Calculate Tension in Connected Masses for 60 N Force

Question:

 

Three blocks of masses m₁=10 kg, m₂=20 Kg and m₃= ___ Kg are connected by strings on a frictionless surface. Horizontal force F= 60 N is applied on mass m₃ as shown below.Calculate the tension in each string.

 

 

Solution: Tension in connected masses of force 60 N

 

Assume a value for mass m₃ and solve for the rest:

 

m₁ = 10 Kg, m₂=20 Kg (Given)

 

choose a reasonable value for mass m₃ (say 30 Kg)

 

 

Relating the forces and accelerations:

 

Since the surface is frictionless, the only horizontal force acting on the system is the applied force 60 N. This force will cause all three blocks to accelerate together.Let’s denote the common acceleration as ‘a‘.

 

Applying Newton’s second law:

 

For each block, write down the equation according to Newton’s second law,which states that the net force acting on an object is equal to its mass times acceleration (F=ma).

 

Determine the total mass(M) of the system:

 

$${M=(m_1+m_2+m_3) \ Kg}$$

 

$${M= 10+20+30= 60 \ Kg}$$

 

Calculate the acceleration of all the blocks and the strings:

 

$${Acceleration = \frac{net \ force}{net \ mass}} $$

 

$${Acceleration = \frac{60 \ N}{60 \ Kg}}$$

 

$${Acceleration = 1 \frac {m}{s^2}}$$

 

Common acceleration a= 1 ms^-2

 

Free body diagram of the 30 Kg block:

 

 

Clearly net force in the horizontal direction is 60 N

 

$${F-T=ma}$$

 

$${60-T=30(1)}$$

 

$${T=60-30 =30 \ N}$$

 

This is the tension between 30 Kg and 20 Kg block.

 

Free body diagram of 10 Kg block:

 

 

Applying Newton’s second law

 

$${T_1=10(1) = 10 \ N}$$

 

This is the tension in the string between 10 Kg and 20 Kg

 

Conclusion:

 

Hence the tension in the string joining 10 Kg and 20 Kg block is 10 N while that in the string joining the 20 Kg and 30 Kg block is 30N.

 

Note that the tension throughout each string will be the same as the mass of the string is zero.