Tensions and pulleys doubt.

[JD7]

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My first question is that my answer is B but apparently the correct answer is D for exercise 14.

http://books.google.com.pr/books?id... in the ramp and pulley system shown?&f=false

And my second question is

tractor pulls a log that has a mass of 500 kg along the ground for 100 m. The rope (between the tractor and the log) makes an angle of 30° with the ground and is acted on by a tensile force of 5,000 N. How much work does the tractor do? (sin 30° = 0.5, cos 30° = 0.866, tan 30° = 0.57)

My answer is 250kJ and apparently the answer is 450kJ



This exercises are from the kaplan book, and I'm being very stubborn with these two exercises and I think Kaplan is wrong...

 

[milski]

Q1: The link does not show the book for me - it says that the page is unavailable or I've reached my limit of pates to view. So not much that I can add.

Q2: The displacement is in horizontal direction. The tension force is at 30 degrees with the horizontal, so the horizontal component will be 5000 N * cos(30) = 5000 N * 0.866 = 4330 N. The work done will be 100 m * 4330 N = 433 kJ which is about 450 kJ when you round it.

 

[chiddler]

since the bottom pulley is moving downward with 100 N force, it's pulling two ropes down so that the F applied to the block is 2*100 = 200 N. 200 - mgsin(theta) = your answer

i think with both the questions you're asking, you are mixing up when to use sin and cos. I know how to use them, but i've no idea how to explain it because it seems to idiosyncratic to me. Sorry I can't help there.

 

[milski]

Are you sure D is the correct answer?

The force to the left is 20 * g * sin(30) = 100 N
The force to the right is 50 N - you have 100 N pulling on two ropes, which makes 100N/2 = 50N for each of them. (Unless I'm misreading how the ropes are attached).
Net force is 100N-50N = 50N to the left.
a = F/m = 50N/20kg = 2.5 m/s, so about 3 m/s to the left, down the ramp => it's A.

 

[JD7]

Q1: The link does not show the book for me - it says that the page is unavailable or I've reached my limit of pates to view. So not much that I can add.

Q2: The displacement is in horizontal direction. The tension force is at 30 degrees with the horizontal, so the horizontal component will be 5000 N * cos(30) = 5000 N * 0.866 = 4330 N. The work done will be 100 m * 4330 N = 433 kJ which is about 450 kJ when you round it.

Wouldn't you need to substract the force acting on the box which is Fincline = Fn*sin(angle)?

The force to the right is 50 N - you have 100 N pulling on two ropes, which makes 100N/2 = 50N for each of them. (Unless I'm misreading how the ropes are attached).

That's what I find confusing, I did the same thing 100/2 instead of how they say to multiply it by 2 which doesn't make sense at all.

 

[JD7]

since the bottom pulley is moving downward with 100 N force, it's pulling two ropes down so that the F applied to the block is 2*100 = 200 N. 200 - mgsin(theta) = your answer

i think with both the questions you're asking, you are mixing up when to use sin and cos. I know how to use them, but i've no idea how to explain it because it seems to idiosyncratic to me. Sorry I can't help there.

So you are telling me that Fa = 100N + 100N != 100N? Since each rope is doing 100N in ur answer.

 

[milski]

Wouldn't you need to substract the force acting on the box which is Fincline = Fn*sin(angle)?

Subtract it from what? F*sin(angle) in that case is the vertical component and has no effect on the work done by the tractor.

That's what I find confusing, I did the same thing 100/2 instead of how they say to multiply it by 2 which doesn't make sense at all.

Yes, I'm with you on the 100/2. It does not make sense otherwise. The drawing is slightly weird but my assumption is that the left end of the rope is attached to the stationary part of the top pulley and the right side just goes over it.

 

[chiddler]

Why does the force down by the pulley cause two 50 N forces and not two 100 N forces?

 

[milski]

Why does the force down by the pulley cause two 50 N forces and not two 100 N forces?

The forces up are result from the normal of the pulley on the rope. Since that is a 100 N, the sum of the two tension forces will be a 100 N. Think about a mass hanging from the ceiling on two ropes.

