Question

A block of mass 5kg is sits on a flat smooth lab bench. It is attached by two cords which are placed over pulley’s to hanging masses m1=8kg and m2=12kg on either side. Determine the acceleration of the system and the tension in each cord (they are different).

Answer 1

Given,

mass, m₁ = 8 kg

mass, m₂ = 12 kg

Let mass kept on the block be m₃, hence, m₃ = 5 kg

Let assume the direction of acceleration as shown in the figure above,

From the figure, we can see

For mass m₁ ,

T₁ - m₁*g = m₁*a

=> T₁ = m₁*a + m₁*g

Now,

For massm₂ ,

m₂*(-g) + T₂ = m₂*(-a)

=> T₂ - m₂*g= -m₂*a

=> T₂ = m₂*g - m₂*a

Now,

For mass m₃ , we have

T₂ - T₁ = m₃*a

Now, put the values of T₁ and T₂ from above equation,

m₂*g - m₂*a - ( m₁*a + m₁*g) = m₃*a

=> m₂*g - m₁*g = ( m_{1 +} m₂ + m₃ ) * a

=> a = ( m₂*g - m₁*g ) / ( m_{1 +} m₂ + m₃ )

Now put the values in the equation

=> a = ( 12*9.8 - 8*9.8 ) / ( 12 + 8 + 5 )

=>    = 19.6 / 25 = 0.784 m/s²

or a = 0.784 m/s²

Now, we have

T₁ = m₁*a + m₁*g

=> T₁ = m₁*( a + g ) = 8 * ( 0.784 + 9.8 )

           = 8 * 10.584 = 84.672 N

or    T₁ = 84.672 N

Now, we have

T₂ = m₂*g - m₂*a

=> T₂ = m₂*( g - a) = 12*( 9.8 - 0.784 )

           = 12 * 9.016 = 108.192 N

or T₂ = 108.192 N

Hence,

acceleration of the system is 0.784 m/s² and Tensions are 84.672 N on 8 kg side and 108.192 N on 12 kg side.