Pulley dynamics problem 12-205 Hibbeler

  • Thread starter Thread starter issacnewton
  • Start date Start date
  • Tags Tags
    Dynamics Pulley
Click For Summary
SUMMARY

The discussion focuses on solving a pulley dynamics problem from Hibbeler's textbook, specifically problem 12-205. The equations derived include 2S_A + S_C = L_1 and (S_B - S_C) + (h - S_C) = L_2, which relate the movements of blocks A and B. The user successfully calculated the velocities and accelerations, concluding that block A moves upwards with a velocity of -1 and an acceleration of 0.5, indicating that the speed of block A is indeed increasing despite the downward acceleration. The interpretation of these results is clarified through the discussion of positive and negative values in the context of motion.

PREREQUISITES
  • Understanding of basic pulley systems and dynamics
  • Familiarity with kinematic equations
  • Knowledge of acceleration and velocity concepts
  • Ability to interpret mathematical signs in physics
NEXT STEPS
  • Study advanced pulley systems in engineering mechanics
  • Learn about kinematic equations in two dimensions
  • Explore the effects of friction in pulley systems
  • Investigate real-world applications of pulley dynamics in mechanical engineering
USEFUL FOR

Students studying mechanical engineering, physics enthusiasts, and anyone tackling problems related to dynamics and kinematics in pulley systems.

issacnewton
Messages
1,035
Reaction score
37

Homework Statement


I have posted the snapshot.

Homework Equations


I have written the distances from the datum line. Since we have two threads, I got two
equations.
[tex]2S_A +S_C=L_1[/tex]

[tex](S_B -S_C)+(h-S_C)=L_2[/tex]

where L1 and L2 are the lengths of the strings excluding the
parts which remain constant in time.

The Attempt at a Solution

I can now relate B and A as

[tex]\dot{S_A}=-\frac{\dot{S_B}}{4}[/tex]

[tex]\ddot{S_A}=-\frac{\ddot{S_B}}{4}[/tex]

So I get [itex]\dot{S_A} =-1[/itex] , which means the block A is going upwards.
Now the problem says that the speed of the cable being pulled at B is decreasing
at the rate of 2 ft/s2. So that means [itex]\ddot{S_B}=-2 ft/s^2[/itex].
So I get [itex]\ddot{S_A}= 0.5[/itex]. I got the first answer right. I have question about
the interpretation of the second answer. Since [itex]\ddot{S_A}[/itex] is positive, does
it mean the speed of block A is increasing ?. My second numerical answer is correct
though.
 

Attachments

  • 1.jpg
    1.jpg
    24.9 KB · Views: 2,056
Physics news on Phys.org
hi IssacNewton! :smile:
IssacNewton said:
… So I get [itex]\dot{S_A} =-1[/itex] , which means the block A is going upwards.

Now the problem says that the speed of the cable being pulled at B is decreasing
at the rate of 2 ft/s2. So that means [itex]\ddot{S_B}=-2 ft/s^2[/itex].
So I get [itex]\ddot{S_A}= 0.5[/itex] …

Since [itex]\ddot{S_A}[/itex] is positive, does it mean the speed of block A is increasing ?

A's velocity is -1 downward, ie 1 upward.

A's acceleration is positive downward, ie negative upward, so the speed of 1 upward is decreasing. :wink:
 
tim, makes perfect sense. these pulley problems sometime throw me...
 

Similar threads

Replies
35
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
14K
Replies
7
Views
5K
Replies
6
Views
4K
Replies
24
Views
4K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
5
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K