# Maximum Tension

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#1
Hi there; I'm in the middle of doing a past paper for Physics OCR A AS-level, and I've come across a confusing question.

Q: 'A lift has a mass of 500kg. It is designed to carry a maximum of 8 people of total mass 560kg. The lift is supported by a steel cable of cross-sectional area 3.8 x 10^-4 m^2. When the lift is at ground floor level the cable is at its maximum length of 140 length. The mass per unit length of the cable is 3.0 kgm^-1.'

It then goes on to get you to work out that the weight of the cable is 420kg, which is fine, but THEN asks you this question:

'The lift with its 8 passengers is stationary at the ground floor level. The initial upward acceleration of the lift and the cable is 1.8 ms^-2. Show that the maximum tension in the cable at point P (the top end of the cable, the bit that's joined to the thing that pulls it) is 1.7 x 10^4 N.'

I have NO idea how to work out maximum tension... I tried 420 + 500 + 560 (total mass of cable/lift/passengers) x 1.8 (acceleration upwards), but I got an answer of magnitude 10^3 not 10^4... what am I doing wrong?? Please help!
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10 years ago
#2
1.8ms^-2 plus f=ma downwards due to weight (1.8+9.8) x by mass
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#3
I see... I don't really understand why you have to add g and the upwards acceleration together, but the answer seems to be right so thanks 0
10 years ago
#4
(Original post by Kyta)
I see... I don't really understand why you have to add g and the upwards acceleration together, but the answer seems to be right so thanks tension opposes the motion, so upwards motion would be more tension, because the lift would be wanting to go downwards even more (inertia?)

(imagine you are in a lift - when you are going upwards, you feel more force on your feet, and going down, less force on the bottom of your feet)
^sorry if that's a little patronising, its just how i always imagine it  0
#5
Ahhh right, yeah I understand now. Thanks again :P
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10 years ago
#6
Here's how I would look at it:

F=ma tells us that the resultant force F equals ma. We know the mass m - it is the sum of the lift (500 kg), the passengers (560 kg) and the cable (420 kg), which comes to 1480 kg. We know a, because we're told that it is 1.8 m/s^2.

So what is F? It is the resultant of the upwards force T, due to the tension in the cable, and the downwards weight W of the lift, passengers and cable, which is 1480*g N. In other words, F = T-W. Putting this together, we get

F = ma -> T-W = ma -> T = W + ma -> T = 1480*g + 1480*1.8 -> T = 17168 N,

which is the answer you want once you've taken significant figures into account.

This isn't really any different from the answer given above, but the point that makes it confusing is that the question talks about calculating the maximum tension in the cable. The thing to notice is that once the lift starts ascending, some of the cable is above the point that we are considering, and so the weight of that portion of the cable no longer counts in our calculation. This will give a smaller value for the tension. Therefore, we get the maximum possible value for the tension at the point where the lift first starts accelerating, when we can still include the entirety of the cable in our calculations.
1
2 years ago
#7
(Original post by Pangol)
Here's how I would look at it:

F=ma tells us that the resultant force F equals ma. We know the mass m - it is the sum of the lift (500 kg), the passengers (560 kg) and the cable (420 kg), which comes to 1480 kg. We know a, because we're told that it is 1.8 m/s^2.

So what is F? It is the resultant of the upwards force T, due to the tension in the cable, and the downwards weight W of the lift, passengers and cable, which is 1480*g N. In other words, F = T-W. Putting this together, we get

F = ma -> T-W = ma -> T = W + ma -> T = 1480*g + 1480*1.8 -> T = 17168 N,

which is the answer you want once you've taken significant figures into account.

