A married couple, Josie and Sean, are returning home after a nice trip to the movies. Sean is recovering from surgery, and currently travels by wheelchair. Josie is pushing Sean up the ramp to their home which is angled at 23º with respect to the ground. (Sean has a mass of 65 kg and the wheelchair itself has a mass of 10 kg. Assume that the coefficient of kinetic friction is \(\mu_{k}=0.4\) and that the wheels of the wheelchair aligned to move forward, so the wheels will not rotate.) How much force does Josie need to exert to push Sean up the ramp at constant velocity?
Make a free body diagram by rotating the coordinate axes so that the \({x-}\)axis is aligned with the direction of motion. Find the normal force, and use this to find the force due to friction which will help you to find the answer.
Newton’s Second Law, written in terms of the the rotated \({y-}\)axis gives:
\begin{equation*}
N – m_s g \cos \theta – m_w g \cos \theta = 0,
\end{equation*}
where \(m_s\) and \(m_w\) are respectively Sean’s mass and the mass of the wheelchair.
In the \({x-}\)direction, assuming the \({x-}\)axis is aligned with the direction of motion, Sean and Josie are traveling at constant velocity which means the net force in the \({x-}\)direction. Newton’s Second Law, written in terms of the rotated \({x-}\)axis, gives:
\begin{equation*}
F_J – f_r – m_s g \sin \theta – m_w g \sin \theta = 0.
\end{equation*}
Solving for \(F_J\), substituting the other variables, and doing some algebra, we get:
\begin{equation*}
F_J = \mu (m_w+m_s)g \cos \theta + (m_w+m_s)g \sin \theta,
\end{equation*}
Plugging in numerical values:
\begin{equation}
F_J = 557.82 \, \text{N}.
\end{equation}
For a more detailed explanation of any of these steps, click on “Detailed Solution”.
We need to find the force Josie must exert to help Sean move up the ramp at a constant speed. To do so, we have to relate that force to the other variables, namely, Sean’s and the wheelchair weight, the angle of the ramp, and the coefficient of kinetic friction. Before we continue, it is important to understand why it is relevant that the wheels of the wheelchair do not rotate. If the wheels were rotating, then there would also be static friction, and this problem would have been more complicated (a wheel that rotates does so due to the static friction, not the kinetic friction).
Now, to find the relation between Josie’s force and the other variables, we’ll have to identify all the forces and use Newton’s second law. Let’s start by using a convenient coordinate system. For inclined planes, it is convenient to use a system where the X axis is parallel to the plane, as illustrated in figure 1.
Figure 1: We choose the coordinate system with the X axis parallel to the inclined plane and Y perpendicular to it.
We can simplify the analysis a lot if we treat Sean and the wheelchair as a single system. If we do so, we only care about their total mass (and their total weight), and we do not need to consider the forces between them (for example, we can ignore the normal force between the wheelchair and Sean’s bottom).
(In many books and sometimes in physics classes, it is not always explicitly explained which objects in a given problem are treated as a single system and which objects are treated as independent objects. Usually this is clear from context, but it is always better to be as explicit as possible).
Now, over the system given by Sean and the wheelchair, there are four forces. One is the normal force produced by the ramp, which is aligned with the Y axis. The other force is the weight (which is the total weight, coming from the sum of Sean’s and the wheelchair’s weight). The weight points downward and vertically. Another force is the one that Josie exerts when pushing the wheelchair, a force that points in the positive X direction. And finally, there is the force of kinetic friction that the floor makes on the wheels, pointing in the negative X direction (see figure 2).
Figure 2: Force diagram for Sean. The forces shown are the contact force with the inclined plane \({N}\), the friction \(f_r\), the weight \(W\) and the force exerted by Josie \(F_J\). The weight is decomposed in its components along the X and Y axis.
