The Box Ghost!

Rotate the coordinate axes to make sure that one axis point will point in the direction of the box’s motion. Determine the acceleration by using a kinematics equation that relates the velocity and acceleration, and the free body diagrams will help you to write the equations needed to solve the problem.

Newton’s Second Law, based on the free body diagram with the rotated \({y-}\)axis gives:

\begin{equation*}
N-mg \cos \theta=0.
\end{equation*}

Newton’s Second Law, based on the rotated \({x-}\)axis gives:

\begin{equation*}
mg \sin \theta – f_r=m a_x,
\end{equation*}

where \(f_r = \mu N = \mu mg \cos \theta\). For \(a_x\) we can use:

\begin{equation*}
a_x = \frac{v_f – v_i}{t}.
\end{equation*}

Substituting \(a\) and \(f_r\) into the equation along the \({x-}\)axis, and solving for \(\mu\) with some algebra, we get:

\begin{equation*}
\mu = – \frac{v_f}{t g \cos \theta} + \tan \theta.
\end{equation*}

Plugging in numerical values yields:

\begin{equation*}
\mu = 0.56.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution” below.

In order to determine the coefficient of kinetic friction, we need to find a relation between that coefficient and the other variables we know. Notice that the coefficient of friction partially determines the force of friction, and the force of friction plays a role in the acceleration of the block along the inclined plane (we can easily find this acceleration because we know the initial and final speeds after two seconds). Thus, in order to find a relation between the coefficient of kinetic friction and the acceleration, we need to consider Newton’s Second Law.

Let’s start by making the force diagram for the box. As is usual for problems with inclined planes, we will use a system of coordinates where the X axis points in the direction of motion, and where the Y axis is perpendicular to the inclined plane (this kind of system has the benefit that most of the forces and the acceleration are aligned with the axes, and so we do not need to find their components).

Figure 1: We choose the coordinate system with the X axis parallel to the ramp and the Y axis perpendicular to it.

There are two forces along the Y axis. One is the normal force, that points in the positive Y direction, and the other one is the Y component of the weight. Along the X axis, we have two forces, the force of friction that opposes the motion of the box (and so points in the negative X direction) and the X-component of the weight, that points in the positive X direction. Thus, the force diagram is shown in figure 2.

Figure 2: Force diagram for the box as it slides down the ramp. The forces shown are the contact force with the ramp \(N\), the friction force \(f_r\), and the weight \(W\) with its components along the X and Y axis.

We now write Newton’s Second Law in the X-axis. As is clear from the force diagram, there are two forces in X, the X-component of the weight (positive in X) and the friction (negative in X), and so we get

\begin{equation}
W_x \, \hat{\textbf{i}} – f_r \, \hat{\textbf{i}} = m a_x \, \hat{\textbf{i}},
\label{CajaFriccion_fuerzasX}
\end{equation}

where \(a_x \, \hat{\textbf{i}}\) is the acceleration of the box, whose direction is along the positive X axis. We have enough information to find this acceleration because we know the change of speed after 2 seconds; however, let’s find its value a bit later.

One can see from the force diagram that \(W_x\) is given by \(W \sin \theta\) (also, recall that \(W=mg\)). And we also know that the magnitude of the kinetic force of friction is given by \(\mu N\), where \(N\) is the magnitude of the normal force and \(\mu\) is the coefficient of kinetic friction that we know. Thus, equation \eqref{CajaFriccion_fuerzasX} becomes

\begin{equation}
{(mg \sin \theta)}\, \hat{\textbf{i}} – {(\mu N)} \, \hat{\textbf{i}}= m a_x \, \hat{\textbf{i}}.
\label{CajaFriccion_fuerzasXReemplazando}
\end{equation}

Since the acceleration in X is constant, it is given by

\begin{equation}
\frac{(v_{f} \, \hat{\textbf{i}} – v_i \, \hat{\textbf{i}})}{t} = a_x \, \hat{\textbf{i}},
\end{equation}

where \(v_f\) is the final speed after time \(t\) and \(v_i\) is the initial speed. Since \(v_i\) is zero because the box starts from rest, we get

\begin{equation}
\frac{v_f}{t} \, \hat{\textbf{i}}= a_x \, \hat{\textbf{i}}.
\end{equation}

Let’s then use this in equation \eqref{CajaFriccion_fuerzasXReemplazando}:

\begin{equation}
mg \sin \theta \, \hat{\textbf{i}} – \mu N \, \hat{\textbf{i}}= m {\frac{v_f}{t}} \, \hat{\textbf{i}}.
\label{CajaFriccion_fuerzasXConAceleracion}
\end{equation}

Now, from this equation we cannot yet find \(\mu\) because we do not know \(N\) (but we know \(t\), \(v_f\), \(m\) and \(\theta\)). Thus, we need to write the force equation along the Y axis.

In the Y axis, the block is at rest (it only moves along X). Also, the normal force is positive and the Y-component of the weight is negative. Thus, Newton’s Second Law along the Y axis is

\begin{equation}
N \, \hat{\textbf{j}} – W_y \, \hat{\textbf{j}} = 0 \, \hat{\textbf{j}}.
\label{CajaFriccion_FuerzasY}
\end{equation}

But from the force diagram, we can see that \(W_y\) is given by \(W\cos \theta\) (and again, \(W=mg\)). So, equation \eqref{CajaFriccion_FuerzasY} becomes

\begin{equation}
N \, \hat{\textbf{j}} – {(mg \cos \theta)} \, \hat{\textbf{j}} = 0 \, \hat{\textbf{j}}.
\end{equation}

If we move \(mg \cos \theta\) to the other side, and if we focus only on the magnitudes, we find that

\begin{equation}
N = mg \cos \theta.
\label{CajaFriccion_Normal}
\end{equation}

Let’s use this result in equation \eqref{CajaFriccion_fuerzasXConAceleracion}. We get

\begin{equation}
mg \sin \theta \, \hat{\textbf{i}} – \mu {(mg \cos \theta )}\, \hat{\textbf{i}} = m \frac{v_f}{t} \, \hat{\textbf{i}}.
\end{equation}

The only unknown variable here is \(\mu\). Thus, let us leave the term with \(\mu\) in the left and move the other terms to the right:

\begin{equation}
– \mu mg \cos \theta \, \hat{\textbf{i}}= m \frac{v_f}{t} \, \hat{\textbf{i}}- mg \sin \theta \, \hat{\textbf{i}}.
\end{equation}

Now, let’s focus on the magnitudes and divide everything by \(-m g \cos \theta\);

\begin{equation}
\mu = – \frac{v_f}{t g \cos \theta} + \tan \theta,
\end{equation}

where the mass cancels everywhere and where we used that \(\sin \theta / \cos \theta= \tan \theta\).

Finally, let’s replace the values of all the terms:

\begin{equation}
\mu = – \frac{{(0.2\; \text{m/s})}}{{(2\; \text{s})} {(9.8\; \text{m/s}^2)} \cos {(30^\circ)}} + \tan {(30^\circ)},
\end{equation}

to get

\begin{equation}
\mu = 0.56.
\end{equation}

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