Answer to Question #189172 in Mechanical Engineering for GOLLA HARISH

Question

Find the acceleration of the 

three weights shown in fig. Assume 

pulleys are frictionless. Also find 

the tension in the cables.

Answer 1

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:

$T-mg= ma_1\\ T- mg= ma_2\\ T-(2m)g =(2m)a_3$

The first two of these equations quickly give a₁ = a₂

Now for the conservation-of-string statement. Let a_p 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 a_p = (a₂ + a₃)/2. But we also have a₁ = -2a_p 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 = -2_p

Combining this with the a_p = (a₂ + a₃)/2 relation gives:

$a_1 = -2(\dfrac{a_2+a_3}{2}) \implies$

$a_1+ a_2+ a_3= 0$

Since from above a₁ = a₂ we obtain a₂ = -2a₃. The last two of the above F = ma equations are then:

$T -mg = ma_2\\ T-(2m)g = (2m)(-2a_2)$

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).