Painter on a Ladder!

a) Make a free body diagram, and sum the forces.

b) With the result in part (a), it is easy to directly solve for the coefficient of friction.

c) Same hint as part (a), and calculate again the force due to friction and the subsequent coefficient of friction.

a) Using a free body diagram, Newton’s Second Law in the \({y-}\)direction can be written as:

\begin{equation*}
N_F – Mg -mg =0,
\end{equation*}

where we can solve for \(N_F\) to get:

\begin{equation*}
N_F = 1000.62 \, \text{N}.
\end{equation*}

Newton’s Second Law in the \({x-}\)direction states:

\begin{equation*}
f_r-N_W = 0.
\end{equation*}

Summing the torques, we have:

\begin{equation*}
-\frac{L}{4}\cos(\theta)mg-\frac{L}{2}\cos(\theta)Mg+L\sin(\theta)N_W=0,
\end{equation*}

where solving for \(N_W\), we get:

\begin{equation*}
N_W=\frac{\cos(\theta)}{2\sin(\theta)}\left(\frac{m}{2}+M\right)g.
\end{equation*}

Since \(f_r = N_W\) (based on Newton’s Second Law in the \({x-}\)direction), we can plug in numerical values to obtain:

\begin{equation*}
f_r \approx 526.7 \, \text{N}.
\end{equation*}

b) Since \(f_r = \mu_s^{\text{min}} N_F\), we can solve for \(\mu\) and plug in numerical values to get:

\begin{equation*}
\mu_s^{\text{min}}\approx 0.53.
\end{equation*}

c) The torques’ sum is:

\begin{equation*}
-\frac{2L}{3}\cos(\theta)mg-\frac{L}{2}\cos(\theta)Mg+L\sin(\theta)N_W=0,
\end{equation*}

Solving for \(N_W\), we get:

\begin{equation*}
N_W=\frac{\cos(\theta)}{2\sin(\theta)}\left(\frac{2m}{3}+\frac{M}{2}\right)g.
\end{equation*}

Since \(f_r = N_W\) (based on Newton’s Second Law in the \({x-}\)direction), we can plug in numerical values to get:

\begin{equation*}
f_r \approx 546.6 \, \text{N},
\end{equation*}

and for the coefficient of friction:

\begin{equation*}
\mu_s^{\text{min}}\approx 0.55.
\end{equation*}

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

a) The first part of the problem is asking us to find the magnitude of the forces that are interacting at the base of the ladder. We must first identify all the forces exerted on the ladder (see figure 1).

In this figure, we identify the weight of the ladder \(\hat{W}=-Mg\,\hat{\textbf{j}}\), located at its center of mass and the weight of the painter \(\hat{N}=-mg\,\hat{\textbf{j}}\) at a quarter of the total length of the ladder \(L\). Here, the \(M\) is the mass of the ladder and \(m\) the mass of the painter, with \(g\) the gravitational acceleration on Earth, both are directed in the negative Y axis.

We also identify three contact forces: the contact force with the wall \(-N_W\,\hat{\textbf{i}}\), the contact force with the floor \(N_F\,\hat{\textbf{j}}\), and the friction force on the floor \(f_r\,\hat{\textbf{i}}\). The directions of each of these forces are given by the coordinate system in the previous figure.

Because we want an expression for the forces at the base of the ladder, we can use Newton’s second law in the equilibrium case, that is null acceleration \(\vec{a}=0\). Writing Newton’s second law explicitly,

\begin{equation}
\sum \vec{F}=m \vec{a},
\end{equation}

and for the static case

\begin{equation}
\sum \vec{F}=\vec{0},
\end{equation}

where \(m\) is the mass of the system and \(\sum\vec{F}\) is the sum of all forces. Using the expressions for all the forces we have identified in Newton’s second law for the static case, we obtain

\begin{equation}
-Mg\,\hat{\textbf{j}}-mg\,\hat{\textbf{j}}-N_W\,\hat{\textbf{i}}+N_F\,\hat{\textbf{j}}+f_r\,\hat{\textbf{i}}=\vec{0}.
\end{equation}

In the equation above, we have terms in the X and Y direction, so we can group the terms in the same direction to obtain

\begin{equation}
\left(-N_W+f_r\right)\,\hat{\textbf{i}}+\left(-Mg-mg+N_F\right)\,\hat{\textbf{j}}=\vec{0}.
\end{equation}

