Newtons Laws EX 3

There is a 100 N gravitational force acting on each block in the above diagram.

Considering the four blocks on the floor we will select a horizontal x-axis and a vertical y-axis pointing up.

We will assume that the blocks will not move up or down. It means that the sum of the vertical components of all forces acting on each block is zero.

We note that we need to consider only y-components of the three forces acting on each block $F_{G}$ ,$F_{N}$ and $F$ because $F_{N}$ has components (0, $F_{N}$).

1. The sum of the y-components of forces acting on the first block is

$-100+50+F_{N}=0$ and therefore $F_{N}=50N$. 

2. The sum of the y-components of forces acting on the second block is

$-100-50+F_{N}=0$ and therefore $F_{N}=150$ N. 

3. The sum of the y-components of forces acting on the third block is

$-100+50sin60^{o}+F_{N}=0$ and therefore $F_{N}=57$ N. 

4. The sum of the y-components of forces acting on the fourth block is

$-100+50sin(-60^{o})+F_{N}=0$ and therefore $F_{N}=143$ N. 

Similarly, the block pushed towards the ceiling has only three forces acting on it.

Focusing only on the y-components of forces we write:

$-100+110+F_{N}=0$ 

and therefore:

$F_{N}=-10$ N. 

It means that the ceiling exerts a downward normal force on the block.

Considering the block that is pushed with a horizontal force pointing towards the wall we will focus on the x-components of forces acting on the block. There are only two such horizontal forces.

We will assume that the block does not move in a horizontal direction.

Therefore the two horizontal forces balance each other and consequently, $F_{N}=50$ N points to right.

We assume that the block on the incline does not move in a direction perpendicular to the incline.

By selecting x-axis parallel to the incline plane and y-axis pointing up from the incline plane, we need to focus on y-components of the gravitational force and the normal force which balance each other since the block does not move in this direction.

Since the gravitational force makes an angle - $120^{o}$ with x-axis we can compute the y-component of the gravitational force to be:

$F_{G}sin(-120^{o})=100x0.867=86.7$ N. 

The y-component of the normal force is $F_{N}$ and it balances the y-component of the gravitational force.

Therefore, the normal force is:

$F_{N}=86.7$ N.