Physics Question : Tension?

The following configuration of the system will be assumed:

1. The 4-kg block is on the left side of the table

2. The 12-kg block is on the right side of the table

3. Movement of the system is to the right

Using Newton's 2nd Law of Motion,

F = ma

where

F = net force acting on a body

m = mass of the body

a = acceleration of the body

For the 12-kg block:

12(9.8) - T1 = 12(a)

where

T1 = tension in the string connected to the 12-kg block

a = acceleration of the system

Solving for T1,

T1 = 12*9.8 - 12(a)

T1 = 12(9.8 - a) --- call this Equation 1

For the 6-kg block on the table,

T2 - T3 = 6(a)

where

T2 = tension in the string connecting the 12-kg and 6-kg blocks

T3 = tension in the string connecting the 6-kg and 4-kg blocks

Since the system is frictionless, T1 = T2 and the above simplifies to

T1 - T3 = 6(a) --- call this Equation 2

For the 4-kg block,

T4 - 4(9.8) = 4a

and again, since the system is frictionless, T3 = T4 and the above becomes

T3 - 4(9.8) = 4a

T3 = 4(9.8) + 4a --- call this Equation 3

From Equation 2, T1 = T3 + 6a

and substituting the above in Equation 1,

T3 + 6a = 12(9.8) - 12a

T3 = 12*9.8 - 18a --- call this Equation 4

NOTE that Equation 3 = Equation 4, hence

4(9.8) + 4a = 12*9.8 - 18a

22a = 9.8(12 - 4)

a = 3.56 m/sec^2

Solving for T1 (using equation 1),

T1 = 12(9.8 - 3.56)

T1 = 74.84 N

Solving for T3 (using Equation 4)

T3 = 12*9.8 - 18a

T3 = 12(9.8) - 18(3.56)

T3 = 53.52 N

ANSWERS:

Tension in the strings:

T1 = T2 = 74.84 N

T3 = T4 = 53.52 N

Acceleration = 3.56 m/sec^2

<< and the 6.0 kg block is on the table... will this block affect the answer?>>

In one simple word, the answer is YES.