Friction and Simple Harmonic Motion

Use Newton's second law to get the acceleration of the blocks as one system. By the equations of the harmonic oscillator, find the maximum acceleration and make equal both accelerations to solve for the amplitude.

Newton's second law states:

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

where for the Y-axis we have:

\begin{equation*}
N_{21}-W_1 = 0.
\end{equation*}

For the X-axis we have:

\begin{equation*}
f_{r1} = m_1 a_{x1},
\end{equation*}

where \(f_{r1} = \mu_s N_{21} \) for the maximum static friction. Then, replacing those variables and solving for the acceleration we get:

\begin{equation*}
a_{x1\text{max}}=\mu_sg.
\end{equation*}

The equation of motion for the oscillating block is:

\begin{equation*}
x(t)=A\sin(\omega t+\delta),
\end{equation*}

whose derivative over time is:

\begin{equation*}
v(t)=-A\omega \cos(\omega t +\delta),
\end{equation*}

And the second derivative over time for the equation of motion is:

\begin{equation*}
a(t)=-A\omega^2 \sin(\omega t +\delta).
\end{equation*}

Then:

\begin{equation*}
a_{x2\text{max}}=A\omega^2.
\end{equation*}

By equalling the accelerations obtained we have:

\begin{equation*}
\mu_s g= A\omega^2,
\end{equation*}

where we can solve for \(A\) so we can write:

\begin{equation*}
A=\frac{\mu_s g}{\omega^2},
\end{equation*}

or:

\begin{equation}
A=\frac{\mu_s g}{\left(\frac{2\pi}{T}\right)^2}.
\end{equation}

Numerically this is:

\begin{equation*}
A\approx 0.0156\,\text{m}.
\end{equation*}

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

We are asked to find the maximum oscillation possible if we want the block on top to not slip. To solve this problem, we must first find the maximum acceleration of the blocks such that the upper block does not slip. Then we will relate this acceleration to the amplitude and period of the harmonic motions performed by the lower block.

Let us first make a free-body diagram of the block on top, for which we will use the sub-index 1 as seen in the following figure.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/dcl1.png}
\caption{Free-body diagram of the block on top}
\label{fig:dcl1}
\end{figure}

To make the free-body diagram we assume that at that instant in time the spring is compressed and both blocks move to the right.

In figure 1, we included the weight of the block \(\vec{W}_1\), the contact force exerted by the block on the bottom \(\vec{N}_{21}\) and the static friction \(\vec{fr}_1\). Notice that in this case, the friction goes in the same direction as the movement of the block, this force is responsible for the block on top to move along the bottom block.

For the block on the bottom, which we will refer now as block 2, we have the free body diagram shown in figure 2.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/DCL2.png}
\caption{Free-body diagram of the block on the bottom}
\label{fig:dcl2}
\end{figure}

In figure 2 we have included the weight of the block \(\vec{W}_2\), the contact force exerted by block 1 \(\vec{N}_{12}\), the friction \(\vec{fr}_2\), the contact force from the floor \(\vec{N}_f\) and the force exerted by the spring \(F_{\text{spring}}\).

Notice that in the free-body diagram shown in Figure the previous figure the friction force opposes the force exerted by the spring, this happens at all times in the harmonic motion.

Since we are required that block 1 do not move over block 2, then their position, velocity and acceleration must be the same. We can now apply Newton's second law

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

for the X and Y axis separately, namely

\begin{equation}
\label{sumx}
\sum F_x=ma_x,
\end{equation}

and

\begin{equation}
\label{sumy}
\sum F_y=ma_y.
\end{equation}

In equations \eqref{sumx} and \eqref{sumy} the left hand side is the sum of all forces in the X and Y axis respectively and \(m\) is the mass of the object. The accelerations of the object in X and in Y are denoted by \(a_x\) and \(a_y\).

We will use Newton's second law in block 1. From the free-body diagram we see that in the X axis the only force is the friction, then using \eqref{sumx}

\begin{equation}
\label{sumx1}
fr_1=m_1a_{x1},
\end{equation}

where \(m_1=3.00\,\text{kg}\) and \(a_{x1}\) is the acceleration of block 1 in the X axis.

