Family Vacation!

a) Draw a free body diagram while rotating the \({x-}\)axis to align with the direction of the car’s velocity. Write Newton’s Second Law, and plug in numerical values to obtain two equations for both the \({x-}\) and \({y-}\)directions. Use these equations to find the normal force and the force due to friction. 

b) Same hint as (a) but solve for acceleration rather than force. Solve for \(a_x\).

c) Same hint as b).

a) Newton’s Second Law in the \({x-}\)direction can be written as:

\begin{equation*}
f_{r}- mg \sin \theta = 0,
\end{equation*}

which, when using numerical values, yields:

\begin{equation*}
f_{r} = 502.8 \, \text{N}.
\end{equation*}

Newton’s Second Law in the \({y-}\)direction is:

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

and the maximum force due to friction is:

\begin{equation*}
f_{r \, max} = \mu N,
\end{equation*}

Plugging in numerical values, we get:

\begin{equation*}
f_{r \, max} = 828.8 \, \text{N}.
\end{equation*}

Since \(f_{r \, max} > f_r \), the luggage will not slide.

b) Newton’s Second Law in the \({x-}\)direction can be written as:

\begin{equation*}
f_{r}- mg \sin \theta = ma,
\end{equation*}

which, with numerical values, gives:

\begin{equation*}
a_x = 2.17 \; \text{m/s}^2.
\end{equation*}

c) Newton’s Second Law in the \({x-}\)direction can be written as:

\begin{equation*}
-f_{r}- mg \sin \theta = -ma,
\end{equation*}

which, with numerical values, is:

\begin{equation*}
a_x = 8.88 \; \text{m/s}^2.
\end{equation*}

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

a) If the car is not accelerating, then we can treat the luggage as if it was placed over a regular inclined plane that is at rest (with no acceleration). In that case, we need to figure out if the static force of friction that the upper surface of the car exerts over the luggage is strong enough to prevent the luggage from sliding.

As with all problems with forces, let’s start by making a force diagram, using a system of coordinates where the X axis is aligned with the slope of the hill and points in the direction of motion of the car (so that the car and the luggage only move along the X axis). Notice that over the luggage there are two forces in the X direction, the X-component of the weight which goes in the negative X direction and the static friction, which points in the positive X direction (this force is what keeps the luggage from sliding). In Y, there are two forces, the Y-component of the weight, and the normal force. Thus, the force diagram is shown in figure 1.

Figure 1: Force diagram for the luggage. We choose the coordinate system with X parallel to the road and Y perpendicular to it. The forces depicted are: the contact force \(N\), the weight \(W\), and the friction \(f_r\). The weight is decomposed along the X and Y axis, using the same angle of inclination for the road \(\theta\).

Now, to determine if the luggage falls, we can simply write Newton’s second law along the X axis:

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

where \(f_r\) is the static friction exerted by the top of the car and \(W_x\) is the X-component of the weight, which is given by \(W \sin \theta\), as we can see from the diagram (and recall that \(W=mg\)). Since the condition is that the luggage does not slide, and since the car has no acceleration, we can set \(a_x=0\) (no acceleration). Thus, we get

\begin{equation}
f_r \, \hat{\textbf{i}}- {(mg \sin \theta)} \, \hat{\textbf{i}}= 0 \, \hat{\textbf{i}}.
\label{CarroMaleta_friccionPeso1}
\end{equation}

If we focus only on the magnitudes, and move the weight term to the other side, we get

\begin{equation}
f_r= {(mg \sin \theta)}.
\label{CarroMaleta_friccionPeso}
\end{equation}

Now, since we have the mass, this equation allows us to determine what the magnitude of the friction must be so that the luggage does not slide. The question, then, is whether the car’s top can exert a force of  static friction of that magnitude. How do we determine this? Well, we know that the maximum force of static friction that the top of the car can exert is given by \(\mu N\), so we can compute this maximum force and then see if it is greater than the force required according to equation \eqref{CarroMaleta_friccionPeso}. That is, we want to see if

\begin{equation}
f_{r\; max} = \mu N
\label{CarroMaleta_friccionNormal}
\end{equation}

is greater than the force given by equation \eqref{CarroMaleta_friccionPeso}.

In order to use this equation, we need \(N\), and to find \(N\) we need the force equations along Y:

\begin{equation}
N \, \hat{\textbf{j}}- W_y \, \hat{\textbf{j}}= m a_y \, \hat{\textbf{j}}.
\end{equation}

But in Y the luggage has no acceleration, and from the force diagram we can see that \(W_y = mg \cos \theta\). Thus, from this equation we get

\begin{equation}
N= {(mg \cos \theta)}.
\label{CarroMaleta_normal}
\end{equation}

We can then use this in equation \eqref{CarroMaleta_friccionNormal}:

\begin{equation}
f_{r\; max} = \mu {(mg \cos \theta)}.
\end{equation}

Let’s insert the numerical values here and in equation \eqref{CarroMaleta_friccionPeso}. We get

\begin{equation}
f_r = 502.8 \, \text{N},
\end{equation}

and

\begin{equation}
f_{r \; max} = 828.8 \, \text{N}.
\end{equation}

Thus, we see that the force of friction that is required to keep the luggage in place is less than the maximum force that the top of the car can exert, and so we infer that the luggage will not slide.

