2—Infinite Series 51m 1 ax Mm 2Two masses are attached by a string of negligible mass and that is wrapped around a pulley of massM so
that it can’t slip on the pulley. Analyze them to determine what is wrong with each. Assume that there is no
friction betweenm 1 and the table and that the string does not slip on the pulley.
(a)ax=m 2 −m 1
m 2 +m 1g (b)ax=m 2
m 2 +m 1 −M/ 2g (c)ax=m 2 −M/ 2
m 2 +m 1 +M/ 2g(a) Ifm 1 m 2 , this is negative, meaning that the motion ofm 1 is being slowed down. But there’s no
friction or other such force to do this.
OR Ifm 1 =m 2 , this is zero, but there are still unbalanced forces causing these masses to accelerate.
(b) If the combination of masses is just right, for examplem 1 = 1kg,m 2 = 1kg, andM = 2kg, the
denominator is zero. The expression foraxblows up — a very serious problem.
OR IfM is very large compared to the other masses, the denominator is negative, meaning thataxis negative
and the acceleration is a braking. Without friction, this is impossible.
(c) IfM m 1 andm 2 , the numerator is mostly−M/ 2 and the denominator is mostly+M/ 2. This makes
the whole expression negative, meaning thatm 1 andm 2 are slowing down. There is no friction to do this, and
all the forces are the direction to cause acceleration toward positivex.
OR Ifm 2 =M/ 2 , this equals zero, saying that there is no acceleration, but in this system,ax will always be
positive.
The same picture, butwithfrictionμkbetweenm 1 and the table.(a)ax=m 2
m 2 +μkm 1 +M/ 2g (b)ax=m 2 −μkm 1
m 2 −M/ 2g (c)ax=m 2
m 2 +μkm 1 −M/ 2g(a) Ifμkis very large, this approaches zero. Large friction should causem 1 to brake to a halt quickly with
very large negativeax.