91Ó°ÊÓ

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$\mathrm{F}=40.0 \mathrm{N}$.

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

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:

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

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

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

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

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