This isn't really any different from the answer given above, but the point that makes it confusing is that the question talks about calculating the maximum tension in the cable. The thing to notice is that once the lift starts ascending, some of the cable is above the point that we are considering, and so the weight of that portion of the cable no longer counts in our calculation. This will give a smaller value for the tension. Therefore, we get the maximum possible value for the tension at the point where the lift first starts accelerating, when we can still include the entirety of the cable in our calculations.
oh thanks a lot this was an amazing reply,,, are u a teacher?
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2 years ago
#8
(Original post by Rasheeed)
oh thanks a lot this was an amazing reply,,, are u a teacher?
Bloody hell, I wrote that eight years ago to the day. If I was doing it again today, I would not have given that much detail. Always best to give hints so that people can put it together themselves!

(And yes.)
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10 months ago
#9
(Original post by Pangol)
Bloody hell, I wrote that eight years ago to the day. If I was doing it again today, I would not have given that much detail. Always best to give hints so that people can put it together themselves!

(And yes.)
Hello, I am on the same question, but I don't know how to find the mass of the cable. Help, please. My question is the same except it says there are 7 people with a total mass of 550kg. The rest is the same.
Last edited by Searcher1; 10 months ago
0
10 months ago
#10
(Original post by Rasheeed)
oh thanks a lot this was an amazing reply,,, are u a teacher?
(Original post by akickinthebutt)
tension opposes the motion, so upwards motion would be more tension, because the lift would be wanting to go downwards even more (inertia?)

(imagine you are in a lift - when you are going upwards, you feel more force on your feet, and going down, less force on the bottom of your feet)
^sorry if that's a little patronising, its just how i always imagine it  Hello, I am on the same question, but I don't know how to find the mass of the cable. Help, please. My question is the same except it says there are 7 people with a total mass of 550kg. The rest is the same.
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10 months ago
#11
(Original post by Kyta)
Hi there; I'm in the middle of doing a past paper for Physics OCR A AS-level, and I've come across a confusing question.

Q: 'A lift has a mass of 500kg. It is designed to carry a maximum of 8 people of total mass 560kg. The lift is supported by a steel cable of cross-sectional area 3.8 x 10^-4 m^2. When the lift is at ground floor level the cable is at its maximum length of 140 length. The mass per unit length of the cable is 3.0 kgm^-1.'

It then goes on to get you to work out that the weight of the cable is 420kg, which is fine, but THEN asks you this question:

'The lift with its 8 passengers is stationary at the ground floor level. The initial upward acceleration of the lift and the cable is 1.8 ms^-2. Show that the maximum tension in the cable at point P (the top end of the cable, the bit that's joined to the thing that pulls it) is 1.7 x 10^4 N.'

I have NO idea how to work out maximum tension... I tried 420 + 500 + 560 (total mass of cable/lift/passengers) x 1.8 (acceleration upwards), but I got an answer of magnitude 10^3 not 10^4... what am I doing wrong?? Please help!
How did you work out the mass of the cable help.
0
10 months ago
#12
(Original post by Searcher1)
Hello, I am on the same question, but I don't know how to find the mass of the cable. Help, please. My question is the same except it says there are 7 people with a total mass of 550kg. The rest is the same.
When you say "the rest is the same", does that include this bit:

(Original post by Kyta)
The lift is supported by a steel cable of cross-sectional area 3.8 x 10^-4 m^2. When the lift is at ground floor level the cable is at its maximum length of 140 length. The mass per unit length of the cable is 3.0 kgm^-1.
If so, you can work out the volume of the cable, and then use the fact that you know the mass of 1 m^3 of the cable to get its mass.
1
10 months ago
#13
(Original post by Pangol)
When you say "the rest is the same", does that include this bit:

