Letting a be the acceleration of m1, and b be the acceleration of m2.
Note that the lower pulley will have an accelerlation of b/2, but this is the acceleration of m2. So
a=b/2
Consider the force on the ceiling holding the upper pulley:
F=m1(g-a)+m2(g+a)
and the tension on the lower rope..
T=m2(g+b)
But the upper force -lower force=system mass* system acceleration
or F-T=m1*a-m2*b
or
m1(g-a)+m2(g+a)-m2(g+b)=m1*a-m2*b
Think on that, it is after all, three am here, and I could be in error.
If correct, you can solve for a, and b.
A string passes over a frictionless, massless pulley attached to the ceiling. A mass m1=15.0 kg hangs from one end of this string, and a second massless, frictionless pulley hangs from the other end of the string. A second string passes over the second pulley and a mass m2=4.0 kg hangs from the end, whereas the other end is attached firmly to the ground. Assume positive accelerations are upward.
1). What is the acceleration of m1?
2). What is the acceleration of m2?
3). What is the acceleration of the second pulley?
WHAT I TRIED:
To find m1 I tried a = (m1-m2)/(m2+m1) but that did not work. ( The value is negative but I entered it as positive that might be a problem)
To find m2 I tried m2a2=T-m2g, finding the tension of then implicating it in the formula to solve for the a2 acceleration but that did not work.
The last part I wrote 0 just because the second pulley does not have mass, so I need help.
1 answer
1 answer