Solved Problem on Dynamics

A box of mass m is on a horizontal surface, the coefficient of kinetic friction between the box and surface is μ. A force \( \vec{F} \) is applied making an angle α with the horizontal.
a) For what value of the angle α, the acceleration of the box is maximum?
b) For what values of α the box remains at rest?

Problem data:

External force applied in the box: \( \vec{F} \);
Box mass: m;
Coefficient of kinetic friction: μ;
Angle between the applied force and horizontal line: α.

Problem diagram:

Drawing a free-body diagram, we have the forces that act on the box.

\( \vec{F} \): external force applied in the box;
\( \vec{F}_{g} \): gravitational force;
\( {\vec F}_{f} \): force of friction;
\( \vec{N} \): normal reaction force.

Figure 1

Solution

Applying Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]

Drawing the forces in a reference frame and decomposing the forces in the x and y directions (Figure 2)

Direction x:

The component of the external force in direction x is given by

\[ \begin{gather} F_{x}=F\cos \alpha \tag{II} \end{gather} \]

the force of friction is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{f}=\mu N} \tag{III} \end{gather} \]

substituting the expressions (II) and (III) into expression (I)

\[ \begin{gather} F_{x} -F_{f}=ma_{x}\\ F\cos \alpha -\mu N=ma_{x} \end{gather} \]

Figure 2

where a_x=a is the acceleration in the x-direction

\[ \begin{gather} F\cos \alpha -\mu N=ma \tag{IV} \end{gather} \]

Direction y:

The component of the external force in the direction y is given by

\[ \begin{gather} F_{y}=F\sin \alpha \tag{V} \end{gather} \]

the gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{VI} \end{gather} \]

substituting the expressions (V), (VI) and the normal reaction force N into expression (I)

\[ \begin{gather} F_{y} +N-F_{g}=ma_{y}\\ F\sin \alpha +N-mg=ma_{y} \end{gather} \]

As there is no motion in y direction, we have a_y=0, and the equation above reduces to

\[ \begin{gather} F\sin\alpha +N-mg=0 \tag{VII} \end{gather} \]

Equations (II) and (III) can be written as a system of two equations with two variables, N and α

\[ \left\{ \begin{array}{l} F\cos \alpha -\mu N=ma\\ F\sin\alpha +N-mg=0 \end{array} \right. \]

solving the second equation for N

\[ N=mg-F\sin\alpha \]

substituting this value in the first equation

\[ \begin{gather} F\cos \alpha -\mu (mg-F\sin\alpha)=ma\\[5pt] ma=F\cos \alpha -\mu (mg-F\sin\alpha)\\[5pt] a=\frac{F\cos \alpha -\mu mg+\mu F\sin\alpha}{m} \end{gather} \]

factoring the force F and canceling the mass m in the numerator and denominator on the right-hand side

\[ \begin{gather} a=\frac{-\mu {\cancel{m}}g}{\cancel{m}}+\frac{F\cos \alpha +\mu F\sin\alpha }{m}\\ a=-\mu g+\frac{F}{m}(\cos \alpha +\mu \sin\alpha) \end{gather} \]

multiplying the numerator and the denominator of the terms in parentheses by \( \frac{\sqrt{1+\mu ^{2}\;}}{\sqrt{1+\mu ^{2}\;}}=1 \) (multiply by one does not change anything)

Note: If we consider one and μ as sides of a riht triangle, the hypotenuse will be given by \( h=\sqrt{1+\mu ^{2}\;} \) (Figure 3), we can define the sine and cosine of the β angle as

\[ \begin{gather} \sin\beta =\frac{1}{\sqrt{1+\mu ^{2}\;}} \tag{VIII-a} \\[10pt] \cos \beta =\frac{\mu }{\sqrt{1+\mu ^{2}\;}} \tag{VIII-b} \end{gather} \]

Figure 3

\[ a=-\mu g+\frac{F}{m}(\cos \alpha +\mu \sin\alpha)\left(\frac{\sqrt{1+\mu ^{2}\;}}{\sqrt{1+\mu ^{2}\;}}\right) \]

applying the distributive property to the denominator and leaving the numerator

\[ \begin{gather} a=-\mu g+\frac{F\sqrt{1+\mu^{2}\;}}{m}\left(\frac{1}{\sqrt{1+\mu ^{2}\;}}\cos \alpha+\frac{\mu }{\sqrt{1+\mu ^{2}\;}}\sin\alpha \right) \tag{IX} \end{gather} \]

Substituting the sine (VIII-a) and cosine (VIII-b) above in the expression (V)

\[ a=-\mu g+\frac{F\sqrt{1+\mu ^{2}\;}}{m}\left(\sin\beta\cos \alpha +\cos \beta \sin\alpha \right) \]

From Trigonometry, the sine of a sum of arcs is given by

\[ \sin(x+y)=(\sin x\cos y+\sin y\cos x) \]

u applying this identity to the term in parentheses

\[ \begin{gather} a=-\mu g+\frac{F\sqrt{1+\mu ^{2}\;}}{m}\sin(\alpha +\beta) \tag{X} \end{gather} \]

This is the expression for the acceleration of the box. Analyzing the result, we see that the maximum value occurs when the sine is equal to 1, this happens when \( \alpha +\beta =\frac{\pi }{2} \) is the angle that produces the maximum acceleration, where α=α_max. In Figure 4 we see angles α_max and β in a graph, the sum of these angles are \( \frac{\pi }{2} \)

\[ \cos \alpha _{max}=\sin\beta \]

Figure 4

substituting the value of the sin β defined in the expression (VIII-A)

\[ \cos \alpha _{max}=\frac{1}{\sqrt{1+\mu ^{2}\;}} \]

\[ \bbox[#FFCCCC,10px] {\alpha_{max}=\arccos\frac{1}{\sqrt{1+\mu ^{2}\;}}} \]

b) For the box to remain at rest, we must have the acceleration equal to zero in the expression (X), we have that the angle α₀ for which the box does not move will be

\[ \begin{gather} 0=-\mu g+\frac{F\sqrt{1+\mu^{2}\;}}{m}\sin(\alpha _{0}+\beta )\\ \frac{F\sqrt{1+\mu^{2}\;}}{m}\sin(\alpha _{0}+\beta )=\mu g\\ \sin(\alpha _{0}+\beta )=\mu g\frac{m}{F\sqrt{1+\mu^{2}\;}}\\ \alpha_{0}+\beta =\arcsin\left(\mu g\frac{m}{F\sqrt{1+\mu ^{2}\;}}\right)\\ \alpha_{0}=\arcsin\left(\mu g\frac{m}{F\sqrt{1+\mu ^{2}\;}}\right)-\beta \end{gather} \]

from the expression (VIII-A)

\[ \sin\beta =\frac{1}{\sqrt{1+\mu ^{2}\;}}\Rightarrow \beta=\arcsin\frac{1}{\sqrt{1+\mu ^{2}\;}} \]

substituting above, the α₀ angle will be given by

\[ \bbox[#FFCCCC,10px] {\alpha _{0}=\arcsin\left(\mu g\frac{m}{F\sqrt{1+\mu^{2}\;}}\right)-\arcsin\left(\frac{1}{\sqrt{1+\mu^{2}\;}}\right)} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .