# Dynamics problem(pulley)

## 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 im having trouble with my coordinate system( direction and signs)

The answer for acceleration is 5.21m/s^2...and the tension is 215N.....I cant 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...

## The Attempt at a Solution

Last edited by a moderator:

Doc Al
Mentor
Please let me know how the FBD looks and I think im 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.

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 completly different tension....when the tensions should obviously be the same

Doc Al
Mentor
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?

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
Mentor
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.)

Doc Al
Mentor
Im not sure how to explain it...but heres an example of an upward Tension on a block going down that has a negative T
Only because they chose down to be positive.

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

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
Mentor
Show me the equations you used.

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

Last edited:
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
Mentor
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?

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 wouldnt 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
Mentor
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?

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....

*T-137 = 15a

Got it!...thanks for all the help 