How do I solve this dynamics problem involving a pulley?

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In summary, the conversation discusses a problem in physics involving forces and tensions on a block. The conversation includes a link to the image of the problem and the attempt at a solution, along with equations and calculations for solving the problem. There is discussion about the direction of tension and how to approach the problem correctly. Ultimately, the conversation ends with the correct solution being found.
  • #1
Jason03
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Homework Statement



http://14.photobucket.com/albums/a322/guitaristx/phy.jpg

Just add the http for the image above..

Homework Equations



Please let me know how the FBD looks and I think I am having trouble with my coordinate system( direction and signs)

The answer for acceleration is 5.21m/s^2...and the tension is 215N...I can't seem to get the equation to work out to the correct answer...any help and direction would be appreciated...

The Attempt at a Solution



http://i14.photobucket.com/albums/a322/guitaristx/Scan10005.jpg

just add the http for the image above...



 
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  • #2
Jason03 said:
Please let me know how the FBD looks and I think I am having trouble with my coordinate system( direction and signs)
Your FBD looks OK. Pick a direction parallel to the plane as positive and stick to it, such as: Down = positive; up = negative.
 
  • #3
I chose up as positive and down as negative and for the equations that I have for the first FBD...and the only way I come up with the correct Tension answer is by making the acceleration negative...and for the second FBD my calculations came up with a completely different tension...when the tensions should obviously be the same
 
  • #4
Jason03 said:
I chose up as positive and down as negative and for the equations that I have for the first FBD...
Then why is tension marked -T?
 
  • #5
The tension is upward...but the problem states that P is a force pulling block A downward...so I figured Tension would be negative because block A is being pulled downward(left) hence the Negative...
 
  • #6
Jason03 said:
The tension is upward...but the problem states that P is a force pulling block A downward...so I figured Tension would be negative because block A is being pulled downward(left) hence the Negative...
I don't quite understand that reasoning. The tension is upward and thus positive. (Ropes can't push--only pull.)
 
  • #8
Jason03 said:
Im not sure how to explain it...but here's an example of an upward Tension on a block going down that has a negative T
Only because they chose down to be positive.
 
  • #9
ok...I think i see now what you mean...let me try some new calculations...
 
  • #10
i changed the T to positive...but when I plug in 5.21 for the acceleration I get a -215 for the Tension??
 
  • #11
Show me the equations you used.

Don't forget that the acceleration will be negative (down the incline).
 
Last edited:
  • #12
heres the equ.

T-P+Fr-mgSin(30) = ma

I tried the neg. accel. and it gave me an answer even further off...
 
  • #13
As long as you have the acceleration negative, that equation should work out OK. What numbers did you plug in? What did you use for Fr?
 
  • #14
Thanks for the help on the first one, it finally worked out...

But for the second FBD I came up with

T-Fr-mgSin30 = ma

But for Friction there are two surfaces. So I have to find both normal forces acting on block A and block B

heres the numbers I came up with

Friction for B = .10(15)(9.8)Cos(30)= 12.7N

Friction for A = .10(30)(9.8)Cos(30)= 25.4N

So

T-12.7N-25.4-73.5N = 15 (5.21)

the acceleration wouldn't be negative now would it?

but now T is not coming out to 215 as it should...do you see any errors in my calculations?
 
  • #15
Jason03 said:
But for the second FBD I came up with

T-Fr-mgSin30 = ma
Looks good.
But for Friction there are two surfaces. So I have to find both normal forces acting on block A and block B

heres the numbers I came up with

Friction for B = .10(15)(9.8)Cos(30)= 12.7N

Friction for A = .10(30)(9.8)Cos(30)= 25.4N
Careful: What's the normal force of the bottom block against the incline?
 
  • #16
ohh.. I did miss that, the total mass of both blocks acting as N on the lower block.

The numbers worked out.

So to solve the system I am trying this

T-372 = 30(-a)
T-177 = 15 (a)


So can I cancel the acceleration to find T?..or is there a better way...
 
  • #17
*T-137 = 15a
 
  • #18
Got it!...thanks for all the help:approve:
 

1. How do I calculate the tension in a pulley system?

To calculate the tension in a pulley system, you need to use the principle of conservation of energy. This means that the total energy in the system (including potential and kinetic energy) must remain constant. By setting up equations for the energy at different points in the system and solving for the unknown tension, you can determine the tension in the pulley system.

2. Can a pulley system change the direction of the applied force?

Yes, a pulley system can change the direction of the applied force. This is one of the main purposes of using a pulley system. By using multiple pulleys and ropes, the direction of the force can be changed to make it easier to lift or move an object.

3. How do I determine the acceleration of a pulley system?

To determine the acceleration of a pulley system, you need to use Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration. By analyzing the forces acting on the objects in the pulley system and setting up equations, you can solve for the acceleration of the system.

4. Can friction affect the motion of a pulley system?

Yes, friction can affect the motion of a pulley system. Friction is a force that opposes motion, so it can slow down the movement of the pulley system. However, if the pulley system is designed properly and lubricated, the effects of friction can be minimized.

5. What is the difference between a fixed and a movable pulley?

A fixed pulley is attached to a solid surface and does not move. It only changes the direction of the applied force. A movable pulley is not fixed and can move with the load. It reduces the amount of force needed to lift an object, but does not change the direction of the force.

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