How do I solve this dynamics problem involving a pulley?

[Jason03]

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...

 

[Doc Al]

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.

 

[Jason03]

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

 

[Doc Al]

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?

 

[Jason03]

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...

 

[Doc Al]

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.)

 

[Jason03]

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...similiar to what i am explaining

http://i14.photobucket.com/albums/a322/guitaristx/Scan10006.jpg

 

[Doc Al]

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.

 

[Jason03]

ok...I think i see now what you mean...let me try some new calculations...

 

[Jason03]

i changed the T to positive...but when I plug in 5.21 for the acceleration I get a -215 for the Tension??

 

[Doc Al]

Show me the equations you used.

Don't forget that the acceleration will be negative (down the incline).

 

[Jason03]

heres the equ.

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

I tried the neg. accel. and it gave me an answer even further off...

 

[Doc Al]

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?

 

[Jason03]

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?

 

[Doc Al]

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?

 

[Jason03]

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...

 

[Jason03]

*T-137 = 15a

 

[Jason03]

Got it!...thanks for all the help