- #1

Rishabh Narula

- 61

- 5

- Homework Statement
- Q) Two particles of masses m and 2m are placed on a smooth horizontal table. A string,

which joins them, hang over the edge supporting a light pulley,

which carries a mass 3m

Two parts of the string on the table are parallel and

perpendicular to the edge of the table. The parts of the string

outside the table are vertical. Show that the acceleration of the

particle of mass 3m is 9g/17.

- Relevant Equations
- F = ma

Let T be the tension in the string, a be the acceleration of

mass 2m, 2a be the acceleration of mass m

T = (m) (2a) ---eq(1)

The mass 3m will come down with acceleration

a’ = (a+2a)/2 = 3a/2

3mg - 2T = 3m . 3a/2

from equation 1

3mg - 2(2ma) = 3m . 3a/2

thus a = 6/17g

thus acceleration of 3m mass = 3/2 a

A' = (3/2) (6/17g) = 9/17 g

i understand the whole solution except for the part where it says -

"The mass 3m will come down with acceleration

a’ = (a+2a)/2 = 3a/2"

Why does 3m come down with accn. = a + 2a / 2 = 3a/2...WHY IS THIS SO?

mass 2m, 2a be the acceleration of mass m

T = (m) (2a) ---eq(1)

The mass 3m will come down with acceleration

a’ = (a+2a)/2 = 3a/2

3mg - 2T = 3m . 3a/2

from equation 1

3mg - 2(2ma) = 3m . 3a/2

thus a = 6/17g

thus acceleration of 3m mass = 3/2 a

A' = (3/2) (6/17g) = 9/17 g

i understand the whole solution except for the part where it says -

"The mass 3m will come down with acceleration

a’ = (a+2a)/2 = 3a/2"

Why does 3m come down with accn. = a + 2a / 2 = 3a/2...WHY IS THIS SO?