As a side-note, notice that we could have solved this problem by finding the critical angle, which is the angle after which an object will start sliding. This angle is given by \(\arctan (\mu_s)\). This angle is clearly greater than the angle of the slope, and so this is another way of figuring out that the luggage does not slide.

b) Here, the reasoning is very similar to the previous case. The maximum acceleration that the car can have uphill before the luggage starts sliding can be found by figuring out what the maximum acceleration the top of the car can give to the luggage (if the car has an acceleration greater than this acceleration, then the luggage will start sliding). And the maximum acceleration that the top of the car can exert on the luggage is given by the maximum force of static friction that the top can exert, which is given by equation \eqref{CarroMaleta_friccionNormal} above.

So, from (a), we know that the maximum force of static friction that the top of the car can exert is:

\begin{equation}
f_{r\; max} =mg \cos \theta.
\end{equation}

Let’s then use equation \eqref{CarroMaleta_sumatoriaFuerzasX} again, but now we won’t set \(a_x=0\) because we want the luggage to accelerate together with the car. If we use the facts that \(W_x=m g \sin \theta\) and \(f_r\) is the maximum force of friction, then the equation \eqref{CarroMaleta_sumatoriaFuerzasX} becomes:

\begin{equation}
{(\mu mg \cos \theta)} – {(mg \sin \theta)} = m a_x. \label{CarroMaleta_aceleracionFuerzas}
\end{equation}

Dividing by \(m\), we get:

\begin{equation}
\mu g \cos \theta – g \sin \theta = a_x.
\label{CarroMaleta_aceleracion}
\end{equation}

Finally, let’s insert the numerical values:

\begin{equation}
{(0.6)} {(9.8 \; \text{m/s}^2)} \cos {(20^\circ)} – {(9.8 \; \text{m/s}^2)} \sin {(20^\circ)} = a_x. \label{CarroMaleta_reemplazando}
\end{equation}

The result is

\begin{equation}
a_x = 2.17 \; \text{m/s}^2.
\end{equation}

This is the maximum acceleration that the top of the car can give the luggage, and so the maximum acceleration that the car can reach when going uphill without making the luggage slide.

c) If the car stops, then the luggage will slide towards the front since it will tend to keep its state of motion (Newton’s first law). The greater the magnitude of the acceleration the car has when stopping, the easier it would be for the luggage to slide towards the front. The reasoning is similar to case (b), except that now the acceleration in question is negative and not positive because the velocity is decreasing (and so \((v_f \, \hat{\textbf{i}} – v_i \, \hat{\textbf{i}})\) is negative).

Since the acceleration is negative, the direction of some of the forces is different as well. In particular, the force of static friction that the top of the car exerts over the luggage goes along the negative X-axis since this force is trying to decrease the luggage’s velocity as the car stops (in case (a), this force was positive because it was helping the luggage increase its speed while the car accelerates). Thus, the force diagram for the luggage in this case is shown in figure 2.

Figure 2: Force diagram for the luggage. We choose the coordinate system with X parallel to the road and Y perpendicular to it. The forces depicted are: the contact force \(N\), the weight \(W\) and the friction \(f_r\) which for this cases changes its direction. The weight is decomposed along the X and Y axis using the same angle of inclination fo the road \(\theta\).

We are interested in finding the maximum braking acceleration that the static friction of the top of the car can give to the luggage. If the car were to have an even greater braking acceleration than this one, then the luggage would slide.

In order to write the equation of forces along X, we need to note that the static friction now has a negative sign because the force is helping the luggage stop, and the acceleration also has a negative sign because the velocity is decreasing. Thus, we get

\begin{equation}
-W_x \, \hat{\textbf{i}}-f_{r} \, \hat{\textbf{i}} = – ma_x. \; \hat{\textbf{i}}
\end{equation}

As before, we know that \(W_x= mg \sin \theta\). Thus,

\begin{equation}
-{(mg \sin \theta)} \, \hat{\textbf{i}}-f_{r} \, \hat{\textbf{i}} = – ma_x. \; \hat{\textbf{i}}
\end{equation}

The maximum braking acceleration for the luggage is given by the maximum static friction (for the static friction is the only force that can help the luggage stop). The magnitude of the maximum static friction is always \(f_{r \; max}= \mu N\).

We already know \(\mu N\) from points (a) and (b), so we can simply plug that result in the equation of forces along X. This gives:

\begin{equation}
-mg \sin \theta – {(\mu m g \cos \theta) }=-ma_x.
\end{equation}

Dividing by \(-m\), we get:

\begin{equation}
g \sin \theta + \mu g \cos \theta = a_x.
\end{equation}

Finally, we insert the numerical values:

\begin{equation}
{(9.8 \; \text{m/s}^2)} \sin {(20^\circ)} + {(0.6)} {(9.8 \; \text{m/s}^2)} \cos {(20^\circ)} = a_x, \label{CarroMaleta_reemplazandoC}
\end{equation}

to get

\begin{equation}
a_x = 8.88 \; \text{m/s}^2.
\end{equation}

This is the magnitude of the maximum breaking acceleration the car can have before the luggage slides. The direction of this acceleration is along the negative X axis, as we explained earlier.

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