Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Mar 5th 2016, 05:04 AM

#1  Member
Join Date: Jan 2015
Posts: 96
 Workenergy theorem problem on a hanging weight
Hi everybody ...
Its been a while since I've been on the site , however I've come across a problem which on first reading appears fairly straightforward but which has got me flummoxed. The problem is as stated below ... and has to be solved using WorkEnergy equations (Kinetic and Potential energies).
Problem  A 120kg mailbag hangs by a vertical rope 3.5 m long. A worker then displaces the bag to a position 2.0 m sideways from it's original position, always keeping the rope taut.
a) What horizontal force is necessary to hold the bag in the new position ?
Now I thought that the bag moves our to the side in an arc and the force applied is at an angle .. so the force will have both an x and a y component and since the force is acting against gravity (mg) then the components of this 'other' force will be ... mg sin (angle) and mg cos (angle). To find the angle I assumed that although the movement caused by the force is an arc .. this could be approximated to a straight line as the distance was small.
I then used the Energy equation as follows .....
K(1) + U(grav1) + W(other) = K(2) + U(grav2)
where .. K(1), K(2) are the kinetic energy values at begining and end of displacement .. U(grav1) and U(grav2) are corresponding Potential energy values ... and W(other) is work done by all forces other than gravity force.
Now K(1), K(2) are bothe zero as weight is stationary at begining and end points also U(grav1) is zero as we are selecting start point here at gthe origin of our coordinate system.
This gives .. W(other) = U(grav2)
and W(other) = Fs (force x distance) = mgh
where F is our target value and s= displacement distance and h is the verical height above the initial hanging position
But no matter how I try and calculate the force .. I get values of between 200 N  400 N and the answer in the book is ... 820 N ???
I am obviously on the wrong track here ... can anyone help ?

 
Mar 6th 2016, 05:50 AM

#2  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

I see no diagram.
I'll repeat that
I see no diagram.
Now why did I say that?
Well first and foremost a horizontal force does no vertical work.
So why have you multiplied a horizontal force by a vertical distance?
The hanging bag is a device for converting horizontal work into vertical work.
So what does the vertical work?
Well this is where a diagram would help.
In your energy balance I see no expression for the work done by the tension in the string.
Secondly is a 2m displacement on a 3.5 m string a small displacement leading to a small angle so that you can make the approximations
arc = chord
and
angle = sin = tan
?
Again a diagram would help.

 
Mar 6th 2016, 06:02 AM

#3  Member
Join Date: Jan 2015
Posts: 96

Hi STUDIOT,
Sorry about the absence of a diagram but I don't know how to 'draw' a diagram within these messages.
I am sure that there must be some way to do it , but I don't have enough knowledge of this media to do so.
However thanks for the help .. although, with apologies again, either I am a bit obtuse or your reply is a little cryptic for me. I'm afraid I find it a bit confusing.
Regards
Jackthehat
PS. .. By the way, I looked through my physics book again, just to check, and the problem in the book does not have a diagram either.

 
Mar 6th 2016, 06:38 AM

#4  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,344

Draw a freebody diagram of the bag. The problem as stated is missing an important piece of information, which is this: you must assume (although not stated) that the force applied by the man is purely horizontal. I think that's why studiot figures there must be a diagram, because without this bit of info it's impossible to answer. You start by calculating the angle tjhe rope makes with the vertical, which is arcsin(2/3.5). Now, with vertical force downward of mg, balanced by the vertical component upward of the rope tension at the angle it is at yields the value for tension in the rope. Then the horizontal force applied by the man balances against the horizontal component of tension in the rope. It turns out to be 820N if you use g=9.81 m/s^2.

 
Mar 6th 2016, 12:03 PM

#5  Member
Join Date: Jan 2015
Posts: 96

Hi ChipB,
I sort of thought there was something missing in the way I saw the problem. I suppose a diagram would have made the problem a little clearer .. however as I said to Studiot there was no diagram with that problem in the book i am using.
Thanks very much for your help .. I don't think I would have got it left to my own devices as I was going around in circles along the same old lines each time I attempted to solve the thing. Thanks again for your help.
Regards,
Jackthehat.

 
Mar 6th 2016, 12:09 PM

#6  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

I think Chip's method was to calculate force balances.
I also think the question explicitly stated that the force was horizontal as in "What horizontal force?"
Your problem required you to work it out using energy balances.
Do you understand the method?
What height did you calculate the bag was raised to ?
Note there are two solutions to this question, one of which must be discarded.
Last edited by studiot; Mar 6th 2016 at 12:17 PM.

 
Mar 6th 2016, 12:40 PM

#7  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 1,035

It is a good idea to sort the geometry before doing the Physics.
In fact the geometry provides some extra constaints (equations).
Everything below is a consequence of pure geometry alone.
Physics is not involved.
I make the height raised 0.628 metres.
The chord length is 2.0963 metres.
The arc of the circle length is 2.1290 metres.
The angle subtended at the centre is 34.8513503 degrees.
The half angle is 17.425675 degrees.
The angle the chord makes with the horizontal axis is 17.4322886 degrees.
Last edited by studiot; Mar 7th 2016 at 12:56 AM.

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