# B A simple pulley two mass system

Tags:
1. Apr 20, 2016

### Sho Kano

In a simple one pulley-two mass system I have two ways of solving for acceleration, but am not sure which one is the correct or proper way.

#1)
assume m2 has the greater mass of the two
T-m1g = m1a
m2g - T = m2a
a = (m2g - m1g) / (m1+m2)

#2)
direction of up is positive, down is negative
T-m1g = m1a
T-m2g = m2a
a = (m1g - m2g) / (m2-m1)

The two accelerations are different. Which one is correct, and why?

2. Apr 20, 2016

### BvU

The rope has a fixed length. If m1 goes up, m2 goes down with the same speed. So $v_1 = -v_2$ and thereby $a_1 = -a_2$ . You only use one a in your second set. Apart from that, the second set is more clear (more consistent in signs of T and mg)

3. Apr 20, 2016

### Sho Kano

Accelerations are equal to zero when summed up,
m1 is moving up at a1
m2 is moving up at -a2

a1 + -a2 = 0
a1 = a2

4. Apr 21, 2016

### BvU

Yes. A consequence of $$y_1+y_2={\rm \ constant} \Rightarrow v_1+v_2= 0 \Rightarrow a_1+a_2= 0$$
It is not wise to avoid negative numbers by redefining/assuming directions. Much better to keep one coordinate system definition and conclude directions from the signs of the calculated variables.

5. Apr 21, 2016

### Sho Kano

I'm confused as to whether the second set is correct or not. Can you check the other part of post #3?

6. Apr 21, 2016

### EddiePhys

#1 would work fine. If the acceleration came negative, then you know that your assumption that m2>m1 is wrong, and it would accelerate with -a in the direction opposite to what you assumed
#2 is incorrect.

7. Apr 22, 2016

### hackhard

this is correct!
this is wrong
specifically the equation
is wrong because 'a' denotes mag of acceleration of m2 and mag of a vector is always positive
since you already know that m2 has accln downwards , so it is known that m2g must be greater than T
so since right side of eqn is +ve , so left side must also be +ve , so it must be (m2g-T) (since m2g > T)
it is always better to use vectors
T*j + mg*(-j)= m2a* (-j) j is unit vector in upward dir

Last edited: Apr 22, 2016
8. Apr 22, 2016

### Sho Kano

It all seems obvious now. These should be the equations paired with the second set.
$T\quad -\quad { m }_{ 1 }g\quad =\quad { m }_{ 1 }{ a }_{ 1 }\\ T\quad -\quad { m }_{ 2 }g\quad =\quad { m }_{ 2 }{ a }_{ 2 }\\ { a }_{ 2 }\quad =\quad { -a }_{ 1 }\\ \\ { a }_{ 1 }\quad =\quad \frac { { m }_{ 2 }g\quad -\quad { m }_{ 1 }g }{ { m }_{ 1 }\quad +\quad { m }_{ 2 } }$
Which gives the correct answer! Thanks guys.