# basic physics - pulley question

Hi

This is the question:

A bucket is being raised by a single pulley by a height of 4m

O\
|     \
|      \
B      F   B=bucket, F=force

The weight of the bucket is 200N. The force exerted by the worker is 210N

How much work is done by the worker?

To calculate work all i know is force * distance travelled = 210 * 4m = 840J which is the answer in the book

Why is F bigger than the weight of the bucket or why is work done on bucket  < work done by worker?
I presume some of the energy is lost as friction. However the worker isn't pulling straight down to raise the bucket. Does this have anything to do with it? The book hasn't covered anything like this yet but i know from experience you can change the force needed by the angle you pull at.

thanks
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Commented:

> why is work done on bucket  < work done by worker?
probably because of friction in the pulley

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Commented:
> However the worker isn't pulling straight down to raise the bucket. Does this have anything to do with it?
It could have something to do with increasing friction, but in an idealized world of frictionless pulleys and massless springs, the change in direction has no effect on the work done.
Commented:
Hi,

If the worker was exerting exactly 200N upward on the bucket, the bucket wouldnt move, ie, the worker would only be supporting it hanging in the air, so to pull the bucket upwards, you have to apply a greater force than that which gravity is applying.
The worker is therefore applying a NET upward force, which is what is lifting the bucket. If you wanted to calculate the acceleration of the bucket, that is what you would use, as it is the NET upward force.
the work done by the worker is calculated using the force they are applying, and the distance moved.
Note: the NET upward force on the bucket exerted by the worker is not 10N. it will depend on the angle between the end they are pulling, and the rope attached to the bucket
Commented:
The bucket is hanging straight down due to gravity.

The worker is not pulling straight down, but this doesn't matter in this case.
The tension in the rope transmits the force from the worker to bucket.
The angle of the rope doesn't matter.

>> Why is F bigger than the weight of the bucket

If the applied force were equal to the weight of the bucket, it would not move.
The extra force may be required to overcome friction or it could accelerate the bucket.
Commented:

> If the applied force were equal to the weight of the bucket, it would not move.
It would not accelerate, but it would continue moving if it was already moving.

> The extra force may be required to overcome friction or it could accelerate the bucket.
or the weight of the rope...
\Commented:
Since there are different answers, do you need a tie-breaker? No! You can't take our words for it.
Here are my words. I agree with post http:#30513968. When you begin to lift a bucket that is stationary (even in a frictionless system), you temporarily have to have a force greater than the bucket's weight to get the bucket moving to some speed. Then you can drop your force down to the bucket's weight. When you do that, the tension force of the rope pulling the bucket upwards is the same as the bucket's weight. So, T - W = 0; that is, there is no net force on the bucket.

Since the bucket was moving upwards, and now the net force on the bucket is 0, then since F=mA, then A is also 0. An acceleration of 0 means that there is no change in speed over time. So, the speed is constant (moving upwards). When you decide that the bucket has reached its destination, you can relax the force below the bucket's weight temporarily (again, assuming a frictionless system) so that now the bucket is slowing down to a speed of 0. Then to hold the bucket stationary at that height, you need to apply a force equal to the weight of the bucket.

Interesting.. If there is friction, then although you needed an extra 10N to keep the bucket moving up at a constant speed, letting the bucket go down at constant speed would require only about 190N since now the friction force is in the opposite direction as when you were pulling the bucket up.
\Commented:
>> Then you can drop your force down to the bucket's weight.
Just to be clearer (because I wasn't) I am still talking about a frictionless system.
Author Commented:
Hi

If the angle has nothing to do with it (and I'm not disagreeing i'm just curious) why when i use a piece of equipment a bit like this at the gym is it easier to pull the weight  at a less steep angle (e.g more outwards rather than downwards). Perhaps that's just to do with how muscles work rather than the actual force needed
Commented:
When you pull down, you get to use more of your weight more efficiently.
But the tension in the rope required to move the weight does not depend on the angle.
Commented:
You are correct in your formula. Work - force over distance. Since it is a single pulley system then for the bucket to rise 4m then the worker had to pull 4m of rope/cable. 410 x 4 = 840J

The Bucket moved an equal distance (4m) and exerted a force of 400N for 800 Jules of work. Where did the other 40 Joules go?

