elevator pulley system

This is the scenario where my mass is $60 kg$, the mass of the elevator is $30kg$, and due to a malfunction, I have to hold myself and the elevator at rest. The question is, if there is a weighing scale under me, (of negligible mass), then what force will it measure (in newtons)?

My approach: The total mass of the system is $90kg$ which is approximately $900 N$ for equilibrium, tension in the rope is $900N$ because I am giving a downward force of $900N$, so according to Newton's 3rd Law, upward force on me by the rope must be $900N$. This means I should go up, and the weighing scale must not show any reading. But that is not among the options to choose from. Guidance rather than a direct answer would be appreciated.


closed as off-topic by Brandon Enright, John Rennie, Kyle Kanos, ACuriousMind, BMS Jan 26 '15 at 19:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Brandon Enright, John Rennie, Kyle Kanos, ACuriousMind, BMS
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ if it's MC, could you give us the options and we may tell you which is correct and why $\endgroup$ – Jim May 2 '13 at 13:28
  • 1
    $\begingroup$ Make separate FBD for each , it can be solved easilly. This approach is correct, but $2T=900$ will be correct. $\endgroup$ – ABC May 2 '13 at 13:29
  • $\begingroup$ @007, why 2t?? isnt Tension in string constant? $\endgroup$ – Saurabh Raje May 2 '13 at 13:52
  • $\begingroup$ a)250N b)200N c)150N d)100N $\endgroup$ – Saurabh Raje May 2 '13 at 13:53
  • 1
    $\begingroup$ Accepting the answer giving just the calculation was a dumb action. Even without FBD that answer can't be understood. Also, i feel sorry for u if you can't even solve a linear equation on your own. Unaccept my answer was really BAD. $\endgroup$ – ABC May 6 '13 at 5:16

we can actually use $300 - T = 0$ (1) for equilibrium of the lift and $600 -T = 0$ (2) for equilibrium of the man as both systems are at rest.

Adding (1) and (2)

$900- 2T = 0$

$T = 450 N$

Taking into account the system of the man, the man exerts a force downwards and tension pulls him upwards.

Thus total force downward in the man's system is $600 N - 450 N = 150 N$

The reading of the weighing machine is $150 N$

  • $\begingroup$ I think it's important to remember that the piezoresistor in the scale will measure the net force that is applied to it by your feet. Yes, you feel 150N down, but being part of the elevator, it feels 150N up. Applying Newton's 3rd law, that means (assuming it isn't moving) that the piezoresistor would experience a net force of 300N applied on top of it, and that's what it would read. $\endgroup$ – Jack Dozer May 8 '13 at 12:59
  • $\begingroup$ No, if I push downward on a scale with a weight of 150N, it reads 150N. Otherwise its a broken scale. $\endgroup$ – Brian Moths May 23 '14 at 14:22


Now these two equations can be solved to get the $N$ , by eliminating $T$ . Remember, only the normal force is reading of the elevator.

  • $\begingroup$ @SaurabhRaje See the FBD and you'll find it's $2T$. Solving you get $c) $ option. $\endgroup$ – ABC May 2 '13 at 14:01
  • $\begingroup$ do i assume that the normal force N there is on the weighing scale? If yes, isnt it cancelling out?? Please make a FBD for the weighing scale, so that I can see what exactly is the resultant force on it. $\endgroup$ – Saurabh Raje May 2 '13 at 14:09
  • $\begingroup$ FBD will include Two normal forces equal to $N$ . On top surface , it acts downward and on bottom surface directed upwards. $\endgroup$ – ABC May 2 '13 at 14:12
  • $\begingroup$ so the weighing scale should experience no net force? $\endgroup$ – Saurabh Raje May 2 '13 at 14:18
  • 1
    $\begingroup$ then wont that be 2N/K?? $\endgroup$ – Saurabh Raje May 2 '13 at 14:52

The total force down is 900N. Thus, the total force up must be 900N. For the rope to be in equilibrium, the force up on each side of the pulley is 450N.

Your force down is 600N, the elevator's force down is 300N. If we carry through the math, that means your net force is 150N down and the elevator's net force is 150N up.

Assuming the weigh scale is considered part of the elevator, it would show 300N. In other words, your mass minus the elevator's mass.


Based on the options provided, I would say you should select the 150N option. However, I believe that is wrong. If we symbolize the masses instead of use numbers, then that answer implies the following:

The force due to gravity of you and the elevator is $-m_1g$ and $-m_2g$ respectively. Thus, in equilibrium, the tension on each side of the pulley would be $g(m_1+m_2)\over2$. And the reading on the scale then is $g(m_1-m_2)\over2$, or the normal force on you as 007 put it.

I believe that this is wrong because in the limit where the mass of the elevator, $m_2$ goes to 0, then it would suggest that the scale reads $gm_1\over2$, or half your mass when, in that limit, it should read your full mass.

What I believe is left out is that due to the tension from the rope, the elevator has a net force upwards and, when combined with your net force downward, the relative normal force the scale would need to provide is a sum of the two, in other words it is $g(m_1-m_2)$, your mass minus the mass of the elevator.

  • $\begingroup$ (-1) $300N$ is not even in options. See my answer You'll get it.Correct it. $\endgroup$ – ABC May 2 '13 at 14:02
  • 1
    $\begingroup$ can I have that -1 taken off now? $\endgroup$ – Jim May 2 '13 at 14:11
  • $\begingroup$ Still, when the mass $m_2$ goes zero, tention does not and $N$ would be half the mass, no doubt in it. $\endgroup$ – ABC May 2 '13 at 14:12
  • $\begingroup$ but the scale must be considered as part of the elevator, so you can't just take into consideration your normal force, but its normal force as well. $\endgroup$ – Jim May 2 '13 at 14:19
  • 1
    $\begingroup$ Reminding you that the commentator doesn't get notified unless you say @commentator. For example, use @Crazy to ping me. A quick advice that -1 don't make your posts look so bad, instead it gets more attention ;-) $\endgroup$ – Waffle's Crazy Peanut May 2 '13 at 15:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.