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Newton's second law along a line

  • Thread starter Jrogers201
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Homework Statement
A lift in a mine shaft takes exactly one minute to descend 500m. It starts from rest, accelerates uniformly for 12.5 seconds to a constant speed which it maintains for some time and then decelerates uniformly to stop at the bottom of the shaft.

The mass of the lift is 5 tonnes and on the day in question it is carrying 12 miners whose average mass is 80kg.
i) Sketch the speed-time graph of the lift.

ii) During the first stage of the motion the tension of the cable is 53 640N. Find the acceleration of the lift during this stage.

iii) Find the length of time for which the lift is travelling at constant speed and find the final deceleration.

iv) What is the maximum value of the tension in the cable?

v) Just before the lift stops one minor experiences an upthrust of 1002N from the floor. What is the mass of the miner?
Homework Equations
Suvat equations.

F=mg + ma
As I mentioned I am self tutoring myself on a subject I studied 20 years ago.

I'm sure I can answer all the questions apart from question (iii). I think i'm supposed to use a simultaneous equation. But I need to answer question (iii) to answer the rest.

Any help please?
 

jbriggs444

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You have to show effort. The sketch from part i and the computed result from part ii will give you part of what you need for answering part iii.

The sketch, especially, can be a helpful visualization tool. The area under a velocity-time curve tells you what?

The scheme I have in mind involves a straightforward series of computations with no simultaneous equations anywhere.
 

Delta2

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if you can answer ii) then you can find (except acceleration) the distance the lift has traveled during the first 12.5seconds. So you know the remaining distance (since you know the total distance is 500m) and the remaining time of travel which will be 60-12.5seconds (since total travel time is 1min). You also know the velocity ##v_0## which the lift has for the part of the trip that its velocity remains constant.
Assume ##t_0## is the time that the velocity remains constant and ##t_d## is the time for the deceleration part of the travel (till lift comes to a stop). Also assume ##a_d## is the deceleration. So we have 3 unknowns. What three equations can you make involving these unknowns.
one equation is $$t_0+t_d=47.5s$$ (that was easy i guess)
what are the other two equations you can make?
One equation will involve these unknowns and the remaining distance and another equation will involve ##v_0##, ##a_d## and ##t_d##.
 
Hi jbriggs

I've definitely shown effort. I've been working on this for a week. What's throwing me is the 4 unknowns tension, distance, time and acceleration. I'm sure the answer has something to do with distance and time. I worked out part ii) quite quickly which was a=(F-mg)/m = 0.8ms -2. The only other part I think I've successfully looked at is SUVAT s=0.5(u+v)*t gives me s=5t.
 

jbriggs444

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I've definitely shown effort.
Expending effort and showing effort are two different things. We need to see your work in order to offer useful assistance.
 
if you can answer ii) then you can find (except acceleration) the distance the lift has traveled during the first 12.5seconds. So you know the remaining distance (since you know the total distance is 500m) and the remaining time of travel which will be 60-12.5seconds (since total travel time is 1min). You also know the velocity ##v_0## which the lift has for the part of the trip that its velocity remains constant.
Assume ##t_0## is the time that the velocity remains constant and ##t_d## is the time for the deceleration part of the travel (till lift comes to a stop). Also assume ##a_d## is the deceleration. So we have 3 unknowns. What three equations can you make involving these unknowns.
one equation is $$t_0+t_d=47.5s$$ (that was easy i guess)
what are the other two equations you can make?
One equation will involve these unknowns and the remaining distance and another equation will involve ##v_0##, ##a_d## and ##t_d##.
Thanks that helps.

I did get to the stage that v=10ms-2 and the distance at 12.5s is 62.5m. From these I worked out the equation for sd=5t. I will keep on plugging at it. The text book I'm using can be a little vague at times.
 
Those are the units for an acceleration.
As @jbriggs444 requested, please post your working (and not as an image).
Sorry I meant to say ν=10ms-1

My working out so far is as follows.

For acceleration in the first 12.5s I used

Total mass (lift + miners) = 5000kg + (12*80kg) = 5960kg

T=mg + ma ⇒ a= (mg-T)/m
= (5960kg*9.8ms-2-53640N)/5960kg
= (58408N-53640N)/5960kg
= 0.8ms-2

Then worked out the speed using

v=u+at
=0ms-1+0.8ms-2*12.5s
=10ms-1

Then worked out the distance using

s=0.5(u+v)*t
=0.5*(0ms-1+10ms-1)*12.5s
=62.5m

As far as I've got is then after the 12.5s the remaining time will be as Delta 2 mentioned above to+td=47.5s

I've also established that the remaining distance will be so+sd=437.5m

and with that that sd=0.5*(u+v)*td which would equal sd=5td

I could also establish that vo=u+ad*td

As I say I've been working on this for a week but the text book suggested in a way with a previous example that simultaneous equations could be used. Also maybe substituting a arbitrary datum point but I think I'm going down the wrong path there.
 

haruspex

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Ok, you have four unknowns, ##s_o, s_d, t_o, t_d##, and three equations relating them. You need a fourth. Consider ##s_o, t_o##.
 

Delta2

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yes, well for the time period of ##t_0## that the lift does uniform motion with constant velocity ##v_0=10m/s## it is easy to find the relationship between ##s_0## ##v_0## and ##t_0##.
 

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