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 A:
{ 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