A chair of mass $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force $\mathbf{F}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{N}$that is directed at an angle of $\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{°}$below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton’s laws to calculate the normal force that the floor exerts on the chair.

The given data can be listed below as,

The mass of a chair is,$\mathrm{m}=12.0\mathrm{kg}$.

The magnitude of an external force is$\mathrm{F}=40.0\hspace{0.33em}\mathrm{N}$.

The angle of inclination of the force is$\mathrm{\theta }=37°$.

Whenever an object lies on a horizontal surface, there would be two forces acting on the object: the weight force and the normal reaction force. The value of the normal reaction force helps to obtain the friction force.

The clearly labeled free-body diagram of the chair is given below.

Here, $\mathrm{N}$is the normal reaction force that the floor exerts on the chair by the horizontal ground surface, $\mathrm{W}$is the weight of the chair and $\mathrm{g}$is the gravitational acceleration whose value is $9.81\mathrm{m}/{\mathrm{s}}^2$.

The expression to calculate the normal force that the floor exerts on the chairis expressed as follows:

$\begin{array}{l}\mathrm{N}=(\mathrm{W}+\mathrm{F}sin(9)\\ =(\mathrm{mm}+\mathrm{F}sin\mathrm{\theta })\end{array}$

Here,$\mathrm{N}$is the normal force that the floor exerts on the chair.

Substituting the values in the above expression, and you get:

$$\begin{gathered} \mathrm{N}=\left[\left(12.0\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right)+\left(40.0\mathrm{N}\right)\sin 37°\right] \\ =\left[\left(117.72\mathrm{N}\right)+\left(40.0\mathrm{N}\times 0.602\right)\mathrm{N}\right] \\ =117.72\mathrm{N}+24.08\mathrm{N} \\ =141.8\mathrm{N} \end{gathered}$$

Thus, the normal force that the floor exerts on the chairis$141.8\mathrm{N}$.