)

Then $\theta = 90 + \phi + d\phi ≈ 90 + \phi$ since $d\phi$ is a differential

Second method:
I believe you could have also found $\theta_2$ from the medium size triangle,

Where in this case
$\theta_2 = 90 - \phi$
$\theta_1 = \phi + d\phi$
$\theta = 90 + \phi$ without having to make the approximation in the first method

Is my proof for $\theta = 90 + \phi$ geometrically correct?

Many thanks!

 

[jbriggs444]

Is my proof for θ=90+ϕ geometrically correct?

Yes. The concern with differentials is distracting and the drawings could be done more clearly -- in particular, labelling the angle you are calling $\theta$. But I agree with the conclusion.

It might make it easier to evaluate the resulting integral if you would write $\cos ( 90 + \phi )$ as $-\sin \phi$.

Personally, I would have looked at the formula for the projection of a vector normal to a [oriented] surface. That involves $\sin \theta$ where $\theta$ is the angle the vector makes with respect to the surface. In this case, the displacement vector makes an angle of $- \phi$ with respect to a horizontal surface. The projection thus involves $\sin -\phi = - \sin \phi$.

You can get the dot product of two vectors by taking the projection of one in the direction of the other and multiplying the magnitude of the one by the [signed] magnitude of the projection of the other.

Actually, I would have been sloppy with the sign convention, ignored the orientation of the surface, looked at $\sin \phi$ and then reasoned that the displacement and gravity are pointing generally away from each other and inverted the sign of the resulting path integral.

Actually, I would have recognized a conservative field, reasoned that all path integrals between a particular pair of endpoints across a conservative field will yield the same result and either picked a trivial path or computed the potential difference.

But we can definitely get to work evaluating the integral.

 

[member 731016]

Yes. The concern with differentials is distracting and the drawings could be done more clearly -- in particular, labelling the angle you are calling $\theta$. But I agree with the conclusion.

It might make it easier to evaluate the resulting integral if you would write $\cos ( 90 + \phi )$ as $-\sin \phi$.

Personally, I would have looked at the formula for the projection of a vector normal to a [oriented] surface. That involves $\sin \theta$ where $\theta$ is the angle the vector makes with respect to the surface. In this case, the displacement vector makes an angle of $- \phi$ with respect to a horizontal surface. The projection thus involves $\sin -\phi = - \sin \phi$.

You can get the dot product of two vectors by taking the projection of one in the direction of the other and multiplying the magnitude of the one by the [signed] magnitude of the projection of the other.

Actually, I would have been sloppy with the sign convention, ignored the orientation of the surface, looked at $\sin \phi$ and then reasoned that the displacement and gravity are pointing generally away from each other and inverted the sign of the resulting path integral.

Actually, I would have recognized a conservative field, reasoned that all path integrals between a particular pair of endpoints across a conservative field will yield the same result and either picked a trivial path or computed the potential difference.

But we can definitely get to work evaluating the integral.

Thank you so much for your reply @jbriggs444 !

Yeah true, there is a much easier method if I had recognized the conservative field! Sorry, what did you find distracting about my concern with differentials?

Was it the fact that first method I had to make an approximation and the second I did not (due to the choice of triangles)

I will post the integral

Many thanks!

 

[member 731016]

Here is the integration @jbriggs444 !

$W = -Fr\int_0^{60} \sin\phi d\phi$
$W = mgr\cos60= \frac {mgr}{2}$ which is correct!

Many thanks!

 

[jbriggs444]

Yeah true, there is a much easier method if I had recognized the conservative field! Sorry, what did you find distracting about my concern with differentials?

Well, a drawing of an isosceles triangle with two right angles is a bit distracting.

 

[member 731016]

Well, a drawing of an isosceles triangle with two right angles is a bit distracting.

Oh true! Thank for letting me know @jbriggs444 !