 

[MedPR]

Wouldn't you need to substract the force acting on the box which is Fincline = Fn*sin(angle)?



That's what I find confusing, I did the same thing 100/2 instead of how they say to multiply it by 2 which doesn't make sense at all.

Work is Fdcostheta where theta is the angle between the force and the displacement, which is 30 degrees in this example.

The force doing work on the box is the horizontal component of the force given, which is Fcos30 which is 5000*.86.

I guess you could think about subtracting the vertical from the horizontal, but since the vertical component of the 5000N force does 0 work (it does not displace the box since the box does not lift up off the ground), you still get 5000*.86 as your answer.

 

[JD7]

Subtract it from what? F*sin(angle) in that case is the vertical component and has no effect on the work done by the tractor.


Yes, I'm with you on the 100/2. It does not make sense otherwise. The drawing is slightly weird but my assumption is that the left end of the rope is attached to the stationary part of the top pulley and the right side just goes over it.

Substract it because the 5,000N is not the NET force. There's always gravity acting upon the box so the tractor's acceleration will decrease thus the force is less than 5,000N.

 

[MedPR]

Substract it because the 5,000N is not the NET force. There's always gravity acting upon the box so the tractor's acceleration will decrease thus the force is less than 5,000N.

Gravity doesn't matter when there is no vertical motion.

The net force on the box in this case is 5000*.86 assuming there is no friction.

 

[milski]

Substract it because the 5,000N is not the NET force. There's always gravity acting upon the box so the tractor's acceleration will decrease thus the force is less than 5,000N.

The questions is how much work does the tractor do. The only force that the tractor exerts is 5000 N in a direction of 30 degrees up from the horizontal. The displacement is horizontal. You can either find the horizontal component of the force and get the work as F*d, which ends up being F*cos(30)*d. Or you can use the F.d formula, where F and d are vectors and . is a dot product. Either way, you'll get the same result.

Any other forces on the log may have effect on how much total work was done on it but do not affect how much work the tractor does on the log.

 

[MedPR]

Your answer of 250kJ would be right if theta was 60.

 

[JD7]

The questions is how much work does the tractor do. The only force that the tractor exerts is 5000 N in a direction of 30 degrees up from the horizontal. The displacement is horizontal. You can either find the horizontal component of the force and get the work as F*d, which ends up being F*cos(30)*d. Or you can use the F.d formula, where F and d are vectors and . is a dot product. Either way, you'll get the same result.

Any other forces on the log may have effect on how much total work was done on it but do not affect how much work the tractor does on the log.

Yeah the tension is the only force acting on the tractor. I think having the mass of the box confused me. I was thinking of it in terms of acquiring the work on the box if there was a non-net tensile force on a frictionless surface where you would be required to calculate the incline force of gravity.

 

[milski]

Yeah the tension is the only force acting on the tractor. I think having the mass of the box confused me. I was thinking of it in terms of acquiring the work on the box if there was a non-net tensile force on a frictionless surface where you would be required to calculate the incline force of gravity.

Yes, it seems that having extra information which is not needed is normal for the MCAT. Which is a fair way to test understanding.



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[MedPR]

Yeah the tension is the only force acting on the tractor. I think having the mass of the box confused me. I was thinking of it in terms of acquiring the work on the box if there was a non-net tensile force on a frictionless surface where you would be required to calculate the incline force of gravity.

There is no incline force of gravity if the box isn't on an incline. The force is at an angle, not the displacement. When there is only x-axis motion, the force of gravity is irrelevent because only the x-component of all forces should be considered and gravity does not have an x-component.

 

[milski]

Even if all this was happening on an incline, the result would still be the same. The tension force is the only force that tractor is exerting on the log. Being on an incline will change how much of this force is opposed by gravity vs friction but if the force and angle between the rope and the surface stay the same, the work will still be the same.

 

[MedPR]

Even if all this was happening on an incline, the result would still be the same. The tension force is the only force that tractor is exerting on the log. Being on an incline will change how much of this force is opposed by gravity vs friction but if the force and angle between the rope and the surface stay the same, the work will still be the same.

