Figure 5-19 gives the free-body diagram for four situations in which an object is pulled by several forces across a frictionless floor, as seen from overhead. In which situations does the acceleration$\overrightarrow{\mathbf{a}}$of the object have (a) an x component and (b) a y component? (c) In each situation, give the direction ofby naming either a quadrant or a direction along an axis. (Don’t reach for the calculator because this can be answered with a few mental calculations.)

The free-body diagram is given in figure 5-19

The problem is based on Newton’s second law of motion. The law states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. Using the concept of net force from Newton's second law of motion, the net force acting on the given object for given conditions can be found.

Formula:

The net force is given by,

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{Ma}$

According to Newton’s second law, the net force is a product of mass and acceleration.

If we want x component acceleration there must be a net force in the x-direction

So

For situation 1

Net force in the x-direction,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=5\mathrm{N}-3\mathrm{N}-2\mathrm{N} \\ =0\mathrm{N} \end{gathered}$$

So there is no x component of acceleration

For situation 2,

Net Force in the x-direction,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=3\mathrm{N}-2\mathrm{N} \\ =1\mathrm{N} \end{gathered}$$

As net force is$1\mathrm{N}$, the x component of acceleration is present

For situation 3,

Net Force in the x-direction,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=3\mathrm{N}-2\mathrm{N} \\ =1\mathrm{N} \end{gathered}$$

As net force is $1\mathrm{N}$, the x component of acceleration is present

For situation 4,

Net Force in the x-direction

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=5\mathrm{N}-4\mathrm{N} \\ =1\mathrm{N} \end{gathered}$$

As net force is$-2\mathrm{N}$ , the x component of acceleration is present

For situation 1,

Net force in the y-direction

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=7\mathrm{N}-4\mathrm{N} \\ =3\mathrm{N} \end{gathered}$$

So the net force is$3\mathrm{N}$,the y component of acceleration

For situation 2,

Net Force in the y-direction

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=6\mathrm{N}-2\mathrm{N}-4\mathrm{N} \\ =0\mathrm{N} \end{gathered}$$

As net force is no y component of acceleration is present

For situation 3,

Net Force in the y-direction,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=6\mathrm{N}-7\hspace{0.17em}\mathrm{N} \\ =-1\mathrm{N} \end{gathered}$$

As net force is$-1\mathrm{N}$, the y component of acceleration is present

For situation 4,

Net Force in the y-direction

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=3\mathrm{N}+2\mathrm{N}-5\mathrm{N}-4\mathrm{N} \\ =-4\mathrm{N} \end{gathered}$$

As net force is $-4\mathrm{N}$, the y component of acceleration is present

The direction of acceleration is in direction of the net force

For situation 1 there is only net force is only in the +y direction so acceleration is also in the +y direction

For situation 2 there is only net force is only +x direction so acceleration is also +x direction

For situation 3 as there is a net force both in the x and y direction and the total net force is in the fourth quadrant.

For situation 4 as there is a net force both in the x and y direction and the total net force is in the third quadrant.