How Do Pulley Velocities and Positions Relate in Dependent Motion?

[jonnycbgood]

The slider block B moves to the right with a constant velocity of 20in/s. Determine (a) the velocity of block A, (b) the velocity of portion D of the cable, (c) the relative velocity of A with respect to B, (d) the relative velocity of portion C of the cable with respect to portion D.

Two blocks rest on a flat frictionless surface. block A is on the left & block B is on the right. A cord (section C) stretches from block B to a pulley attached to block A, the cord (section D) continues back to another pulley attached to block B and ends in between both blocks at a fixed point. Sorry this the best description of the picture I can do.

Since block B has a constant velocity the acceleration is zero.

X = Xo + Vt

I am trying to apply the theory of dependent motions with the equation
Xa + 2Xb = constant. But I am at a complete lost. Any assistance is greatly appreciated.

 

[Valkarie]

Thank you. a) The velocity of block A is equal to the velocity of block B, which is 20in/s. b) The velocity of section D of the cable is also equal to the velocity of block B, which is 20in/s. c) The relative velocity of A with respect to B is zero since both blocks have the same velocity. d) The relative velocity of section C of the cable with respect to section D is zero since both sections of the cable have the same velocity.

 

[Mene]

I would approach this problem by first analyzing the system and identifying all the relevant variables and equations that describe the motion of the pulleys and blocks. Let's start by defining some variables:

- Xa: position of block A
- Xb: position of block B
- V: velocity of block B
- Vc: velocity of portion C of the cable
- Vd: velocity of portion D of the cable

Using the equation X = Xo + Vt, we can determine the position of block B at any given time. Since the velocity is constant, the position of block B will also change at a constant rate. Therefore, we can say that:

Xb = Xbo + Vt

where Xbo is the initial position of block B.

Next, we can use the concept of dependent motion to determine the position of block A. The equation Xa + 2Xb = constant tells us that the sum of the positions of block A and twice the position of block B will remain constant. This means that as block B moves to the right, block A will move to the left at a rate twice that of block B. Therefore, we can say that:

Xa = Xao - 2Vt

where Xao is the initial position of block A.

Now, let's move on to the velocities. The velocity of block A can be found by taking the derivative of the position equation:

Va = -2V

This tells us that block A is moving to the left with a velocity equal to twice the velocity of block B.

To find the velocity of portion D of the cable, we can use the fact that the length of the cable remains constant. This means that the sum of the lengths of portions C and D will remain constant. Therefore, we can say that:

Lc + Ld = constant

Taking the derivative of this equation, we get:

Vc + Vd = 0

Since we know that the velocity of block B is constant, we can substitute V for Vd in the above equation and solve for Vc:

Vc = -V

This tells us that portion C of the cable is moving to the left with the same velocity as block B.

Finally, to determine the relative velocities of block A and portion C of the cable with respect to block B and portion D, we can simply subtract their respective velocities:

Relative velocity of A with respect to B =