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Work example wrong in my text? (it's driving me bonkers!)

by User1247
Tags: bonkers, driving, text, work
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User1247
#1
May5-07, 04:16 PM
P: 17
1. The problem statement, all variables and given/known data
A textbook shows an example of work. Either my understanding of Work and how to calculate it is wrong, or the text is wrong, I am going nutty here and not sure which is wrong, I think it is the text but, then again, I am biased.

The example goes as follows: (between the -----)
(I also included a photo of the example as an attachment)
-----
A rollerskater of 45kg coasts down a frictionless hill. The hill has an angle of 10 degrees to the horizontal. The horizontal distance from top to bottom is 40m.

A Free Body Diagram Shows Fg vertically downwards and Fgh, perpendicular to Fg, as a horizontal vector. Fgh is here defined as "the component of force of gravity acting in the direction of motion."

Calculations follow under the diagram:
Theta = 10 degrees, Δd = 40m, g=9.8N/kg, m=45kg

W=Fgh x Δd
W=Fg Cos(Theta) x Δd
W=mg Cos(Theta) x Δd
W=45kg x 9.8N/kg x Cos10 x 40m
W=1.7x10^4 J
-----

2. Relevant equations
W=FΔd


3. The attempt at a solution
I would address the problem like this:

The only two forces acting are(for this problem, no friction or air)
Fg is force of gravity, downwards
Fn is normal force, perpendicular to and upwards from hill surface

Split Fg into two components:
Fgperp is the component of Fg that is perpendicular to the hill surface
Fgpar is the component of Fg that is parallel to the hill surface and downhill

|Fgperp|=|Fn| and they are in opposite directions so they cancel out.
This leaves Fgpar as the only remaining contributor to the net force.

Finding Fgpar:
Fgpar=Fg x Sin(10)
Fgpar=(45kg)(9.8N/kg) x Sin(10)
Fgpar=76.56N

Finding Displacement:
The question clearly says "inclination of the hill is 10 degrees and the horizontal distance from top to bottom is 40m"
So we can say that the hill surface is the hypotenuse of a right-angled triangle with theta=10 and adjacent or x length is 40m. Δd is the hill surface and hypotenuse side.

Since:
Cos(theta)=x/r
rCos(theta)=x
r=x/Cos(theta)

Therefore:
Δd=40m/Cos(10)
Δd=40.62m

Now to find the work done:
Fgpar is in the direction of the displacement.
W=FΔd
W=Fgpar x Δd
W=76.56N x 40.62m
W=3109.87Nm

Is my solution right and the book's wrong? Am I totally missing the boat somehow? Please detail to me what I have misunderstood, if anything.

Thanks to anybody who answers, and please address the conceptual side, I have no difficulty with the calculation parts (apart from thousands of careless mistakes).
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denverdoc
#2
May5-07, 04:31 PM
P: 1,350
It could have been found more directly from the difference in potential energy since gravity is a conservative force and the path is irrelevant, but still looks fine to me.

as in mg(delta Y)=3110.4

Edit: Actually I misread the problem;
the "horizontal distance is 40"--what a crazy way to phrase it, but no nevermind, it means
h/40=sin 10, not tan 10 as I supposed= Result, now 3063, in agreement with the last post
User1247
#3
May5-07, 06:28 PM
P: 17
Ok, so I take it that the book was just out to lunch and I should not injure my brain by devising various twisted understandings of work and interpretations of the question in an attempt to reconcile my understanding of work with what the book says? I'm just now learning about work so when the example in my text that illustrates to me how work works is wrong it's confusing.

denverdoc
#4
May5-07, 11:07 PM
P: 1,350
Work example wrong in my text? (it's driving me bonkers!)

Edit: the approach of the book is ok, for frictional work its the only way to solve the problem. When you learn about the difference between conservative and non-conservative forces, you will know when you can take short cuts.
husky88
#5
May5-07, 11:40 PM
P: 79
The complete definition of work is
W= F*d*cos theta
where F is the magnitude of the constant force
d is the magnitude of the displacement of the object
and theta is the angle between the directions of the force and the displacement

In a straight line, work is defined as F*d because cos theta = cos 0 = 1

Your reasoning is correct, but more complicated. The easier way to do it is W=F*d*cos theta, just use this formula.
For an incline actually, W = mgh. (Where h is the change in elevation, equal to d*cos theta, same thing.)

The mistake the book does is that it says the angle of inclination of the hill is 10. Then is uses this 10 to calculate the value of cos from our formula. But instead, they should take the angle between work and distance, which is 90-10=80.
So W= F*d*cos theta
=mg*d*cos theta
=(45)*(9.80)*(40)*cos 80
= 3063.15

Which is close to your result. You are correct. :)

I don't think there should be a negative value for the result because d is positive. d is defined from the problem, motion from left to right as positive. In general, the work done by a force is negative whenever the force acts in the direction opposite to the direction of motion.
User1247
#6
May6-07, 12:35 PM
P: 17
Hmm, lol, I am now possibly more confused than I was at the beginning.

R1[Denverdoc]: "It could have been found more directly ... but still looks fine to me."
My response: "it" is my solution?

