# An atwood machine(consisting of mass m attached to the left side of the string and mass 2m attached to the right side of the string)has a third mass m attached to mass m by a limp string.After being released,the 2m mass falls a distance x before the limp string becomes taut.Thereafter both the masses on the left rise with the same speed.What is the final speed?Assume that pulley is ideal.

# An atwood machine(consisting of mass m attached to the left side of the string and mass 2m attached to the right side of the string)has a third mass m attached to mass m by a limp string.After being released,the 2m mass falls a distance x before the limp string becomes taut.Thereafter both the masses on the left rise with the same speed.What is the final speed?Assume that pulley is ideal.

## 1 Answers

Dear Anushka,

**Case 1.** Assume that the hanger mass is included in the value *M*, and the pulley has negligible mass and moment of inertia. Obviously the left hanger accelerates upward and the right hanger downward. Call this value *a*. The tension on the string joining the two hangers is considered uniform throughout the string.

Taking signs as positive in the direction of the acceleration, the force equations are:

Left side: *T - Mg = Ma*

Right side: *(M+m)g - T = (M+m)a*

When one simply wishes to find the acceleration, these equations are solved to eliminate the variable T. The result is:

* mg = (2M+m)a *, or

*.*

**a = mg/(2M+m)**This is an intuitively appealing form of the resuslt, for it says that the unbalanced force accelerating the system, *mg* equals the total mass *(2M+m)* times the systems acceleration, *a*. We could have used this idea to calculate the acceleration very simply. This tempts one to try a similar intuitive shortcut with other rope and pulley problems. But this can cause serious blunders when the two parts of the system dont have the same acceleration, as when theres more than one rope in the problem, or when a rope is attached to something that is not accelerating, like a solid floor or ceiling.

When we want to know *T* we eliminate the variable *a*. Again, the result isnt pretty:

* T = 2g(M+m)M) /(2M+m)*>.

The spring balance reading is twice as much,

* T = 4g(M+m)M) /(2M+m)* .

One should always check such results with a known case, for its very easy to make sign or notation errors in the mathematical steps of derivation. The test case is this: When either mass is zero, the tension should drop to zero and the acceleration of the other mass should be *g*.

But this brute force mathematical derivation isnt necessary to obtain an intuitive answer to the question. The motion of the masses simply moves the center of the mass of the system. The center of mass accelerates downward.

The scale exerts an upward force *F _{scale} = (2M+m)a* where a is the downward acceleration of the center of mass.

If the masses werent accelerating, the spring balance would exert a force of just*(2M+m)g*, the total weight of the suspended masses. When they are accelerating, *M*moving up, and *(M+m)* moving down at the same instantaneous speed, the center of mass of the two amounts *M* remains fixed. But the center of mass of the small additional mass m moves downward with acceleration *a*. The spring scale reading is therefore reduced by amount *D F = m(g-a)*. In the above derivations I took the unorthodox strategy of labeling the masses on the atwood machine as *M* and *(M+m)*. This expresses results in terms of the unbalance amount, m. Heres an example of where that strategy pays off. Lets find the force, *S* registered by the spring balance without finding the tension in the string, by looking at what happens to the center of mass of the system.

The center of mass of the two masses of size *M* does not move. The unbalanced extra mass m accelerates upward at value *a*. So the spring balance reading, *S*, is an amount*ma* less than in the static case, ie.

*S = (2M+m)g - ma*

This reduces to

** S = 4M(M+m)g/(2M+m)** .

This was a good way to check the previously derived answer for blunders. In fact, it is (I think) the method Mach used in the excerpt above. But we could only get away with it here because of the relatively simple system we are dealing with. As we noted above, when a system has more than one string, or a string attached to a fixed object, its best not to try sneaky short-cuts unless you have justified every step.

**Case 2.** As a consequence of the above analysis, we conclude that the left end of the balance will move up. Experiment easily confirms this.

**Case 3.** This one is a bit tricky. When the motion is established, the tension is nearly uniform all along the string. so since the left side has the additional unmoving mass *m*, the left side should move down.

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