# Homework Help: [Math Verification] Massless Pulley - Acceleration of Block A & B?

1. Jul 25, 2012

### ksm2288

1. The problem statement, all variables and given/known data

http://img152.imageshack.us/img152/8286/imag0103l.jpg [Broken]

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

F=m*a

3. 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

Last edited by a moderator: May 6, 2017
2. Jul 25, 2012

### PeterO

I am sure you realise 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]

Last edited by a moderator: May 6, 2017
3. Jul 26, 2012

### ksm2288

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)

4. Jul 26, 2012

### 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.
FN is pointing the wrong way.
TB@A ... just call it T (for tension) and it's the same for the upwards force on mB. The FBD for mB 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.

5. Jul 26, 2012

### PeterO

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.

6. Jul 26, 2012

### PeterO

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.

7. Jul 26, 2012

### 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:
... @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.)