R3[Denverdoc]: "Edit: the approach of the book is ok, for frictional work its the only way to solve the problem."
My response: The book says that the hill is frictionless, so I'm still lost. There was no coefficient of friction given or anything so how did they incorporate friction?

Husky88: Ok, the way you did the problem makes sense to me; finding the component of the force parallel to the displacement by Cos(theta) where theta is the angle between the applied force and the displacement direction. One thing I still do not understand is how my original method in my original post does not end up giving the same answer as yours. As far as I can see, my method should also have given the net force (this net force is in the same direction as the displacement). Why is this net force not the same as the force found by using F*d*Cos(Theta)?

Thanks to all.
denverdoc
#7
May6-07, 01:25 PM
P: 1,350
Sorry, I was unclear as I hadn't looked at the thumbnails the first pass thru, what I meant to say, the books answer appears to be wrong, that the approach the book had taken was ok, if somewhat indirect in the case of a conserved force, and that it should have used the sine of 10 degrees to compute fgh. The small discrepancy between your answer and huskies owes to the same mistake I made, misreading what is meant by the horizontal distance--the hypotenuse is 40.
AngeloG
#8
May6-07, 01:37 PM
P: 103
No, you use the inclination of the hill =p. The work book is correct.

d*cos(theta) = height. mgh -> (45)(-9.8)*[0-40cos(10)] = 1.7x10^4 J (two sig figs)
nrqed
#9
May6-07, 02:21 PM
Sci Advisor
HW Helper
P: 2,957
Quote Quote by User1247 View Post
1. The problem statement, all variables and given/known data
A textbook shows an example of work. Either my understanding of Work and how to calculate it is wrong, or the text is wrong, I am going nutty here and not sure which is wrong, I think it is the text but, then again, I am biased.

The example goes as follows: (between the -----)
(I also included a photo of the example as an attachment)
-----
A rollerskater of 45kg coasts down a frictionless hill. The hill has an angle of 10 degrees to the horizontal. The horizontal distance from top to bottom is 40m.

A Free Body Diagram Shows Fg vertically downwards and Fgh, perpendicular to Fg, as a horizontal vector. Fgh is here defined as "the component of force of gravity acting in the direction of motion."

Calculations follow under the diagram:
Theta = 10 degrees, Δd = 40m, g=9.8N/kg, m=45kg

W=Fgh x Δd
W=Fg Cos(Theta) x Δd
W=mg Cos(Theta) x Δd
W=45kg x 9.8N/kg x Cos10 x 40m
W=1.7x10^4 J
-----

2. Relevant equations
W=FΔd


3. The attempt at a solution
I would address the problem like this:

The only two forces acting are(for this problem, no friction or air)
Fg is force of gravity, downwards
Fn is normal force, perpendicular to and upwards from hill surface

Split Fg into two components:
Fgperp is the component of Fg that is perpendicular to the hill surface
Fgpar is the component of Fg that is parallel to the hill surface and downhill

|Fgperp|=|Fn| and they are in opposite directions so they cancel out.
This leaves Fgpar as the only remaining contributor to the net force.

Finding Fgpar:
Fgpar=Fg x Sin(10)
Fgpar=(45kg)(9.8N/kg) x Sin(10)
Fgpar=76.56N

Finding Displacement:
The question clearly says "inclination of the hill is 10 degrees and the horizontal distance from top to bottom is 40m"
So we can say that the hill surface is the hypotenuse of a right-angled triangle with theta=10 and adjacent or x length is 40m. Δd is the hill surface and hypotenuse side.

Since:
Cos(theta)=x/r
rCos(theta)=x
r=x/Cos(theta)

Therefore:
Δd=40m/Cos(10)
Δd=40.62m

Now to find the work done:
Fgpar is in the direction of the displacement.
W=FΔd
W=Fgpar x Δd
W=76.56N x 40.62m
W=3109.87Nm

Is my solution right and the book's wrong? Am I totally missing the boat somehow? Please detail to me what I have misunderstood, if anything.

Thanks to anybody who answers, and please address the conceptual side, I have no difficulty with the calculation parts (apart from thousands of careless mistakes).
Your solution is perfectly good if we assume that the 40 meters is indeed the horizontal distance.

The book is completely wrong.

(First, the "free-body diagram" is completely confused. If the force of gravity is shown as down, it's impossible to have a component straight to the right!!)

I can't make sense of their calculation. I am trying to guess what they had in mind when they calculated [itex] mg cos \theta ~\Delta d [/itex] but I can't make sense of it. Even assuming that they are using [itex] \Delta d [/itex] to be the length along the plane, their solution makes no sense, because then the solution would be [itex] mg sin \theta ~\Delta d [/itex]. I think that whoever wrote the solution was just completely confused.

Notice that a second way to do the calculation is to use the work to be the change of potential energy, [itex] mg \Delta h [/itex] where [itex] \Delta h [/itex] is the change of vertical height. Again, using the 40 meters to be purely horizontal, the change of vertical height is [itex] \Delta h = tan(\theta) \Delta d [/itex] and one gets back your result.