Because on the other side we have a \(\vec{0}\), the equation will hold only if the two terms in parenthesis above are zero, thus obtaining a pair of equations; namely

\begin{equation}
\label{fx}
-N_W+f_r=0,
\end{equation}

\begin{equation}
\label{fy}
-Mg-mg+N_F=0.
\end{equation}

From equation \eqref{fy}, we can solve for \(N_F\) immediately to obtain

\begin{equation}
N_F=mg+Mg,
\end{equation}

which numerically is

\begin{equation}
N_F=(80\,\text{kg})(9.81\,\text{m/s}^2)+(22\,\text{kg})(9.81\,\text{m/s}^2),
\end{equation}

\begin{equation}
\label{Nf}
N_F=1000.62\,\text{N}.
\end{equation}

Then, we must use an additional equation to solve for the other unknowns \(N_W\) and \(f_r\). We will then demand that the sum of torques \(\tau\) is also zero for the static condition to be complete, namely

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

The torque that a force exerts on the ladder can be calculated using the definition

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

where \(\vec{r}\) is the distance vector from the point where the torque is being calculated to the point where the force \(\vec{F}\) is exerted.

We choose a convenient point to calculate the sum of torques: the point where the ladder touches the floor. It’s convenient because the contact force with the floor and the friction with the floor are applied at this point, meaning that \(\vec{r}=0\) and according to \eqref{torque} the torque they exert is null. We are only concerned with calculating the torques exerted by the weight of the ladder, the weight of the painter, and the contact force with the wall.

Let’s start by calculating the torque created by the weight of the painter. For this purpose, we’ll need the vector \(\vec{r}\), which in this case has a magnitude of \(L/4\), and the direction is given by the ladder’s inclination. We can then write

\begin{equation}
\label{rm}
\vec{r}_m=\frac{L}{4}\cos(\theta)\,\hat{\textbf{i}}+\frac{L}{4}\sin(\theta)\,\hat{\textbf{j}},
\end{equation}

where we denote by \( \vec{r}_m\) the distance vector associated to the weight of the painter of mass \(m\). Thus, using equations \eqref{torque}, \eqref{rm} and the expression for the weight, we can find the torque exerted by the weight of the painter \(\vec{\tau}_m\) as

\begin{equation}
\vec{\tau}_m=\left(\frac{L}{4}\cos(\theta)\,\hat{\textbf{i}}+\frac{L}{4}\sin(\theta)\,\hat{\textbf{j}}\right)\times (-mg\,\hat{\textbf{j}}).
\end{equation}

Using the distributive law to expand the expression above, we get

\begin{equation}
\vec{\tau}_m=-\frac{L}{4}\cos(\theta)mg\,\hat{\textbf{i}}\times\hat{\textbf{j}}-\frac{L}{4}\sin(\theta)mg\,\hat{\textbf{j}}\times\hat{\textbf{j}},
\end{equation}

\begin{equation}
\label{tm1}
\vec{\tau}_m=-\frac{L}{4}\cos(\theta)mg\,\hat{\textbf{k}},
\end{equation}

where we have used in the last line the fact that \(\hat{\textbf{i}}\times\hat{\textbf{j}}=\hat{\textbf{k}}\) and \(\hat{\textbf{j}}\times\hat{\textbf{j}}=\vec{0}\).

Now, let’s calculate the torque exerted by the weight of the ladder. Similarly to the distance for the painter, the distance vector for the ladder \(\vec{r}_M\) has magnitude \(L/2\) and the direction is given by the inclination angle of the ladder, namely

\begin{equation}
\label{RM2}
\vec{r}_M=\frac{L}{2}\cos(\theta)\,\hat{\textbf{i}}+\frac{L}{2}\sin(\theta)\,\hat{\textbf{j}}.
\end{equation}

We can then calculate the torque exerted by the weight of the ladder using equations \eqref{torque}, \eqref{RM2} and the given expression for the weight; explicitly,

\begin{equation}
\vec{\tau}_M=\left(\frac{L}{2}\cos(\theta)\,\hat{\textbf{i}}+\frac{L}{2}\sin(\theta)\,\hat{\textbf{j}}\right)\times\left(-Mg\,\hat{\textbf{j}}\right).
\end{equation}

Using the distributive law to expand the expression above, we get

\begin{equation}
\vec{\tau}_M=-\frac{L}{2}\cos(\theta)Mg\,\hat{\textbf{i}}\times\hat{\textbf{j}}-\frac{L}{2}\sin(\theta)Mg\,\hat{\textbf{j}}\times\hat{\textbf{j}},
\end{equation}

\begin{equation}
\label{tM2}
\vec{\tau}_M=-\frac{L}{2}\cos(\theta)Mg\,\hat{\textbf{k}},
\end{equation}

where we have used in the last line the fact that \(\hat{\textbf{i}}\times\hat{\textbf{j}}=\hat{\textbf{k}}\) and \(\hat{\textbf{j}}\times\hat{\textbf{j}}=\vec{0}\).

