Two horizontal forces $\overrightarrow{{\mathit{F}}_{\mathbf{1}}}$and $\overrightarrow{{\mathit{F}}_{\mathbf{2}}}$act on a $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$disk that slides over frictionless ice, on which an x-ycoordinate system is laid out. Force $\overrightarrow{{\mathit{F}}_{\mathbf{1}}}$is in the positive direction of the xaxis and has a magnitude of $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. Force $\overrightarrow{{\mathbf{F}}_{\mathbf{2}}}$has a magnitude of$\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. Figure gives the xcomponent ${\mathit{V}}_{\mathbf{x}}$of the velocity of the disk as a function of time tduring the sliding. What is the angle between the constant directions of forces $\overrightarrow{{\mathit{F}}_{\mathbf{1}}}$and $\overrightarrow{{\mathbf{F}}_{\mathbf{2}}}$?

• Mass of the disk, $m=4.0\mathrm{kg}$.

The acceleration is the rate of change of velocity with respect to time. If we have a graph of velocity against time, then the slope of that graph represents the acceleration.

Using a graph, we can find the acceleration of the disk. As we are given that the 1st force is in the horizontal direction, so the 2nd force must be at some angle with x-direction. Using this information and the net acceleration, we can find the angle between them.

Formula:

The force of a body according to Newton’s second law,

$F=ma$

localid="1657016494854" $\left(\mathrm{i}\right)$

Here, F is the force acting on the body, m is the mass of the body and a is the acceleration of the body.

The acceleration of a body in motion, $a=\frac{dv}{dt}$ (ii)

From the graph

$$\begin{gathered} \mathrm{slope}\mathrm{of}\mathrm{graph}=\frac{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)} \\ \mathrm{since},\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(0,4\right)\mathrm{and}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(3,5\right) \\ =\frac{\left(5-\left(-4\right)\right)}{\left(3-0\right)} \\ =\frac{9}{3} \\ =3 \end{gathered}$$

Slope of graph = Acceleration of the motion (from equation (ii))

$=3\mathrm{m}/\mathrm{s}$$\left(\because \mathrm{The}\mathrm{graph}\mathrm{is}\mathrm{a}\mathrm{displacement}\mathrm{vs}\mathrm{time}\mathrm{graph}\right)$

Again, substituting the values in equation (i), we get force in x-direction. Substitute the x component of forces, the mass and acceleration values in equation (i) to calculate the angle.

role="math" localid="1657017376723" $F_x=m{a}_x$

$F_1+F_2\cos \theta =4\mathrm{kg}\times 3\mathrm{m}/{\mathrm{s}}^2$

$7.0\mathrm{N}+9\mathrm{N}\times \mathrm{cos\theta }=12\mathrm{N}$

$9\mathrm{Ncos\theta }=12\mathrm{N}-7\mathrm{N}$

$\cos \theta =\left(\frac{5}{9}\right)$

$\theta ={\cos }^{-1}\left(0.55\right)$

$=56.25$

$\approx 56°$

Hence, the angle of motion is $56°$.