[Math Verification] Massless Pulley - Acceleration of Block A & B?

[ksm2288]

Homework Statement

http://img152.imageshack.us/img152/8286/imag0103l.jpg

*I'm aware my free body diagram is off and that normal force should be in the left quadrant.

Homework Equations

F=m*a

The Attempt at a Solution

Would I be correct in saying..

Fg of A = mg * sin (20)
= 2kg * 9.8 * sin (20)
=17.9N

Fg of B= mg
= (3 kg) (9.8 m/s^2)
=29.4N

Fnet = 29.4N - 17.9N = 11.5 N (tension of massless rope)
F= m*a
a= F/m
a of (A) = 11.5/2 = 5.75 m/s^2
a of (B) = 11.5/3 = 3.83 m/s^2

 

[PeterO]

Homework Statement

http://img152.imageshack.us/img152/8286/imag0103l.jpg

*I'm aware my free body diagram is off and that normal force should be in the left quadrant.

Homework Equations

F=m*a

The Attempt at a Solution

Would I be correct in saying..

Fg of A = mg * sin (20)
= 2kg * 9.8 * sin (20)
=17.9N

Fg of B= mg
= (3 kg) (9.8 m/s^2)
=29.4N

Fnet = 29.4N - 17.9N = 11.5 N (tension of massless rope)
F= m*a
a= F/m
a of (A) = 11.5/2 = 5.75 m/s^2
a of (B) = 11.5/3 = 3.83 m/s^2

I am sure you realize that since the masses are tied together, they will have the same magnitude of acceleration, and that no doubt brings about your query.

That 11.5 N is the net force on the "system" in the direction of motion. It will result in a certain acceleration of the 5kg system.

tension will be greater than that: Big enough to give each mass the acceleration above - you will find there is one value that does that.

eg:
If the tension was 25N, the net force on the 2kg mass would be 7.1N up the slope
If the tension was 25N, the net force on the 3kg mass would be 4.4N up
That value wouldn't work either, but there is one such value [between 17.9 and 25]

 

[ksm2288]

I am sure you realize that since the masses are tied together, they will have the same magnitude of acceleration, and that no doubt brings about your query.

That 11.5 N is the net force on the "system" in the direction of motion. It will result in a certain acceleration of the 5kg system.

tension will be greater than that: Big enough to give each mass the acceleration above - you will find there is one value that does that.

eg:
If the tension was 25N, the net force on the 2kg mass would be 7.1N up the slope
If the tension was 25N, the net force on the 3kg mass would be 4.4N up
That value wouldn't work either, but there is one such value [between 17.9 and 25]

Ah, yes. A narrow-minded approach for me. I need this class to graduate so any help/verification is incredibly appreciated.

therefore... if they have the same acceleration
a = aA = aB

a = (FgA + FgB) / (mA+ mB)
= (17.9N + 29.4N) / ( 5kg)
= 9.4 m/s^2 North of East (to the right)

T = 5kg * 9.4 m/s^2
= 47 kg*m/s^2 (Newtons)

 

[Simon Bridge]

Immediately apparent: how can the two masses have a different acceleration? That would suggest that the string stretches.

Just to make sure: off the fbd's.
F_N is pointing the wrong way.
T_B@A ... just call it T (for tension) and it's the same for the upwards force on m_B. The FBD for m_B has too few forces on it.

Formally, the analysis should treat each mass in isolation ... so you'd do:
\Sigma F = ma for each one ... and then solve the simultaneous equations.

 

[PeterO]

Ah, yes. A narrow-minded approach for me. I need this class to graduate so any help/verification is incredibly appreciated.

therefore... if they have the same acceleration
a = aA = aB

a = (FgA + FgB) / (mA+ mB)
= (17.9N + 29.4N) / ( 5kg)
= 9.4 m/s^2 North of East (to the right)

T = 5kg * 9.4 m/s^2
= 47 kg*m/s^2 (Newtons)

9.4 ms^-2 if waaaay too big. If the string broke, the acceleration of the 3 kg mass would only just exceed that value.

The 29.4 N is attempting to accelerate the 3 kg mass DOWN, and thus the 2kg mass UP the slope.
The 17.9 N - which in itself is far too big [re-check your 2.0 * 9.8 * sin(20) calculation] is attempting to accelerate the 2 kg mass DOWN the slope, and the 3 kg mass UP.

note: 2.0 * 9.8 * sin(30) = 9.8 [since sin(30) = 0.5] so 2.0 * 9.8 * sin(20) must be less than 9.8 ? Perhaps it is 7.9 and you put in a spurious 1??
You must learn to recognise answers / values that are clearly in correct.

 

[PeterO]

Immediately apparent: how can the two masses have a different acceleration? That would suggest that the string stretches.

Just to make sure: off the fbd's.
F_N is pointing the wrong way.
T_B@A ... just call it T (for tension) and it's the same for the upwards force on m_B. The FBD for m_B has too few forces on it.

Formally, the analysis should treat each mass in isolation ... so you'd do:
\Sigma F = ma for each one ... and then solve the simultaneous equations.

Sentence in red is not strictly true. It is far easier to analyse the two masses, plus string, as a single entity. The Tension thus becomes an internal force and does not come into play when applying Newton's First and Second Law which refer to External forces.

 

[Simon Bridge]

Depends on the method that is being used "formally" :) In this case, it looked like there was some confusion about how to treat the tension (twice) and advising a formal "separate the masses" approach clears that up... and works in general, in a readily transparent way that is easy for the student to check.

Anyway - the latest error is in this line:

therefore... if they have the same acceleration
a = aA = aB

a = (FgA + FgB) / (mA+ mB)[/color]
= (17.9N + 29.4N) / ( 5kg)
= 9.4 m/s^2 North of East (to the right)

... @ksm2288: do these two forces act in the same direction?

I'm also a little iffy about using "north" and "east" for these directions ... I don't think gravity acts to the south! How we think about a problem can affect the way we do the problem. Presumably the acceleration is "up the ramp" for mass A and "down" for mass B?

(I think we've been sloppy there: the masses have the same magnitude of acceleration - but different directions.)