Acceleration of elevator and force in cable

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SUMMARY

The discussion focuses on calculating the acceleration of an elevator and the force in cable C2, given a mass of 500 kg for the elevator and a counterweight of 134 kg. The force in cable C1 is specified as 1.2 kN. The acceleration of the elevator is determined to be 0.85 m/s² using the equation a = (F - C1) / m, where F is the net force acting on the counterweight. The final force in cable C2 is calculated to be 4.13 kN, taking into account the weight of the elevator and the forces acting on it.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of free body diagrams (FBD)
  • Basic principles of forces in mechanical systems
  • Familiarity with units of force and acceleration (N and m/s²)
NEXT STEPS
  • Study the application of free body diagrams in mechanical systems
  • Learn about the dynamics of pulleys and their effect on connected masses
  • Explore advanced topics in elevator mechanics and safety systems
  • Investigate the implications of varying counterweight configurations on elevator performance
USEFUL FOR

Engineering students, mechanical engineers, and anyone involved in elevator design or analysis will benefit from this discussion, particularly those interested in the dynamics of mechanical systems.

Wilson123
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Homework Statement


upload_2015-12-17_20-25-21.png


Mass of elevator = 500kg
Counterweight = 134kg
Force in cable C1 is 1.2kn

Determine the acceleration of the elevator and the force in cable C2

Homework Equations

The Attempt at a Solution



a = f / m

a = (C2 - C1) / m2

a = ((m2 * g) - C1) / m2

a = ((134 * 9.81) - 1200) / 134 = 0.86m/s2

Not sure if I need to be taking the mass of the elevator into account somewhere?
 
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Ignoring the elevator and motor for the moment, draw a free body diagram for the 134 kg counterweight alone. What forces are acting on it? Which ones do you know? What can you conclude about the motion of the counterweight?
 
gneill said:
Ignoring the elevator and motor for the moment, draw a free body diagram for the 134 kg counterweight alone. What forces are acting on it? Which ones do you know? What can you conclude about the motion of the counterweight?

The counterweight has a force of 1.314kN acting downwards and 1.2 kN acting upwards. Therefore the counterweight is going down and pulling the elevator upwards?
 
gneill said:
Ignoring the elevator and motor for the moment, draw a free body diagram for the 134 kg counterweight alone. What forces are acting on it? Which ones do you know? What can you conclude about the motion of the counterweight?
That appears to be what OP has done already. That takes care of the motion of the counterweight.

The next step would be to see what that implies about the motion of the elevator.
 
jbriggs444 said:
That appears to be what OP has done already. That takes care of the motion of the counterweight.
I saw the C2 in the acceleration equation (second line) and had my doubts as to what the OP had in mind.
 
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Wilson123 said:
The counterweight has a force of 1.314kN acting downwards and 1.2 kN acting upwards. Therefore the counterweight is going down and pulling the elevator upwards?
Yes. And what's more, you can find the acceleration of the counterweight, which you actually did when you replaced the "C2" term in your equation with the weight of the counterweight. I couldn't tell if this was a fortuitous mistake or intentional... ... which is why I asked you to concentrate on the counterweight alone.
 
I am confused now as to whether or not I have used C2 in the right application?
 
Wilson123 said:
I am confused now as to whether or not I have used C2 in the right application?
If you concentrate on the counterweight alone, C2 is nowhere in the picture. What appears in the FBD for the counterweight?

We'll get to C2 later when we look at the elevator.
 
Disregarding C2. The acceleration on the counterweight alone is:

F=ma
a = F/ m
a = (1314N - 1200N) / 134
a = 0.85 ??
 
  • #10
That's right. Be sure to include units on results, and indicate the direction of the acceleration.

Now, what does that tell you about the direction of motion of the elevator car and its acceleration?
 
  • #11
Units are in m/s^2 for acceleration? The elevator car will be moving upwards as the counterweight is moving down, I'm not sure what this suggests about the acceleration..
 
  • #12
Wilson123 said:
Units are in m/s^2 for acceleration? The elevator car will be moving upwards as the counterweight is moving down, I'm not sure what this suggests about the acceleration..
Yes, units of acceleration are m/s2.

How are the counterweight and elevator car connected? Can they have different velocities or accelerations?
 
  • #13
They are connected through the same pulley therefore will have the same velocity and acceleration. Which would mean the acceleration of the elevator is the same as the counter block... Is that the logic you were trying to make me realize?
 
  • #14
Wilson123 said:
They are connected through the same pulley therefore will have the same velocity and acceleration. Which would mean the acceleration of the elevator is the same as the counter block... Is that the logic you were trying to make me realize?
It is indeed :smile:

So with that fact in mind, can you use the known forces operating on the elevator and its acceleration to find the "missing" force C2? Think Newton's 2nd law. An FBD for the isolated elevator car is in order.
 
  • #15
Is it as simple as doing F=ma, f=500*0.85, giving me an answer of 0.425kN. Or do I need to take the 1.2 kN and weight of the elevator into account as well?
 
  • #16
Wilson123 said:
Is it as simple as doing F=ma, f=500*0.85, giving me an answer of 0.425kN. Or do I need to take the 1.2 kN and weight of the elevator into account as well?
Yup. Hence the FBD for the elevator car. F = MA relates the mass, acceleration, and NET force acting.
 
  • #17
(m1*a) + (m1*g) - C1 = C2
(500 * 0.85) + (500 * 9.81) - 1200 = 4.13kN

Is this correct taking all variables into account?
 
  • #18
Looks good to me! :oldbiggrin:
 
  • #19
Thanks for your help! Got there in the end...
 
  • #20
So far i believe this is correct, however I'm
Wilson123 said:
Thanks for your help! Got there in the end...
Well Done! o_O
 

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