Lift relative velocity problem.

[Legendon]

Homework Statement

motor is mounted on top of lift.
a)if p moves by y, what is the displacement of Pt A? Prove that V_p=-3V_a.
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 V_a 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.

 

[Legendon]

anyone ?

 

[Legendon]

this is a better picture

 

[Simon Bridge]

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?

 

[asteriatic]

Hi there,

Thank you for your question. It seems like you are having trouble understanding how to approach part b of the problem. I will try to provide a clear explanation and solution below.

First, let's define the variables in the problem:

- P: the motor
- A: point on the lift where the cable is attached
- Vp: velocity of the motor (or point P)
- Va: velocity of point A
- y: displacement of point P
- Vc: velocity of the cable with respect to the lift
- Vl: velocity of the lift

Now, let's break down the problem into smaller steps:

1. Determine the relationship between Vp and Va.

From the problem statement, we know that if P moves by y, then the displacement of point A will be equal to -3y (because Vp = -3Va). This means that for every 1 unit of displacement for P, there will be a displacement of -3 units for A. This also means that Vp is always three times larger than Va in the opposite direction.

2. Find the velocity of the cable with respect to the lift (Vc).

Since the cable is attached to both P and A, the velocity of the cable will be a combination of the velocities of these two points. We can use the formula Vc = Vp + Va to find the velocity of the cable with respect to the lift.

Substituting the relationship we found in step 1, we get Vc = -3Va + Va = -2Va. This means that the velocity of the cable with respect to the lift is always two times larger than Va in the opposite direction.

3. Find the velocity of the lift (Vl).

Since the lift is attached to point A, the velocity of the lift will be equal to Va. From the problem statement, we know that the motor unwinds the cable at a constant rate of 2m/s. This means that Va = -2m/s (since Vp = -3Va). Therefore, the velocity of the lift is also equal to -2m/s.

4. Prove that Vc = 0.5m/s.

To prove this, we simply substitute the value we found for Vl into the equation Vc = -2Va. This gives us Vc = -2(-2m/s) = 4m/s. Since the velocity of the cable with