Having identified the forces, we can write Newton’s second law to find a relation between Josie’s force and the other variables. As we explained above, in X there are three forces, Josie’s, which is positive, the X component of the weight, which is negative, and the kinetic friction, also negative. Hence, we get
\begin{equation}
\label{Wheelchair_fuerzasX}
F_J \, \hat{\textbf{i}} – f_r \, \hat{\textbf{i}} – W_x \, \hat{\textbf{i}} = m a_x \, \hat{\textbf{i}},
\end{equation}
where \(W_x\) is the x component of the total weight of the Sean-Wheelchair system and \(a_x\) is the magnitude of the acceleration in X. But we want the force that Josie exerts to move the wheelchair (and Sean) with constant velocity. Hence, the acceleration in equation \eqref{Wheelchair_fuerzasX} is zero
\begin{equation}
F_J \, \hat{\textbf{i}} – f_r \, \hat{\textbf{i}} – W_x \, \hat{\textbf{i}} = 0 \, \hat{\textbf{i}}.
\end{equation}
If we move \(f_r\) and \(W_x\) to the other side, and focus on the magnitudes only, we get
\begin{equation}
\label{Wheelchair_FuerzaJosie}
F_J = f_r + W_x.
\end{equation}
So to find \(F_J\), we need to find \(W_x\) and \(f_r\). \(W_x\) is easy to find, since from the force diagram it is clear that
\begin{equation}
W_x = W \sin \theta.
\end{equation}
where \(\theta\) is the ramp’s angle (which is known). Also, recall that \(W=mg\), where \(m\) is the total mass (that is, \(m=m_w+m_s\), the sum of the wheelchair’s mass and Sean’s mass). So we have
\begin{equation}
W_x = (m_w+m_s)g \sin \theta.
\end{equation}
Let’s then insert this in equation \eqref{Wheelchair_FuerzaJosie} to get
\begin{equation}
\label{Wheelchair_fuerzaJosieParaReemplazar}
F_J = f_r + (m_w+m_s)g \sin \theta.
\end{equation}
All we need to find now is the kinetic friction. Recall that the magnitude of the kinetic friction is given by
\begin{equation}
\label{Wheelchair_friccion}
f_r = \mu N,
\end{equation}
where \(N\) is the magnitude of the normal produced by the surface. To find \(N\), we need to consider Newton’s second law along Y. According to our coordinate system, in Y there are only two forces, the Y-component of the weight, which is negative, and the normal, which is positive. So we get
\begin{equation}
N \, \hat{\textbf{j}} – W_y \, \hat{\textbf{j}} = m a_y \, \hat{\textbf{j}}.
\end{equation}
However, the Sean-wheelchair system does not move along Y, and so \(a_y\) is zero.
\begin{equation}
\label{Wheelchair_fuerzasY}
N \, \hat{\textbf{j}} – W_y \, \hat{\textbf{j}} = 0 \, \hat{\textbf{j}}.
\end{equation}
Let’s now use again the force diagram to write
\begin{equation}
W_y = W \cos \theta.
\end{equation}
And use again that \(W = (m_w+m_s)g\). So
\begin{equation}
W_x = (m_w+m_s)g \cos \theta.
\end{equation}
If we use this result in equation \eqref{Wheelchair_fuerzasY} and focus on the magnitudes only, we get
\begin{equation}
N – (m_w+m_s)g \cos \theta = 0.
\end{equation}
Rearrange the terms to get
\begin{equation}
N = (m_w+m_s)g \cos \theta.
\end{equation}
Now, let’s use this in equation \eqref{Wheelchair_friccion}
\begin{equation}
f_r = \mu ( (m_w+m_s)g \cos \theta ).
\end{equation}
All we need to do now is insert this in equation \eqref{Wheelchair_fuerzaJosieParaReemplazar} to get an expression for Josie’s force:
\begin{equation}
F_J = \mu (m_w+m_s)g \cos \theta + (m_w+m_s)g \sin \theta.
\end{equation}
If we insert the numerical values for the different variables
\begin{equation}
F_J = (0.4) [(10 \, \text{kg})+(65 \, \text{kg})] (9.8 \, \text{m/s}^2) \cos (23^\circ) + [(10 \, \text{kg})+(65 \, \text{kg})] (9.8 \, \text{m/s}^2) \sin (23^\circ),
\end{equation}
we get
\begin{equation}
F_J = 557.82 \, \text{N}.
\end{equation}
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