Physics : case 1 for 1st year

CASE: 1

Consider that two bodies masses m1 and m2 are attached end of the string passing over frictionless pulley both a hanging vertically

Consider body A

There are two forces on acting on body A

1. ·         Tension of string
2. ·         Weight of body

Since the body A move downward

W1 > T

W1 – T = Net force

W1 – T  =  F1

M1g – T = m1a --------- (1)

Consider body B

There are two forces acting on body B

• ·         Tension of string “T”
• ·         Weight of body “W2”

Since the body B move upward

T > W2

T – W2 = Net force

T – W2 = F2

T – m2 g = m1a ------ (2)

For Acceleration

Add 1 and 2

M1g – T = m1 a --------- (1)

T – m2 g = m2 a --------- (2)

Result:

M1 g - m2 g = m1 a + m2 a

G ( m1 – m2 ) = a ( m1 + m2 )

A = G ( m1 – m2 ) / ( m1 + m2 )

Tension:

Divide 1 and 2

M1g – T = m1a

T – m2 g = m2 a

Acceleration cancel

M1g – T = m1

T – m2 g = m2

Then cross multiply

m2 (M1g – T) = m1 ( T – m2 g )

m2 m1 g – m2 T = m1 T – m1 m2 g

m1 T + m2 T = m2 m1g –m1 m2g

T= 2m1 m2g / m1 + m2