Standing on a Uniform Bar!

a) Sum the torques, and relate the variables to solve for the distance, \({x}\).

b) Same hint as part (a). Note that the distance is different from that given in part (a).

a) At equilibrium, the sum of the torques is:

\begin{equation*}
\sum \vec{\tau} = \vec{0}.
\end{equation*}

In this case:

\begin{equation*}
(\ell_1 + \ell_2 – L/2) Mg – xmg = 0,
\end{equation*}

Solving for \(x\) and plugging in numerical values, we get:

\begin{equation*}
x = 1.2 \, \text{m}.
\end{equation*}

b) The sum of the torques in this case is:

\begin{equation*}
(\ell_1 + \ell_2 {}^{\prime} – L/2) Mg – Lmg = 0,
\end{equation*}

where solving for \(\ell_2 {}^{\prime}\), we get:

\begin{equation*}
\ell_2 {}^{\prime}=L\frac{\left(1+\frac{M}{2m}\right)}{\left(1+\frac{M}{m}\right)}-\ell_1,
\end{equation*}

which, with numerical values, yields:

\begin{equation*}
\ell_2 {}^{\prime} \approx 10.62 \, \text{m}.
\end{equation*}

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

a) To approach the solution of this problem, we must demand rotational equilibrium around point B and find the maximum distance \(x\) such that the bar is still in equilibrium. For this purpose, we first draw the force diagram with the given distances in order to calculate the torque, as seen in figure 1.

Figure 1: Free-body diagram for the bar. The coordinate axis is chosen so that the Y axis is positive in the upwards direction and its origin is at B. The four forces exerted on the bar are shown: the contact forces from the triangular supports \(\vec{N}_A\) and \(\vec{N}_B\), the contact force exerted by the person on the bar \(\vec{N}_p\), and the weight of the bar \(\vec{W}=-Mg\,\hat{\textbf{j}}\) located at the middle of the bar.

We’ll now analyze the forces exerted on the bar and write them explicitly using the given coordinate system. As seen in the figure above, we have the weight of the bar with total mass \(M\) given by \(\vec{W}=-Mg\,\hat{\textbf{j}}\) located at its center of mass. The term \(g\) is the gravitational acceleration on Earth. We have also the force exerted by the person \(\vec{N}_p=-mg\,\hat{\textbf{j}}\), where \(m\) is the mass of the person. We have two contact forces, one at point A \(N_A\,\hat{\textbf{j}}\) and one at point B \(N_B\,\hat{\textbf{j}}\).

As the person walks towards the right of point B, the bar tends to rotate clockwise around point B. The maximum distance the person could walk would be such that \(N_A=0\) and the bar loses contact at point B and starts to rotate. Let’s then consider that exact situation and write Newton’s second law for rotations

\begin{equation}
\label{newton}
\sum\vec{\tau}=I\vec{\alpha},
\end{equation}

where \(I\) is the moment of inertia, \(\vec{\alpha}\) the angular acceleration, and \(\sum\vec{\tau}\) the sum of all the torques around a certain point. In our static case, the angular acceleration is zero, then equation \eqref{newton} becomes

\begin{equation}
\label{newton2}
\sum \vec{\tau}=\vec{0}.
\end{equation}

We may now calculate the torque on point B exerted by each force. In order to do this, we use the definition of torque

\begin{equation}
\label{torque}
\vec{\tau}=\vec{r}\times \vec{F},
\end{equation}

where \(\vec{r}\) is the distance vector from point B to the point where the force \(\vec{F}\) is applied. Let’s then calculate the torque for all the forces.

For the contact force at point B, the distance vector \(\vec{r}\) is zero, so this force does not produce torque. Because \(N_A=0\) then the contact force at point A exerts no torque on point B. We are only left to calculate the torque produced by the weight of the bar and the torque produced by the weight of the person.

In the case of the weight of the bar, the distance vector will have magnitude \(\ell_1+\ell_2-L/2\) (the distance from point B to the center of the bar) and is directed to the left. Hence, using equation \eqref{torque}, we have, for the torque generated by the bar’s weight,

\begin{equation}
\vec{\tau}_M=(-(\ell_1+\ell_2-L/2)\,\hat{\textbf{i}})\times (-Mg\,\hat{\textbf{j}}),
\end{equation}

which is

\begin{equation}
\vec{\tau}_M=(\ell_1+\ell_2-L/2)Mg\,\hat{\textbf{i}}\times \hat{\textbf{j}},
\end{equation}

\begin{equation}
\label{taum1}
\vec{\tau}_M=(\ell_1+\ell_2-L/2)Mg\,\hat{\textbf{k}},
\end{equation}

where in the last line, we have used the fact that \(\hat{\textbf{i}}\times\hat{\textbf{j}}=\hat{\textbf{k}}\).

