Horizontal Blocks in Contact

Horizontal blocks in contact is a classic problem that considers both Newton’s 2nd and 3rd laws. Consider two objects A and B, with masses of 5kg and 3kg respectively, that are being pushed with a force of 40N to the right as shown below:

Newtons 3rd Law predicts that A exerts a force on B and B exerts an equal but opposite force on B:

{ F }_{ AB }=-F_{ BA }

We can consider a few aspects of this problem:

If the total force acting on both objects is 40N and they have a combined mass of 8kg, according to Newton’s 2nd law the acceleration of both objects:

a=\cfrac { F }{ m }

a=\cfrac { 40 }{ 8 }

a=5\cfrac { m }{ { s }^{ 2 } }

We can now determine the net force acting on each object:

Object A: F=ma

F=5\times 5

F=25\quad N

Object B: F=ma

F=3\times 5

F=15\quad N

Based on Newtons 3rd we can see that the reaction force of B on A; F_{ BA }=15N

We can now observe the net force acting on A, F_{ net }=40-15=25N which agrees with our working above. 

Solving these problems requires an understanding of the forces analysed above.

Example 1:

In the diagram below, box A has a mass of 8kg and box B has a mass of 5kg. A force of 78N acts on the boxes to the right. Calculate the following:

a) the acceleration of the boxes

b) the force acting on box A and B

c) the reaction force of B on A

a) Using Newtons 2nd law: F=ma

a=\cfrac { F }{ m }

a=\cfrac { 78 }{ (8+5) }

a=6\quad \cfrac { m }{ { s }^{ 2 } }

b) Again, using F=ma and the answer to part a:

Box A: F=ma

F=8\times 6

F=48\quad N right

Box B: F=ma

F=5\times 6

F=30\quad N right

c) The reaction force of B on A is equal and opposite to the force of A on B:

F_{ BA }=30N left



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