Finally, we’ll calculate the torque due to the contact force with the wall. In this case, the distance vector, which we’ll denote by \(\vec{r}_W\), has length \(L\) and the direction is that of the ladder, given by angle \(\theta\). Therefore, we can write

\begin{equation}
\label{rw}
\vec{r}_W=L\cos(\theta)\,\hat{\textbf{i}}+L\sin(\theta)\,\hat{\textbf{j}}.
\end{equation}

Using the definition of the torque given in equation \eqref{torque}, and together with the expressions of equation \eqref{rw} and the contact force, we can write and expression for the torque \(\vec{\tau}_W\) exerted by this contact force, that is

\begin{equation}
\vec{\tau}_W=\left(L\cos(\theta)\,\hat{\textbf{i}}+L\sin(\theta)\,\hat{\textbf{j}}\right)\times\left(-N_W\,\hat{\textbf{i}}\right).
\end{equation}

Using the distributive law to expand the expression above, we get

\begin{equation}
\vec{\tau}_W=-L\cos(\theta)N_W\,\hat{\textbf{i}}\times\hat{\textbf{i}}-L\sin(\theta)N_W\,\hat{\textbf{j}}\times\hat{\textbf{i}},
\end{equation}

\begin{equation}
\label{tw}
\vec{\tau}_W=L\sin(\theta)N_W\,\hat{\textbf{k}},
\end{equation}

where we have used in the last line the fact that \(\hat{\textbf{j}}\times\hat{\textbf{i}}=-\hat{\textbf{k}}\) and \(\hat{\textbf{i}}\times\hat{\textbf{i}}=\vec{0}\).

Using the expressions for the torques given by equations \eqref{tm1}, \eqref{tM2} and \eqref{tw} into equation \eqref{sumt}, we obtain

\begin{equation}
\vec{\tau}_m+\vec{\tau}_M+\vec{\tau}_W=\vec{0},
\end{equation}

\begin{equation}
-\frac{L}{4}\cos(\theta)mg\,\hat{\textbf{k}}-\frac{L}{2}\cos(\theta)Mg\,\hat{\textbf{k}}+L\sin(\theta)N_W\,\hat{\textbf{k}}=\vec{0}.
\end{equation}

Because all the terms are directed in the same axis, we can drop the vector notation in the result above to obtain

\begin{equation}
-\frac{L}{4}\cos(\theta)mg-\frac{L}{2}\cos(\theta)Mg+L\sin(\theta)N_W=0,
\end{equation}

where we can solve for \(N_W\) as follows

\begin{equation}
L\sin(\theta)N_W= \frac{L}{4}\cos(\theta)mg+\frac{L}{2}\cos(\theta)Mg,
\end{equation}

and factorizing

\begin{equation}
L\sin(\theta)N_W=\frac{L}{2}\cos(\theta)\left(\frac{m}{2}+M\right)g.
\end{equation}

Cancelling out the \(L\) and dividing both sides by \(\sin(\theta)\) in the equation above, we obtain

\begin{equation}
\label{nwresult}
N_W=\frac{\cos(\theta)}{2\sin(\theta)}\left(\frac{m}{2}+M\right)g.
\end{equation}

From equation \eqref{fx}, we deduce that

\begin{equation}
f_r=N_W,
\end{equation}

which implies, from the result given in equation \eqref{nwresult} that

\begin{equation}
f_r=\frac{\cos(\theta)}{2\sin(\theta)}\left(\frac{m}{2}+M\right)g.
\end{equation}

Using the numerical values, we get

\begin{equation}
\label{friction}
f_r=\frac{\cos(30^{\circ})}{2\sin(30^{\circ})}\left(\frac{80\,\text{kg}}{2}+22\,\text{kg}\right)(9.81\,\text{m/s}^2),
\end{equation}

\begin{equation}
f_r\approx 526.7\,\text{N}.
\end{equation}

b) We need to find the minimum static coefficient of friction for the ladder to not slip. Let’s start by assuming that the static friction force takes its maximum value, which is

\begin{equation}
f_r{}_\text{max}=\mu_s N_F,
\end{equation}

where \(\mu_s\) is the static friction coefficient and \(N_F\) the contact force at the same point where the friction is exerted, that is the contact force with the floor. Assuming that the friction force found in equation \eqref{friction} is the maximum friction force, we can then write

\begin{equation}
f_r=\mu_s^{\text{min}}N_F,
\end{equation}

where \(\mu_s^{\text{min}}\) is the minimum static friction coefficient between the ladder and the floor such that the system remains in equilibrium. Solving for \(\mu_s^{\text{min}}\) in the equation above, we get

\begin{equation}
\label{mumin}
\mu_s^{\text{min}}=\frac{f_r}{N_F},
\end{equation}

which using the numerical values given in equations \eqref{Nf} and \eqref{friction} is

