Horizontal Blocks in Contact - Learn - ScienceFlip

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