If so, you can work out the volume of the cable, and then use the fact that you know the mass of 1 m^3 of the cable to get its mass.
Yes, indeed, and I have worked out the mas of the cable by now. What I am having a problem with is the time taken for the lift to reach the top. I tried D/S and SUVAT equations, but I could not get the answer. I would appreciate the help.
0
10 months ago
#14
(Original post by Searcher1)
Yes, indeed, and I have worked out the mas of the cable by now. What I am having a problem with is the time taken for the lift to reach the top. I tried D/S and SUVAT equations, but I could not get the answer. I would appreciate the help.
You're going to have to post the full text of your version of the question - there isn't enough information here to solve this.
0
10 months ago
#15
A lift has a mass of 500kg, inside are 7 people with a total mass of 550kg. The lift is supported by a cable which has a cross-sectional area of 3.8x10^-4m^2. The maximum of the length of the cable (when the lift is on the ground floor) is 140m. The mass per unit length of the cable is 3.0kgm^-1.
I have worked out: the mass of the cable: 420kg
ii) The lift is stationary on the bottom floor. The initial acceleration of the lift is 1.8ms^-2 for 3s. What is the maximum tension in the cable? (Which I worked out to be 17066.7N) iii)How fast is the lift travelling after 3s? - (5.4 ms^-1) correct. iv)Then the question: How long will it take the lift to reach the top? (the answer is 27.4s but I do not get it). Thanks so I need help on (iv)
0
10 months ago
#16
(Original post by Pangol)
You're going to have to post the full text of your version of the question - there isn't enough information here to solve this.
A lift has a mass of 500kg, inside are 7 people with a total mass of 550kg. The lift is supported by a cable which has a cross-sectional area of 3.8x10^-4m^2. The maximum of the length of the cable (when the lift is on the ground floor) is 140m. The mass per unit length of the cable is 3.0kgm^-1.
I have worked out: the mass of the cable: 420kg
ii) The lift is stationary on the bottom floor. The initial acceleration of the lift is 1.8ms^-2 for 3s. What is the maximum tension in the cable? (Which I worked out to be 17066.7N) iii)How fast is the lift travelling after 3s? - (5.4 ms^-1) correct. iv)Then the question: How long will it take the lift to reach the top? (the answer is 27.4s but I do not get it). Thanks so I need help on (iv)
0
10 months ago
#17
(Original post by Pangol)
You're going to have to post the full text of your version of the question - there isn't enough information here to solve this.
A lift has a mass of 500kg, inside are 7 people with a total mass of 550kg. The lift is supported by a cable which has a cross-sectional area of 3.8x10^-4m^2. The maximum of the length of the cable (when the lift is on the ground floor) is 140m. The mass per unit length of the cable is 3.0kgm^-1.
I have worked out: the mass of the cable: 420kg
ii) The lift is stationary on the bottom floor. The initial acceleration of the lift is 1.8ms^-2 for 3s. What is the maximum tension in the cable? (Which I worked out to be 17066.7N) iii)How fast is the lift travelling after 3s? - (5.4 ms^-1) correct. iv)Then the question: How long will it take the lift to reach the top? (the answer is 27.4s but I do not get it). Thanks so I need help on (iv)
0
10 months ago
#18
(Original post by Searcher1)
A lift has a mass of 500kg, inside are 7 people with a total mass of 550kg. The lift is supported by a cable which has a cross-sectional area of 3.8x10^-4m^2. The maximum of the length of the cable (when the lift is on the ground floor) is 140m. The mass per unit length of the cable is 3.0kgm^-1.
I have worked out: the mass of the cable: 420kg
ii) The lift is stationary on the bottom floor. The initial acceleration of the lift is 1.8ms^-2 for 3s. What is the maximum tension in the cable? (Which I worked out to be 17066.7N) iii)How fast is the lift travelling after 3s? - (5.4 ms^-1) correct. iv)Then the question: How long will it take the lift to reach the top? (the answer is 27.4s but I do not get it). Thanks so I need help on (iv)
You need to do one suvat calculation for the first three seconds to work out how far the lift has travelled in that time, and then subtract that from the 140 m and do another suvat with constant velocity this time to find the time the second part of the journey has taken, which you can then add to the three seconds. Although to be realistic, there should also be a deceleration phase at the top.

Is this what you have tried? I haven't actually done it myself so I don't know if it gives their answer.
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