\begin{equation}
\mu_s^{\text{min}}\approx \frac{526.7\,\text{N}}{1000.62\,\text{N}}\approx 0.53.
\end{equation}

c) For the last part of the problem, we need to calculate the static coefficient of friction if the painter goes up 2/3rds of the length of the ladder. When the painter moves to \(2L/3\), its weight does not change, which implies that \(N_F\) does not change; however, the vector \(\vec{r}_m\) changes, and as a consequence, the sum of torques must be changed. The new distance vector is given by

\begin{equation}
\label{rm2}
\vec{r}_m=\frac{2L}{3}\cos(\theta)\,\hat{\textbf{i}}+\frac{2L}{3}\sin(\theta)\,\hat{\textbf{j}},
\end{equation}

Thus, using equations \eqref{torque}, \eqref{rm2} and the expression for the weight, we can find the new value for the torque exerted by the weight of the painter \(\vec{\tau}_m\) as

\begin{equation}
\vec{\tau}_m=\left(\frac{2L}{3}\cos(\theta)\,\hat{\textbf{i}}+\frac{2L}{3}\sin(\theta)\,\hat{\textbf{j}}\right)\times (-mg\,\hat{\textbf{j}}).
\end{equation}

Using the distributive law to expand the expression above, we get

\begin{equation}
\vec{\tau}_m=-\frac{2L}{3}\cos(\theta)mg\,\hat{\textbf{i}}\times\hat{\textbf{j}}-\frac{2L}{3}\sin(\theta)mg\,\hat{\textbf{j}}\times\hat{\textbf{j}},
\end{equation}

\begin{equation}
\label{tm11}
\vec{\tau}_m=-\frac{2L}{3}\cos(\theta)mg\,\hat{\textbf{k}},
\end{equation}

where we have used in the last line the fact that \(\hat{\textbf{i}}\times\hat{\textbf{j}}=\hat{\textbf{k}}\) and \(\hat{\textbf{j}}\times\hat{\textbf{j}}=\vec{0}\).

We can now perform the sum of torques again using the expressions for the torques given by equations \eqref{tm11}, \eqref{tM2} and \eqref{tw} into equation \eqref{sumt}; explicitly,

\begin{equation}
\vec{\tau}_m+\vec{\tau}_M+\vec{\tau}_W=\vec{0},
\end{equation}

\begin{equation}
-\frac{2L}{3}\cos(\theta)mg\,\hat{\textbf{k}}-\frac{L}{2}\cos(\theta)Mg\,\hat{\textbf{k}}+L\sin(\theta)N_W\,\hat{\textbf{k}}=\vec{0}.
\end{equation}

Because all the terms are directed in the same axis, we can drop the vector notation in the result above to obtain

\begin{equation}
-\frac{2L}{3}\cos(\theta)mg-\frac{L}{2}\cos(\theta)Mg+L\sin(\theta)N_W=0,
\end{equation}

where we can solve for \(N_W\) as follows

\begin{equation}
L\sin(\theta)N_W= \frac{2L}{3}\cos(\theta)mg+\frac{L}{2}\cos(\theta)Mg,
\end{equation}

and factorizing

\begin{equation}
L\sin(\theta)N_W=L\cos(\theta)\left(\frac{2m}{3}+\frac{M}{2}\right)g.
\end{equation}

Cancelling out the \(L\) and dividing both sides by \(\sin(\theta)\) in the equation above, we obtain

\begin{equation}
\label{nwresult2}
N_W=\frac{\cos(\theta)}{2\sin(\theta)}\left(\frac{2m}{3}+\frac{M}{2}\right)g.
\end{equation}

From equation \eqref{fx}, we deduce that

\begin{equation}
f_r=N_W,
\end{equation}

which implies, from the result given in equation \eqref{nwresult2} that

\begin{equation}
f_r=\frac{\cos(\theta)}{2\sin(\theta)}\left(\frac{2m}{3}+\frac{M}{2}\right)g.
\end{equation}

Using the numerical values, we get

\begin{equation}
f_r=\frac{\cos(30^{\circ})}{2\sin(30^{\circ})}\left(\frac{2(80\,\text{kg})}{3}+\frac{22\,\text{kg}}{2}\right)(9.81\,\text{m/s}^2),
\end{equation}

\begin{equation}
f_r\approx546.6\,\text{N}.
\end{equation}

Using once more the expression to calculate \(\mu_s^{\text{min}}\) given by equation \eqref{mumin}, we have, with the new value for \(f_r\) given above, that

\begin{equation}
\mu_s^{\text{min}}=\frac{546.6\,\text{N}}{1000.62\,\text{N}}\approx 0.55.
\end{equation}

Hence, in order for the ladder and painter to be static, the static coefficient of friction can take any value larger than 0.55 in this second configuration.

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