If the tractor was pulling the box up a 30 degree incline, wouldn't the work be Fdsintheta?

 

[milski]

If the tractor was pulling the box up a 30 degree incline, wouldn't the work be Fdsintheta?

No, it will still be F*d*cos(30). Note that now the displacement is not horizontal but parallel to the 30 degree incline. The force parallel to the incline is F*cos(30), so you end up with the same result.

That should be fairly easy to see from the dot product formula for the work: F.d=|F|*|d|*cos(angle(F,d)). It shows that the work depends only on the magnitude of the force, the magnitude of displacement and the angel between them.

 

[MedPR]

No, it will still be F*d*cos(30). Note that now the displacement is not horizontal but parallel to the 30 degree incline. The force parallel to the incline is F*cos(30), so you end up with the same result.

That should be fairly easy to see from the dot product formula for the work: F.d=|F|*|d|*cos(angle(F,d)). It shows that the work depends only on the magnitude of the force, the magnitude of displacement and the angel between them.

What..?

 

[milski]

T is given to you already - it's 5000N * cos(30). Displacement is along the x-axis as well and is given to be 100 m. That's all you need to calculate the work.

The rest of the diagram would be useful if you were given the incline, friction, the mass of the block hanging from the right. That would allow you to calculate T. But in this problem T is already given as constant, so none of these matter.

What program did you use to draw this? I would love to be able to post diagrams, since they can be much clearer than a few paragraphs of text. But it takes forever to draw even something remotely nice looking.

Edit: When I talk about x-axis, I mean the coordinate system draw on the left of the diagram. I did not notice that there were two of them. 🙂

 

[MedPR]

T is given to you already - it's 5000N * cos(30). Displacement is along the x-axis as well and is given to be 100 m. That's all you need to calculate the work.

The rest of the diagram would be useful if you were given the incline, friction, the mass of the block hanging from the right. That would allow you to calculate T. But in this problem T is already given as constant, so none of these matter.

What program did you use to draw this? I would love to be able to post diagrams, since they can be much clearer than a few paragraphs of text. But it takes forever to draw even something remotely nice looking.

Edit: When I talk about x-axis, I mean the coordinate system draw on the left of the diagram. I did not notice that there were two of them. 🙂

I just googled for an incline plane 🙂

 

[milski]

I just googled for an incline plane 🙂

That explains why your "tractor" went off the cliff. 😀

 

[Pktgresch]

since the bottom pulley is moving downward with 100 N force, it's pulling two ropes down so that the F applied to the block is 2*100 = 200 N. 200 - mgsin(theta) = your answer

i think with both the questions you're asking, you are mixing up when to use sin and cos. I know how to use them, but i've no idea how to explain it because it seems to idiosyncratic to me. Sorry I can't help there.

I am still confused by this problem...

 

[Ssina]

I am still confused by this problem...

same here...

shouldn't the formula be something like this:

Fgravity down - Tension up = ma
m(g)sinθ - T = m(a)sinθ
20(10)(.5)-50 = 20 (x) (.5)
so x=a= 5m/s2 going down

I think the acceleration should be still sinθ since if there was no incline plane, we would be just lifting the box or lowering it... but here we are also moving in the x direction so it will take some of the force and it will be lower and moves to the left.

with that said, why 50N for the tension.... 100N down on the first rope, then splits into 2 (50N on each rope), and then back to one rope again!? I guess you can say one of the ropes is extended to the other side, so only 50N, but where did the other 50N go!? to pull the first pulley up!?

 

[milski]

same here...

shouldn't the formula be something like this:

Fgravity down - Tension up = ma
m(g)sinθ - T = m(a)sinθ
20(10)(.5)-50 = 20 (x) (.5)
so x=a= 5m/s2 going down

I think the acceleration should be still sinθ since if there was no incline plane, we would be just lifting the box or lowering it... but here we are also moving in the x direction so it will take some of the force and it will be lower and moves to the left.