On the Y axis we have, using equation \eqref{sumy}

\begin{equation}
\label{sumy1}
N_{21}-W_{1}=m_1a_{y1},
\end{equation}

where \(a_{y1}\) is the acceleration of the block in the Y axis. This acceleration is zero because the blocks move exclusively in the X axis, then equation \eqref{sumy1} becomes

\begin{equation}
\label{sumy2}
N_{21}-W_{1}=0,
\end{equation}

which implies

\begin{equation}
\label{n21}
N_{21}=W_1.
\end{equation}

Examining the motion in the X axis through equation \eqref{sumx1} we see that the acceleration can be solved in terms of the friction force as

\begin{equation}
\label{ax1}
a_{x1}=\frac{fr_{1}}{m_1}.
\end{equation}

The static friction \(fr_{1}\) can take values from zero to a maximum \(fr_{1\text{max}}\)given by

\begin{equation}
\label{frmax}
fr_{1\text{max}}=\mu_s N_{21},
\end{equation}

where \(\mu_s\) is the static friction coefficient given by the problem and \(N_{21}\) is the contact force between blocks 1 and 2. From equation \eqref{ax1} we can write an expression for the maximum value of the acceleration as

\begin{equation}
a_{x1\text{max}}=\frac{fr_{1\text{max}}}{m_1},
\end{equation}

which using the result of \eqref{frmax} becomes

\begin{equation}
\label{ax12}
a_{x1\text{max}}=\frac{\mu_s N_{21}}{m_1}.
\end{equation}

Using the result of the analysis in the Y axis given by equation \eqref{n21} into equation \eqref{ax12} we get

\begin{equation}
\label{ax13}
a_{x1\text{max}}=\frac{\mu_sW_1}{m_1}.
\end{equation}

Using the definition for the weight

\begin{equation}
W_{1}=m_1g,
\end{equation}

into equation \eqref{ax13} we arrive to the expression

\begin{equation}
\label{ax1max}
a_{x1\text{max}}=\frac{\mu_s m_1g}{m_1}=\mu_sg,
\end{equation}

where \(g=9.81\,\text{m/s}^2\) is the gravitational acceleration on earth. Now we have an expression for the maximum acceleration of block 1 such that the friction is always static, that is, the block does not move relative to block 2.

Now let us write an expression for the maximum acceleration reached by block 2 in the harmonic motion. For a harmonic motion of amplitude \(A\) the position of block 2 \(x\) can be described by the equation dependent on time \(t\) as

\begin{equation}
\label{xt}
x(t)=A\sin(\omega t+\delta),
\end{equation}

where \(\omega\) is the angular frequency of the motion and \(\delta\) is the phase angle, determined by the initial conditions of the problem. Since the \(\ \sin\) function has values between -1 and 1, then the greatest displacement from the origin is precisely the amplitude \(A\).

To find the velocity we must derive with respect to time the expression in \eqref{xt} to obtain

\begin{equation}
v(t)=\frac{dx(t)}{dt},
\end{equation}

\begin{equation}
v(t)=\dfrac{d}{dt}\left(A\sin(\omega t +\delta)\right)
\end{equation}

\begin{equation}
\label{vt}
v(t)=-A\omega \cos(\omega t +\delta).
\end{equation}

To find the acceleration we must derive with respect to time the expression in \eqref{vt} to obtain

\begin{equation}
a(t)=\frac{dv(t)}{dt},
\end{equation}

\begin{equation}
a(t)=\dfrac{d}{dt}\left(-A\omega\cos(\omega t +\delta)\right)
\end{equation}

\begin{equation}
\label{at}
a(t)=-A\omega^2 \sin(\omega t +\delta).
\end{equation}

From the expression in equation \eqref{at} we can see that the maximum value of the acceleration is \(A\omega^2\) explicitly

\begin{equation}
\label{a2xmax}
a_{x2\text{max}}=A\omega^2.
\end{equation}

We seek that the blocks move as a whole, that is, the block 1 does not slip. Then, as we have said before, the accelerations of both blocks must be the same. In particular, the maximum accelerations must be the same

\begin{equation}
\label{equal}
a_{x1\text{max}}=a_{x2\text{max}}.
\end{equation}

Using the results from equations \eqref{ax1max} and \eqref{a2xmax} into equation \eqref{equal} we get

\begin{equation}
\mu_s g= A\omega^2,
\end{equation}

where we can solve for \(A\) so we can write

\begin{equation}
\label{ampli}
A=\frac{\mu_s g}{\omega^2}.
\end{equation}

The only unknown is \(\omega\) which can be expressed in terms of the period \(T\), a known quantity in the problem, as

\begin{equation}
\label{omega}
\omega=\frac{2\pi}{T}.
\end{equation}

Using the expression for \(\omega\) given in equation \eqref{omega} into equation \eqref{ampli} we finally obtain the maximum amplitude such that the block on top does not slip

\begin{equation}
A=\frac{\mu_s g}{\left(\frac{2\pi}{T}\right)^2}.
\end{equation}

which numerically is

\begin{equation}
A=\frac{0.7 (9.81\,\text{m/s}^2)}{\left(\frac{2\pi}{0.3\,\text{s}}\right)^2},
\end{equation}

\begin{equation}
A\approx 0.0156\,\text{m}=1.56\,\text{cm}.
\end{equation}