In the case of the weight of the person, the distance vector will have magnitude \(x\) and is directed to the right. Thus, using equation \eqref{torque}, we have for the torque generated by the person’s weight

\begin{equation}
\vec{\tau}_m=(x\,\hat{\textbf{i}})\times (-mg\,\hat{\textbf{j}}),
\end{equation}

which is

\begin{equation}
\vec{\tau}_m=-xmg\,\hat{\textbf{i}}\times \hat{\textbf{j}},
\end{equation}

\begin{equation}
\label{taum2}
\vec{\tau}_m=-xmg\,\hat{\textbf{k}},
\end{equation}

where in the last line, we have used the fact that \(\hat{\textbf{i}}\times\hat{\textbf{j}}=\hat{\textbf{k}}\).

Using the explicit expressions for the torques given in equations \eqref{taum1} and \eqref{taum2} into equation \eqref{newton2},
we get

\begin{equation}
(\ell_1+\ell_2-L/2)Mg\,\hat{\textbf{k}}-xmg\,\hat{\textbf{k}}=\vec{0}.
\end{equation}

We can drop the vector notation in the equation above since all the quantities are on the same axis, so we can write

\begin{equation}
(\ell_1+\ell_2-L/2)Mg-xmg=0.
\end{equation}

Solving for \(x\) in the equation above, we get

\begin{equation}
xmg=(\ell_1+\ell_2-L/2)Mg,
\end{equation}

\begin{equation}
x=\frac{(\ell_1+\ell_2-L/2)Mg}{mg},
\end{equation}

and cancelling out \(g\), we finally arrive to

\begin{equation}
\label{equis}
x=\frac{(\ell_1+\ell_2-L/2)M}{m}.
\end{equation}

Using the numerical values, we get

\begin{equation}
x=\frac{(4\,\text{m}+7\,\text{m}-(18\,\text{m})/2)(45\,\text{kg})}{75\,\text{kg}},
\end{equation}

\begin{equation}
x=1.2\,\text{m}.
\end{equation}

Thus, the person can be at any distance \(x\) less than or equal to \(1.2\,\text{m}\) from point B and the bar will be at equilibrium.

b) Now we should consider the scenario where the person is standing at the end of the bar. We need to find the new position for pivot B so that the system maintains equilibrium. The situation proposed in this part of the problem is equivalent to saying that

\begin{equation}
\label{condition}
\ell_1+\ell_2’+x=L,
\end{equation}

where \(\ell_2’\) is the new distance from point A to point B and \(x\) is still the distance between point B and the person, as seen in figure 2. Thus, if we calculate \(\ell_2’\), we’ll have the new position of point B with respect to point A.

Figure 2: Coordinate system for the second case.

Because none of the forces changed, we can still use our result for \(x\) by changing \(\ell_2\) with \(\ell_2’\), explicitly, from equation \eqref{equis}

\begin{equation}
x=\frac{(\ell_1+\ell_2 {}^{\prime}-L/2)M}{m}.
\end{equation}
Using the result above into equation \eqref{condition}, we obtain

\begin{equation}
\ell_1+\ell_2 {}^{\prime}+\frac{(\ell_1+\ell_2 {}^{\prime}-L/2)M}{m}=L.
\end{equation}

The expression above can be used to solve for \(\ell_2’\). Let’s start by factorizing the terms with \(\ell_2’\)

\begin{equation}
\ell_1+\ell_2 {}^{\prime}+\frac{M}{m}\ell_1+\frac{M}{m}\ell_2 {}^{\prime}-\frac{M}{2m}L=L,
\end{equation}

which is equivalent to

\begin{equation}
\ell_1\left(1+\frac{M}{m}\right)+\ell_2 {}^{\prime}\left(1+\frac{M}{m}\right)=L\left(1+\frac{M}{2m}\right).
\end{equation}

Dividing all terms in the equation above by \(\left(1+\frac{M}{m}\right)\), we get

\begin{equation}
\ell_1+\ell_2 {}^{\prime}=L\frac{\left(1+\frac{M}{2m}\right)}{\left(1+\frac{M}{m}\right)},
\end{equation}

and solving for \(\ell_2’\), we obtain

\begin{equation}
\ell_2 {}^{\prime}=L\frac{\left(1+\frac{M}{2m}\right)}{\left(1+\frac{M}{m}\right)}-\ell_1.
\end{equation}

Using the numerical values, we get

\begin{equation}
\ell_2 {}^{\prime}=(18\,\text{m})\frac{\left(1+\frac{45\,\text{kg}}{2(75\,\text{kg})}\right)}{\left(1+\frac{45\,\text{kg}}{75\,\text{kg}}\right)}-4\,\text{m},
\end{equation}

\begin{equation}
\ell_2 {}^{\prime}\approx 10.62\,\text{m}.
\end{equation}

The position of the pivot B will change by 3.62 m.

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