2 examples are an accelerating car and a pulley system where one of the loads are accelerating.
In the first example:
As you can see, there are 4 forces acting on the car.
If there is a total net force/resultant force of 0 (##F_(R) = 0##) both vertically and horizontally, the car will either be at rest or moving at a constant velocity in the direction of the engine thrust.
##F_e-F_{f} = 0##
##F_r-F_{g} = 0##
If there is a total net force/resultant force which is greater than zero in the direction of the engine thrust (##F_(R)>0 ##) horizontally, the car will be accelerating at a constant rate in the direction of the engine thrust as there is a force that makes it go faster.
##F_e-F_{f} > 0##
##F_r-F_{g} = 0##
There is net force only when there is an which is explained by ,
##F=ma##
where ##F## is the net force, ##m## is the mass and ##a## is the acceleration.
When ##a## is 0, ##F## will be 0 regardless of the value of ##m##
Similarly, in the second example:
The force from the weight of ##m_2## will be larger than that from the weight of ##m_1##, therefore there will be a net force acting in the downwards direction, pulling the whole system down.
Let’s do a simple calculation of the acceleration of ##m_2## using Newton’s Second Law:
Assuming ##m_2## to have a mass of 5 kg and ##m_1## to have a mass of 3 kg:
Their weights would be 50N and 30N respectively.
The 50N weight will overcome the 30N weight, leaving a net force of 20N in the direction of the 50N weight.
##F_R##
##=50N-30N##
##=20N##
This net force of 20N is pulling the whole system which consists of both weights, which has a mass of 8kg altogether.
Applying Newton’s Second Law,
##F=ma##
##20=(5+3)a##
##20=8a##
##a=2.5m s^-2##
In summary, there is a net force in the direction of acceleration when an object is accelerating.
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