Hope this helps.

Patrick

EDIT: I just read Husky88's post and he made me understand where the book made its mistake. Indeed, they use the wrong angle: the way they did it, they should have used 80 degrees. Their "free-body diagram" (which is not a free-body diagram at all) is still completely wrong.
AngeloG
#10
May6-07, 02:25 PM
P: 103
Horizontal distance from top to bottom; refers to the distance that the object travels on the incline. Rather than the bottom of the triangle. The actual distance it travels is 40m, however, we want the height. So we just do the cos(theta) of it, cos(theta) = adjacent/hypotenuse. adjacent = cos(theta)*hypotenuse.
nrqed
#11
May6-07, 02:30 PM
Sci Advisor
HW Helper
P: 2,957
Quote Quote by AngeloG View Post
Horizontal distance from top to bottom; refers to the distance that the object travels on the incline. Rather than the bottom of the triangle. The actual distance it travels is 40m, however, we want the height. So we just do the cos(theta) of it, cos(theta) = adjacent/hypotenuse. adjacent = cos(theta)*hypotenuse.
Well, it's a question of interpretation but I would say that the " HORIZONTAL distance from top to bottom" is the horizontal distance between two points aligned with the initial and final points.

But EVEN if that was the correct interpretation, the final result would still be wrong!!

Your mistake is that the height is NOT the adjacent side, it's the OPPOSITE side, so a sine would have to be used.

I really think that Husky put his finger on the mistake of the book: they used 10 degrees where they should have used 80 degrees.
AngeloG
#12
May6-07, 02:51 PM
P: 103
Through similar triangles.

{Fx = mgsin(theta) = ma
{Fy = N - Wcos(theta) = 0 (no accel in the y); N= Wcos(theta). The theta is 10 degrees.
User1247
#13
May6-07, 03:23 PM
P: 17
nrged: Glad to hear someone say that the book is totally lost.
I keep looking at the problem and it seems more obvious that the book is totally off.

As for interpretation of the wording of the problem regarding what the 40m is:
The book says: "the -horizontal- distance from top to bottom is 40m". Which makes it seem very clear to me that 40m is the horizontal distance, and so, x or adjacent side of a triangle with theta of 10 degrees. This means that the path along the hypotenuse, r, or down the face of the hill side has to be 40m/Cos(10)=40.617m.

If we try to claim that 40m is the total displacement, that is, the distance of the line segment on the 10 degree angle or the actual path followed by the skater, then how can we say "40m is the -horizontal- distance from top to bottom" if the line is at 10 degrees to horizontal? A line at 10 degrees to horizontal is not horizontal!, so this interpretation does not make sense at all.

So, if we say that total displacement (call it r on the triangle) is 40m/cos10, I am convinced, more than of anything else, that the work done is:
x=40m
r=40m/cos10
y=40m * tan10
y=Δh (change in height)

W=mgΔh

W=(45kg)(9.8N/kg)(40m*tan10)
W=3110.4J

AngeloG: I'm not following you; "(no accel in the y)". If something is rolling down a frictionless hill then it will be accelerating in x and y directions, how did you get no acceleration in y?

Thanks again everyone.
AngeloG
#14
May6-07, 04:28 PM
P: 103
\ is the Y, / is the x. There is no acceleration in the Y, as this object is fixed on the ramp. If it had acceleration in the Y, it would be moving up or down. However, it is not.

Through similar triangles, the {Fy = N = mgcos(theta).
User1247
#15
May6-07, 05:09 PM
P: 17
AngeloG: Thanks for trying to help but I think you are reading the wrong question or something. The object is not fixed to anything, it is moving to the side and down. It is also accelerating to the side and down. Since there is no friction and there is a net force, it is indeed accelerating in the direction of the hill's surface and downwards. If you set up your solution more carefully instead of skipping 90% of the steps you would also find this to be true.
AngeloG
#16
May6-07, 05:30 PM
P: 103
Calling the X and Y on the diagonals; there is no acceleration in the Y.

User1247
#17
May6-07, 06:42 PM
P: 17
AngeloG: Okay, if we choose to define x to be parallel to the hill's surface I now see that there is no acceleration or movement in the y direction from that perspective. I am new to work but here is what I see wrong with your solution:

You use:
ΣFy=0 (which I agree with given your definition of x direction)
W=ΣFΔd (which is true for a force in the same direction as the displacement)

Then you go:
W=ΣFyΔd

and that is where you made your mistake, work is only done in the same direction as the displacement, when a force acting is perpendicular to the displacement, it does no work at all. Also, that normal force is not the net force in your y direction, that normal force is exactly canceled out by a component of the gravity vector parallel and in the opposite direction of the normal.
husky88
#18
May6-07, 10:25 PM
P: 79
As someone already pointed out, the difference between our results is due to the sentence: "The horizontal distance from top to bottom is 40m."

A You could say that it is horizontal, so it has to be the straight line at the bottom of the hill.
B You could also say that, going from top to bottom the horizontal for the top-bottom is 40.

That is confusing, but not all problems are that bad. :)


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