But the force that you calculated is already parallel to the incline . In other words, it's parallel to the direction of the acceleration - there is no need to multiply the right side by sinθ.

with that said, why 50N for the tension.... 100N down on the first rope, then splits into 2 (50N on each rope), and then back to one rope again!? I guess you can say one of the ropes is extended to the other side, so only 50N, but where did the other 50N go!? to pull the first pulley up!?

Yes, the diagram sort of sucks. The left side of the rope stops at one of the stationary parts of the top pulley. The 50 N associated with it are exerted on whatever is supporting the pulley in place.

 

[SaintJude]

So--let me just verify. Basically, the answer for the pulley question is

F(applied) = 100 N = 20 kg (a). So a = 5 m/s^2

 

[Ssina]

But the force that you calculated is already parallel to the incline . In other words, it's parallel to the direction of the acceleration - there is no need to multiply the right side by sinθ.

you're right, I didn't draw the forces!

 

[milski]

So--let me just verify. Basically, the answer for the pulley question is

F(applied) = 100 N = 20 kg (a). So a = 5 m/s^2

No, 50 N = 20 kg * a, a = 2.5 m/s^2, to the left, down the ramp. The sin(theta) on the right side should not be there.

you're right, I didn't draw the forces!

MedPR's drawing from a few posts above works quite well for this problem too. 😉

 

[SaintJude]

So you're saying, Kaplan is wrong? I mean, of course it's possible, but the net force the box feels is 100 N since the rope of the pulley "condenses" the two tensions together. Isn't it? Maybe I should go ask my kaplan teacher....

 

[milski]

So you're saying, Kaplan is wrong? I mean, of course it's possible, but the net force the box feels is 100 N since the rope of the pulley "condenses" the two tensions together. Isn't it? Maybe I should go ask my kaplan teacher....

Well, that's the thing. The way that the pulley is drawn it is not very clear what they mean. The only logical arrangement is that the rope from the right side goes over the top of the pulley while the left end is attached to a stationary part of the pulley. If that is the case, the tension is only 50 N.

For the tension from both parts of the rope to translate to the right you'll need another pulley to change the direction.

But yes, if Kaplan claims that the answer is different, I would love to hear their reasoning.

 

[chiddler]

pretty sure you are correct. i asked an engineer to verify the answer :3

 

[JD7]

"12. D To displace the box a distance x, a length of rope equal to 2x must be pulled through the pulley. If a force of 100 N is applied on the rope, the pulley system will magnify the force on the box by a factor of 2. Therefore, a force of 200 N is exerted on the box. Now you can set up the equations of motion. The forces acting on the box are 200 N up the ramp and mg sin 30° down the ramp. So ma = 200 N - mg sin 30° implies that a = 200 N/20 kg - g sin 30° = 5 m/s2 upward."

-Kaplan's illogical answer.

 

[milski]

"12. D To displace the box a distance x, a length of rope equal to 2x must be pulled through the pulley. If a force of 100 N is applied on the rope, the pulley system will magnify the force on the box by a factor of 2. Therefore, a force of 200 N is exerted on the box. Now you can set up the equations of motion. The forces acting on the box are 200 N up the ramp and mg sin 30° down the ramp. So ma = 200 N - mg sin 30° implies that a = 200 N/20 kg - g sin 30° = 5 m/s2 upward."

-Kaplan's illogical answer.

Yes, that's pretty far out there. 👎 The box is directly tied to the rope - you have to displace x length of rope for x displacement of the box. You have to displace the bottom pulley by x/2 to move the box for displacement of x, which is consistent with halving the force, not doubling it.

Do they have some sort of errata for their books? I would expect typos and similar erors but they're outright wrong here.

 

[MedPR]

I still don't understand how the work done by pushing a box up an incline is defined by f=mgcostheta and not f=mgsintheta.. The force that does work is the force parallel to the displacement. mgsintheta is the force parallel to the displacement up or down an incline, while fgcostheta is perpendicular...

 

[milski]

I still don't understand how the work done by pushing a box up an incline is defined by f=mgcostheta and not f=mgsintheta.. The force that does work is the force parallel to the displacement. mgsintheta is the force parallel to the displacement up or down an incline, while fgcostheta is perpendicular...