The additional 10N of force applied to the aprox 20.34 KG of mass in the bucket (assuming massless and frictionless pully and ropes) accelerated the bucket. (Assuming zero initial velocity on the bucket.) That acceleration allowed the bucket to travel the necessary 4 m in aproximately 4 seconds (

I'll leave the math for calculationg mass, acceleration, and final velocity to the reader as an exercise). The end system had extra energy imparted into the bucket as potential energy (the work necessary to get to the new height) and also kinetic energy from motion. (40J or 1/2(M * V squared) This accounts for all the work done by the worker.

Without making assumptions about the system, as I did, this calculation can not be made. Add friction and inerta in the pully and some energy would be lost due to friction and some energy would be transfered to the pulley. A single 'true' answer can't exist without assumptions. However, In the end, all the extra 40J must go to kinetic energy, additional potential energy as the system slows, and/or heat from friction.  Laws of conservation of motion and energy.

BTW - the angle of pull has no effect on the equation. The vector forces of the bucket, the worker, and the support all add to zero. The counter force of the bracket is static so it does not move over a distance, ergo... no work.

Hope this helps

Author Commented:
many thanks everyone
Commented:
"This is the question:

A bucket is being raised by a single pulley by a height of 4m

How much work is done by the worker?

To calculate work all i know is force * distance travelled = 210 * 4m = 840J which is the answer in the book"

And that is all you need to know for this problem.

"Why is F bigger than the weight of the bucket "
Because that is the number given by the author

"why is work done on bucket  < work done by worker?"
The work on on the bucket is not less than the work done by the worker. The bucket has increased its KE as well as its PE.

Problems of this type in books of this type almost always assume frictionless pulleys unless friction is explicitly stated.
In this case the weight of the bucket is entirely extraneous.

-

Just for my curiosity, I would appreciate some information about the book you are using. Is it a textbook.? for what level? what is its reputation?
\Commented:
If your book is a practice book for an exam (especially a multiple choice exam, no essay or qualifying assumptions allowed), then aburr is right on. The work done by the person is his 410N over 4 m; and the effective work done on the bucket is 400N over 4m. In that case all our other answers based on various assumptions, e.g., friction or not, constant speed or accelerating, whether or not the speed of the bucket was 0 or not at the beginning and end of this 4m journey do provide you with additional insight into the laws of physics; but in a sense are throwing you off track if this is for a timed test.

In a timed test, you already have gone through and hopefully mastered the insightful issues, and then can focus on the question, and realize that sometimes, assumptions about the system do not need to be made.

If that is the nature of the book, then I would recommend using a true learning book first, and then go over practice exam questions second.

>> "why is work done on bucket  < work done by worker?"
>>>>The work on on the bucket is not less than the work done by the worker. The bucket has increased its KE as well as its PE.
As aburr says "Problems of this type in books of this type almost always assume frictionless pulleys unless friction is explicitly stated". Maybe that is true - it's been a long time since I've read this book; my own recollection is that the assumptions were either not ambiguous, or we had to state our assumptions explicitly; and in some problems, the stated assumptions had to be reasonable and non-trivial. You should review your book about what it's "tone" is - does it expect assumptions or not.

In this latter case with a constant force on a frictionless system, then you probably understand that the bucket is still moving upwards at the 4m mark with the same acceleration as it had at the beginning of its journey (unless you make the assumption that the person stops pulling at the 4m mark, in which case the bucket continues upwards while decelerating and then drops back down to the 4m mark).
\Commented:
>> The work done by the person is his 410N over 4 m; and the effective work done on the bucket is 400N over 4m.
Sorry, just flip-flopped about the "assumptions" again! :(
Going with frictionless system, and constant 410N force, then the force on the bucket is 410N and the bucket's delta (PE + KE) over the 4m will be the work done on it.
Commented:
Clarification:  I did not explicitly say this in my answer but I want to make sure it is clear. The total work was force applied to the system times distance over which that force was applied. 840J. No other work went into the system.  total answer.

Yes, that  work went into and was stored by the system in various ways (PE and KE if frictionless, PE, KE, and Heat if not frictionless) While intellectually interesting...answering  that was not part of the book question and we are correct to just stop at that point in the answer.  (Although it is Grand Fun to look into the details as it were and figure out where everything went.)  In the end we did not even really need to know the force on the bucket, its mass or even initial velocity. Those would just affect how much of the work would go into PE, how much into KE, etc.   Aburr was quite right and in the end had the most elegant solution. Although maybe not the most Fun solution :-)  bye all
\Commented:
correction: >> 410N over 4 m ; 400N over 4m
I meant 210 and 200 over 4m - a sort of a typo
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