Dynamics problem(pulley)

  • Thread starter Jason03
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  • #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...

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #2
Doc Al
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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
Jason03
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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
Doc Al
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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
Jason03
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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
Doc Al
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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
Doc Al
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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
Jason03
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ok...I think i see now what you mean...let me try some new calculations...
 
  • #10
Jason03
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i changed the T to positive...but when I plug in 5.21 for the acceleration I get a -215 for the Tension??
 
  • #11
Doc Al
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Show me the equations you used.

Don't forget that the acceleration will be negative (down the incline).
 
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  • #12
Jason03
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heres the equ.

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

I tried the neg. accel. and it gave me an answer even further off...
 
  • #13
Doc Al
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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
Jason03
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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
Doc Al
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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
Jason03
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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
Jason03
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*T-137 = 15a
 
  • #18
Jason03
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Got it!...thanks for all the help:approve:
 

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