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F E B R U A R Y 1 2 , 2 0 1 4
D E P E N D E NT
M OT I O N S
The motion of one particle will
depend on the motion of another
particle.
Commonly occurs when the particle
are interconnected by inextensible
cords which are wrapped around
pulleys.
DEPENDENT MOTIONS
In consideration, a path coordinate can be specified by
1. Reference from a fixed point or datum
2. Measured along the plane in the direction of
motion
3. Have a positive sense from C to A and D to B
Since the total length is constant, the coordinates can
be related by the equation
SA + SB = Lt – LCD = L
Where: LCD is the length of cord passing over arc CD
To get the velocities of the blocks, derive this given
length with respect to time, yielding
SA + SB = Lt – LCD = L
dSA/dt + dSB/dt = 0
OR
A = - B
Note: sign indicates direction
Similarly, time differential of velocities yields
acceleration,
aA = -aB
Effective Cord Length
2SB + h + SA = L
2 B + 0 + A = 0
2 B = - A
2aB = -aA
Effective Cord Length
2(h – SB) + h + SA = L
– 2SB + 2h + h + SA = L
– 2SB + 3h + SA = L
-2 B + 0 + A = 0
2 B = A
2aB = aA
Block B has an upward speed of 6 ft/s. Determine
the speed of block A.
EXAMPLE # 1
Block B has an upward speed of 6 ft/s. Determine
the speed of block A.
EXAMPLE # 2

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Dependent motions

  • 1. F E B R U A R Y 1 2 , 2 0 1 4 D E P E N D E NT M OT I O N S
  • 2. The motion of one particle will depend on the motion of another particle. Commonly occurs when the particle are interconnected by inextensible cords which are wrapped around pulleys. DEPENDENT MOTIONS
  • 3. In consideration, a path coordinate can be specified by 1. Reference from a fixed point or datum 2. Measured along the plane in the direction of motion 3. Have a positive sense from C to A and D to B
  • 4. Since the total length is constant, the coordinates can be related by the equation SA + SB = Lt – LCD = L Where: LCD is the length of cord passing over arc CD
  • 5. To get the velocities of the blocks, derive this given length with respect to time, yielding SA + SB = Lt – LCD = L dSA/dt + dSB/dt = 0 OR A = - B Note: sign indicates direction Similarly, time differential of velocities yields acceleration, aA = -aB
  • 6. Effective Cord Length 2SB + h + SA = L 2 B + 0 + A = 0 2 B = - A 2aB = -aA
  • 7. Effective Cord Length 2(h – SB) + h + SA = L – 2SB + 2h + h + SA = L – 2SB + 3h + SA = L -2 B + 0 + A = 0 2 B = A 2aB = aA
  • 8. Block B has an upward speed of 6 ft/s. Determine the speed of block A. EXAMPLE # 1
  • 9. Block B has an upward speed of 6 ft/s. Determine the speed of block A. EXAMPLE # 2