Question

An $\mathbf{8}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{\text{kg}}$object slides down a fixed, frictionless, inclined plane. Use a computer to determine and tabulate (a) the normal force exerted on the object and (b) its acceleration for a series of incline angles (measured from the horizontal) ranging from ${\mathbf{0}}^{\mathbf{\circ }}$to ${\mathbf{90}}^{\mathbf{\circ }}$in ${\mathbf{5}}^{\mathbf{\circ }}$ increments. (c) Plot a graph of the normal force and the acceleration as functions of the incline angle. (d) In the limiting cases of ${\mathbf{0}}^{\mathbf{\circ }}$and ${\mathbf{90}}^{\mathbf{\circ }}$, are your results consistent with the known behavior?

Short answer

(a) The tabulation of the normal force exerted on the object.

$\theta (\text{indeg)}$	$\theta$
0	82.32
5	82.00
10	81.06
15	79.51
20	77.35
25	75.61
30	71.29
35	67.43
40	63.06

45	58.21
50	52.92
55	47.22
60	41.17
65	34.81
70	28.18
75	21.32
80	14.31
85	7.18
90	0

(b) The tabulation of the acceleration.

localid="1663697324586" $\theta (\text{indeg)}$	localid="1663696414745" $a_x=9.8\sin \theta {\text{m/s}}^2$
0	0
5	0.85
10	1.70
15	2.54
20	3.34
25	4.14
30	4.90
35	5.62
40	6.30

45	6.92
50	7.50
55	8.02
60	8.48
65	8.88
70	9.20
75	9.46
80	9.65
85	9.75
90	9.8

(c) The graph of normal force and acceleration vs incline angle are shown below.

(d) The graph obtained results are consistent with the known behaviour.

Step-by-step solution

Write the given data from the question:

The mass of object, $m=8.40\mathrm{kg}$

The inclined plane is frictional less.

 Step 2: The formulas to calculate the normal force and acceleration:

The expression to calculate the force on the object is given as follows.

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, $\mathit{F}$ is the working acting on object, $\mathit{m}$is the mass of the object and $\mathit{a}$ is the acceleration of the object.

Consider the free body diagram shown below.

(a) Determine the value for the normal force exerted on the object.

Consider the equilibrium along the $y-$axis.

$\begin{array}{c}\sum F_y=0\\ n-mg\cos \theta =0\\ n=mg\cos \theta \end{array}$

Substitute $84\text{kg}$for $m$and $9.81{\text{m/s}}^2$for $g$into above equation.

$\begin{array}{c}n=8.4\times 9.8\times \cos \theta \\ =82.32\cos \theta \end{array}$

(a) The tabulation of the normal force exerted on the object.

0	82.32
5	82.00
10	81.06
15	79.51
20	77.35
25	74.61
30	71.29
35	67.43
40	63.06

45	58.21
50	52.92
55	47.22
60	41.17
65	34.81
70	28.18
75	21.32
80	14.31
85	7.18
90	0

(b) Determine the value for the acceleration:

Consider the equilibrium along the axis.

$\begin{array}{rcl}\sum F_{}& =& m{a}_x\\ mg\sin \theta & =& m{a}_x\\ g\sin \theta & =& a_x\end{array}$

Substitute $9.81{\text{m/s}}^2$for $g$ into above equation.

$a_x=9.8\sin \theta$

$\theta \left(\text{indeg}\right)$	$a_x=9.8\sin \theta {\text{m/s}}^2$
0	0
5	0.85
10	1.70
15	2.54
20	3.34
25	4.14
30	4.90
35	5.62
40	6.30

45	6.92
50	7.50
55	8.02
60	8.48
65	8.88
70	9.20
75	9.46
80	9.65
85	9.75
90	9.8

(c) Plot a graph of the normal force and the acceleration as functions of the incline angle:

The graph between the normal force and angle is shown below.

The graph between the acceleration and angle is shown below.

(d) Determine that results consistent with the known behaviour:

From the graph we can see that,

$\begin{array}{rcl}a\left(0\right)& =& 0{\text{m/s}}^2\\ n\left(0\right)& =& 82.32\mathrm{N}\end{array}$

And

$\begin{array}{l}a\left(90\right)=9.8{\text{m/s}}^2\\ n\left(90\right)=0\mathrm{N}\end{array}$

Hence, the graph obtained results are consistent with the known behaviour.