Lift relative velocity problem.

Click For Summary
SUMMARY

The discussion focuses on the relative velocity problem involving a motor mounted on a lift. It establishes that if point P moves by y, the displacement of point A is such that Vp = -3Va. Additionally, when the motor unwinds the cable at a constant rate of 2 m/s, the velocity of the lift is determined to be 0.5 m/s. The confusion arises from the relationship between the cable's unwinding rate and the lift's movement, emphasizing that the motor's movement affects the overall velocity calculations.

PREREQUISITES
  • Understanding of relative velocity concepts
  • Familiarity with basic principles of mechanics
  • Knowledge of conservation of energy principles
  • Ability to analyze motion in systems with pulleys
NEXT STEPS
  • Study the principles of relative velocity in mechanical systems
  • Learn about the mechanics of pulleys and cable systems
  • Explore conservation of energy in dynamic systems
  • Review kinematic equations related to motion and velocity
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, engineers working with lift systems, and anyone interested in understanding the dynamics of moving systems involving cables and motors.

Legendon
Messages
21
Reaction score
0

Homework Statement


motor is mounted on top of lift.
a)if p moves by y, what is the displacement of Pt A? Prove that Vp=-3Va.
b)If motor unwidns cable at const rate 2m/s, what is vel of cable wrt lift.
Prove vel of lift is 0.5m/s


Homework Equations





The Attempt at a Solution


I can do part a. I have prob with part b. If motor releases cable at 2m/s, can i take it that pt p moves 2m/s. so Va is 2/-3=-0.667. Since pt a is connected to the lift, the vel of cable wrt lift is 2+0.667=2.667.
But velocity of lift is 0.5 so the vel of cable wrt to lift is 2.5. This is getting confusing.
 

Attachments

  • Untitled-1.jpg
    Untitled-1.jpg
    11.2 KB · Views: 489
Physics news on Phys.org
anyone ?
 
this is a better picture
 

Attachments

  • cmd.jpg
    cmd.jpg
    15.7 KB · Views: 480
Conservation of energy is your friend here.

What is the effective load on the motor?

What is the force exerted to move P a distance y?
Thus - what is the work done to do this?

What is the work done to raise the lift, and thus point A, a distance x?

The answer has a clue, and another method:
Unwinding the cable at 2m/s does not mean that P moves at 2m/s.
Remember that the motor moves with the lift. 2m/s is the rate the total length of cable increases.

You'll note that the answer, 0.5ms, is exactly a quarter of 2m/s.
How many cables are pulling on the lift?
If the total length of cable increases by 2m/s, how much does each length of cable increase by?
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
2
Views
1K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
5
Views
8K
Replies
2
Views
8K