That's for the tractor problem, right? The theta in the answer is the angle between the rope and the incline. The angle of the incline is not part of the solution at all.

In my (ugly) picture the black line is the rope and the blue one below is just parallel to the incline, to help you measure the angle. The attach point on the tractor is higher than the attach point of the log, so that the tension force is not strictly parallel to the incline.

 

[MedPR]

That's for the tractor problem, right? The theta in the answer is the angle between the rope and the incline. The angle of the incline is not part of the solution at all.

In my (ugly) picture the black line is the rope and the blue one below is just parallel to the incline, to help you measure the angle. The attach point on the tractor is higher than the attach point of the log, so that the tension force is not strictly parallel to the incline.

Oh I see, we were talking about two different problems 🙂 I understand how it's costheta based on that drawing, thanks 🙂!


The problem I was talking about was if the box was on a 30 degree incline and the rope between the box and the tractor was parallel to the incline.

 

[milski]

Oh I see, we were talking about two different problems 🙂 I understand how it's costheta based on that drawing, thanks 🙂!

It was your last post that finally tipped me that we're talking about different things. 🙂

The problem I was talking about was if the box was on a 30 degree incline and the rope between the box and the tractor was parallel to the incline.

If the tensions is 5000 N, the work in that case would be 5000 N * 100 m = 500kJ since both force and displacement are parallel to the incline.

 

[MedPR]

It was your last post that finally tipped me that we're talking about different things. 🙂



If the tensions is 5000 N, the work in that case would be 5000 N * 100 m = 500kJ since both force and displacement are parallel to the incline.

Oh.. Yea that makes more sense 😀

 

[JD7]

That image is wrong. There was never an incline haha that's what had me confused.

I was confused at first, but then I noticed that the angle is not reference to the ground but the rope.

 

[milski]

That image is wrong. There was never an incline haha that's what had me confused.

I was confused at first, but then I noticed that the angle is not reference to the ground but the rope.

I know, we were not talking about the initial problem anymore. Or at least I was not. We got sidetracked, talking about the fact that even on an incline the result would be the same.

 

[MedPR]

That image is wrong. There was never an incline haha that's what had me confused.

I was confused at first, but then I noticed that the angle is not reference to the ground but the rope.

Well the angle given is in reference to the ground. It's the angle between the rope and the horizontal motion (the ground).

 

[JD7]

Well the angle given is in reference to the ground. It's the angle between the rope and the horizontal motion (the ground).

incline - horizontal vs inclined ground angle

rope angle - horizontal vs angle of the rope

It's all matter of perspective, don't get all technical 😛

 

[MedPR]

incline - horizontal vs inclined ground angle

rope angle - horizontal vs angle of the rope

It's all matter of perspective, don't get all technical 😛

I just wanted to make sure you understood all the explanations in the thread.

🙂

 

[MedPR]

I'm sure someone can clear this up for me. From TBR:

If we have a short wavelength, then the frequency of that wave must be large. Converseley, if the wavelength is long, then the frequency of the wave must be short.

That makes sense, based on the equation, but I don't understand what short/long frequencies are.

So Frequency is the number of vibrations that occur during a specified amount of time. So does a "long" frequency mean a lot of vibrations occur over time? Or a few vibrations occur over the same time?

I've always thought of frequency as high or low.. Low frequency = less per unit time, high frequency = more per unit time. I'm not sure how to apply short/long..

 

[JD7]

Long = High, Short = Low. TBR used crappy and misleading analogies IMO. I don't think they meant anything complicated.

 

[MedPR]

Long = High, Short = Low. TBR used crappy and misleading analogies IMO. I don't think they meant anything complicated.

Ok thanks, that's what I figured. If velocity is high, but wavelength is short, that means the frequency must be high/long/fast.. I know fast probably isn't a great term to apply to frequency, but it makes it more logical to me 🙂 Thank you.

 

[MedPR]

I just realized that I didn't make a new thread for this like I thought I did.. Sorry.

 

[milski]

Will this thread ever die? 😀

And yes, using 'long frequency' for 'high frequency' should be considered abuse of the reader. Faster is much easier to swallow.