Why do the masses in an Atwood's machine move?

The masses you see on the following picture are $m_1=8$kg and $m_2=12$kg. Also the coefficient of friction between $m_2$ and the floor is $0.25$.

What i'm concerned with is that although $m_2 >m_1$ the system will move. $m_1$ will move downwards and $m_2$ to the left. Why do they move? Also, should the tension of the string $T_1$ and $T_2$ be the same (because we have only one string)? And finally when the bodies will move should their accelerations be the same, or not? Thanks a lot in advance.

2 Answers

So first, yes the tensions at $T_1$ and $T_2$ are equal. Knowing this, you can now set up two force equations. One for the $m_1$ block and one for the $m_2$ block. What you will find is that you will be able to related the force of gravity applied on $m_1$ to the frictional force on $m_2$. Now, if the force of gravity on $m_2$ is greater than the frictional force, your system will accelerate and the two masses will accelerate at the same rate, though in different directions.

The main note is to think of the forces, not the mass. For example, if your frictional force was 0, then your system would always accelerate regardless of the mass of $m_2$.

• Note, this is assuming a frictionless pully. There is also a distinction between static and kinetic friction which is not listed in the problem statement. – Greg Petersen Jan 25 '16 at 18:46

$\mu$ is coefficient of friction

Both masses move because $F_\mathrm{friction} = \mu F_\mathrm{normal}$ and as only $m_2$ is lying on the floor its maximum resistance to the motin is $F_\mathrm{friction}= 0.25*(12*9.81)$ which is 29.43 newton (N) and $m_1$ weight is $8*9.81$ = 78.48 newton(N) so

as 78.74 N > 29.43 $m_1$ will fall taking $m_2$ with it however net resultant force will be $78.48-29.43$ which is 49.05 N.

The tension is the same in the string since its only one string connected to both masses. however to calculate this tension we break Tension into $T_1$ and $T_2$ as in the diagram for calculation but $T_1 = T_2$

this will help you in calculation

now as both bodies are connected they move at same acceleration and velocity with the net resultant force calculated above but remember now with net resultant force you have to use net mass too since both objects are associated with motion. $a=F_\mathrm{net}/(m_1+m_2) = 49.05/20 = 2.45$ m/s${}^2$

• So when there's only one string the tension is always the same? – KeyC0de Jan 25 '16 at 19:11
• The tension at one side of the string is the same as the other only if the string is massless (or negligibly small), or the string is at rest. – garyp Jan 25 '16 at 19:34
• @garyp And the moment of inertia of the pulley and the pulley friction are negligible. – Bill N Jan 25 '16 at 23:27