Find the acceleration of the
three weights shown in fig. Assume
pulleys are frictionless. Also find
the tension in the cables.
If T is the tension in the lowest string, then the tensions in the other strings are shown in the fig. above. Let all of the accelerations he defined with upward being positive. Then the three F = ma equations are:
The first two of these equations quickly give a1 = a2
Now for the conservation-of-string statement. Let ap be the acceleration of the upper right pulley (which is the same as the acceleration of the lower right pulley). The average height of the two right masses always remains the same distance below this pulley. Therefore ap = (a2 + a3)/2. But we also have a1 = -2ap because if the pulley goes up a distance d, then a length d of string disappears from both segments above the pulley, so 2d of string appears above the left mass. This means that it goes down by 2d; hence a1 = -2p
Combining this with the ap = (a2 + a3)/2 relation gives:
Since from above a1 = a2 we obtain a2 = -2a3. The last two of the above F = ma equations are then:
Taking the difference of these equations yields mg = 5ma2, so a2 = g/5.
The desired acceleration of the mass 2 is then a3 = -a2 = -2g/5. This is negative, so the mass 2 goes downward (which makes sense. If all three masses are equal to m. you can quickly show that T = mg and all three accelerations are zero. Increasing the right mass to 2m therefore makes it go downward).