Q29E - University Physics with Modern Physics

Question

 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.

Step-by-Step Solution

Answer

(a) The free-body diagram for the chair is

                         

(b) The normal force that the floor exerts on the chair is $141.8 \mathrm{N}$.

Step 1: Identification of the given data:

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 $\mathrm{\theta }=37°$.

Step 2: Significance of normal reaction force

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.  

Step 3: (a) Draw the free-body diagram for the chair

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 $9.81 \mathrm{m}/{\mathrm{s}}^2$.

Step 4: (b) Determination of the normal force that the floor exerts on the chair

The expression to calculate the normal force that the floor exerts on the chair is 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 chair is$141.8 \mathrm{N}$.