- #1

Sho Kano

- 372

- 3

## Homework Statement

I need help with number 4.

Also I am altogether unsure of this problem... if you catch any mistakes, point them out! :)

## Homework Equations

F = ma

Here are the variables I used in my attempt at the solution

mx = m subscript 1, 2, or 3

±a1 = acceleration of either mass on smaller pulley

T1 = tension of string on smaller pully

±a2 = acceleration of either mass (one is a pulley) on larger pully

T2 = tension of string on larger pully

## The Attempt at a Solution

[/B]

(a)

m2*g - T1 = m2*a1 ---> from the free body diagram of mass 2

T1 - m1*g = m1*a1 ---> from the free body diagram of mass 1

m3*g - T2 = m3*a2 ---> from the free body diagram of mass 3

T2 - m1*g - m2*g = (m1+m2)*a2 ---> from the free body diagram of small pulley

substituting T1 from the second equation above in the first equation

m2*g - m2*a1 = m1*a1 + m1*g

a1 = (m2*g - m1*g) / (m1+m2)

so if downward is the positive direction,

mass 1's acceleration = -a1

mass 2's acceleration = +a2

substituting T2 for the last two equations,

m3*g - m3*a2 = (m2+m1)*a2 + m1*g + m2*g

a2 = (m3*g - m1*g - m2*g) / (m3+m2+m1)

mass 3's acceleration is +a2

(b)

a1 = 0 now, because the numerator of a1 would end out equal to zero.

This means the smaller pulley's masses would not experience any acceleration

(c)

since we want a2 to be zero, the numerator of a2 must equal 0, so

m3 = m1 + m2

also,

m2>m1

(d)

I'm confused, the expression won't change for a1? It will still be ±a1 for either masses.