# Atwood Machine - with pulley attached to one mass. What's the constraint?

2 problems here: What is the force pulling $m_2$ when $m_1$ falls? I can only think of tension. When $m_1$ does fall (maybe a height $H$), does $m_2$ move $H/2$ to the right? Since the other end of the rope is attached to the support of the pulley, and thus can't move.

Also, the coefficient of static/dynamic friction between $m_1$ and the table is $\mu_s$ and $\mu_k$

Usually these beginner Atwood Machine problems were easy, since the ones I did were two masses attached to a taut rope, (then their displacement would be equal) but this is something entirely different.

• What about friction? Commented Aug 11, 2017 at 20:59

Try a static analysis: the tension in the rope is $m_1g$--as this is required to keep the mass hanging there. Since there a 2 pieces of rope pulling on mass 2, the total force is $2m_1g$.

The answer to the second question is "yes": Mass 2 moves half the distance that Mass 1 does.

Moving over a distance, the work does is the same, since $(m_1g)(\Delta x) = (2m_1g)(\frac{1}{2}\Delta x)$--and that is indeed, the point of a simple machine: the work done is the same, but you get some multiplier increasing your applied force while requiring more distance traveled (or visa versa).

• Is a static analysis appropriate? If $m_1$ falls then the situation is not static. If it accelerates the tension in the string is not $m_1g$. Commented Aug 11, 2017 at 20:58
• @sammygerbil I think the assumption is made that m1 falls infinitely slowly. Commented Aug 11, 2017 at 22:44
• I look at this differently, by considering the kinematics: The total length of rope is constant. So if m1 moves down x, the two lengths of rope between m2 and the right pulley must decrease in length by x. But since they are always equal in length, each one must decrease by x/2. Commented Aug 12, 2017 at 0:46
• @ The assumption is made... Who is making that assumption? There is no mentionn of it in the question. Commented Aug 12, 2017 at 3:40

I think the most easy way to solve these problem is by power method . In these whole system the net power will be equal to zero.
so, for block m1 let the tension force on it be T and velocity be v1 , and for m2 the tension force will be 2T and let velocity be v2.
now the net power will be equal to zero

Tv1=2Tv2

so,

v1=2v2

By integrating we get,

s1=2s2 ,where s1 and s2 are distance traveled by m1 and m2

A more complete analysis would start with a Lagrangian as a function of height (+z up):

$L(z) = T(z) - V(z)$

$L(z)= \frac{1}{2}(m_1 + \frac{1}{2}m_2)\dot{z}^2 - m_1gz$

Or, considering friction, it becomes a driven oscillator (with no restoring force) and a static driving force:

$Ma + F_{friction} \ \ \ + F_{restore}\ \ = F_{drive}$

$(m_1+\frac{1}{2}m_2)\ddot - \mu m_2g \dot z = -m1g$

(you may want to verify the signs on those terms...).

• -1. Unclear and misleading. How do you get the force pulling $m_2$ from this? And the relation between the distances moved by $m_1$ and $m_2$? An oscillator without a restoring force is not an oscillator. Commented Aug 12, 2017 at 3:45
• Like I said in the comment: friction--which if I remember correctly is weight times velocity times some coefficient. One then has a forced damped oscillator--which is an equation that has been solved a gazillion times, and any physicist should recognize it, regardless of some terms not being present.
– JEB
Commented Aug 12, 2017 at 6:21