## Section6.18Forces Bootcamp

### Subsection6.18.13Miscellaneous

A $30\text{-kg}$ box on a smooth floor is attached to a spring of spring constant $100\text{ N/m}$ and pulled horizontally by an increasing force $F$ until the point when the box starts to slide. At that instant, $F = 75\text{ N}\text{.}$ The coefficient of static friction between the bottom of the box and the floor surface is $0.2\text{.}$

At the instant when box is about to slide, by how much is the spring stretched?

Hint

Balance forces acting on the block.

$0.161\text{ m}.$

Solution

We start by identifying forces on the block as in Figure 6.18.51. Note that static friction has its maximum value at the instant of interest.

Since acceleration of the block is zero, forces on the block are balanced. Since every force is along one or the other axis, it is very easy to write equations of motion along the two axes directly. Lets also just use $F_{s}^{\text{max}} = \mu_s F_N \text{.}$ Also, lets use $F_{sp} = k \Delta l\text{.}$ We need to find $\Delta l\text{.}$

\begin{align*} \amp F - \mu_s F_N - k\Delta l = 0, \\ \amp F_N - mg = 0, \end{align*}

Solve the second equation for $F_N$ and use that in the first equation, which we solve for $\Delta l\text{.}$

\begin{equation*} \Delta l = \dfrac{1}{k} \left( F -\mu_smg \right). \end{equation*}

Putting the numerical values in this equation we get

\begin{equation*} \Delta l = \dfrac{1}{100} \left( 75 -0.2\times 30 \times 9.81 \right) = 0.161\text{ m}. \end{equation*}

A $100\text{-kg}$ athlete pulls a rope attached to a strong pole at an angle of $15^{\circ}$ from horizontal such that the tension in the rope devlops to $800\text{ N}\text{.}$

(a) Find nomal and frictional forces on the athlete by the floor.

(b) What must be minimum value of the coefficient of static friction between the sole of the shoes of the athlete and the floor so that the athlete does not slip on the floor while he is pulling on the rope?

Hint

Draw a Free-Body Diagram of the athlete.

(a) Friction = $773\text{ N}\text{,}$ Normal = $774\text{ N}\text{.}$ (b) $1.0 \text{..}$

Solution 1 (a)

(a) We start by identifying forces on the block in Figure 6.18.54, where I have used symbols for the forces: $W$ for weight, $T$ for weight, $N_L$ for the normal on the left foot, $N_R$ for the normal on the right foot, $S_L$ for the static friction on the left foot, and $S_R$ for the static friction on the right foot.

In the translational motion, we can combine the normal forces into one and call that $N=N_L+N_R\text{,}$ and similary for the static friction, $S=S_L+S_R\text{.}$ We will have to keep them separate when we look at rotation.

Since acceleration of the block is zero, the forces are balanced. Since only $T$ is not along one of the axes, we will need to work out its components. Let $\theta$ stand for angle $15^{\circ}\text{.}$

\begin{equation*} T_x = T\,\cos\,\theta,\ \ T_y = T\,\sin\,\theta. \end{equation*}

Now, we can write the equations of motion along the two axes.

\begin{align*} \amp T\,\cos\,\theta - S = 0, \\ \amp T\,\sin\,\theta + N - W = 0, \end{align*}

Here, the values of $T \text{,}$ $\theta\text{,}$ and $W=mg\text{,}$ are known. Therefore, we solve the first equation to get $S \text{,}$ and the second one for $N \text{.}$

\begin{align*} \amp S = T\,\cos\,\theta = 800 \times \cos\,15^{\circ} = 773\text{ N}, \\ \amp N = W - T\,\sin\,\theta = 100\times 9.81 - 800 \times \sin\,15^{\circ} = 774\text{ N}. \end{align*}
Solution 2 (b)

(b)The definition of $\mu_s$ requires the maximum static friction.

\begin{equation*} \mu_s = \dfrac{S^{\text{max}}}{N}. \end{equation*}

From the problem description, we need the minimum $\mu_s \text{.}$ That means, if we assume the $S$ we found was the $S^{\text{max}}\text{,}$ then the $\mu_s$ we will get will be the minimum required for the static condition to hold.

\begin{equation*} \mu_s^{\text{min}} = \dfrac{S^{\text{found}}}{N} = \dfrac{773}{774} = 1.0. \end{equation*}

Figure 6.18.56 shows a two pulley two block system. While pulley $\text{P}_1$ only rotates with its center remaining fixed, $\text{P}_2$ rotates as well as moves. Assume both pulleys to be ideal (massless and frictionless) so that tension on the two sides of each pulley have the same magnitude. Do not assume tensions in the two strings are equal but you can assume that the length of strings do not change during the motion. Find the tensions in the two strings and accelerations of the two blocks.

Hint

Use an upward pointed $y$ axis and write lengths of the two strings in terms of the $y$ coordinates. From there deduce relation among accelerations of the two blocks.

Partial: $a_{1y} = \left( \frac{2m_2 - m_1}{m_1 + 4 m_2} \right)\; g$ if positive $y$ axis pointed up.

Solution

Let us use upward pointed $y$ axis as in Figure 6.18.57 to analyze the motion. Let $y_1\text{,}$ $y_p\text{,}$ $y_2$ denote the positions of the centers of block 1, pulley $P_2\text{,}$ and block 2 respectively. We will need $y_p$ to figure out the relation between changes in $y_1$ and $y_2\text{.}$

Let us look at each string to find relations among changes in these $y$'s. To facilitate this, let $y_\text{top}$ be the $y$ of the fixed pulley. Let us also denote lengths of the strings by $l_1$ and $l_2$ respectively. Then, we have

\begin{align*} \amp \left( y_\text{top} - y_1 \right) + ( y_\text{top} - y_p ) = l_1 = \text{constant}\\ \amp \left( y_p \right) + ( y_p - y_2 ) = l_2 = \text{constant} \end{align*}

Therefore, we have the following relations among the changes in the coordinates.

\begin{align*} \amp \Delta y_1 + \Delta y_p = 0\\ \amp 2 \Delta y_p - y_2 = 0 \end{align*}

From this, we get

$$v_{2y} = -2v_{1y},\ \ a_{2y} = -2a_{1y}.\label{eq-one-pulley-two-blocks-on-one-pulley-constraint}\tag{6.18.1}$$

This relation between the accelerations and $y$ equations of motion of the two blocks and the moving pulley (zero mass) will be sufficient to answer the questions about accelerations of the two blocks and the tensions. Using the force diagrams of each block we have the following equations.

\begin{align*} \amp T_1 - m_1 g = m_1 a_{1y}\\ \amp T_2 - m_2 g = m_2 a_{2y}\\ \amp T_1 - 2T_2 = 0 \end{align*}

Using Eq. (6.18.1), we find

\begin{align*} \amp a_{1y} = \left( \frac{2m_2 - m_1}{m_1 + 4 m_2} \right)\; g,\ \ \ a_{2y} = -2a_{1y}. \\ \amp T_1 = \left( \frac{6m_1m_2}{m_1 + 4 m_2} \right)\; g,\ \ \ T_2 = \frac{1}{2}T_1. \end{align*}

Figure 6.18.59 shows a two pulley three block system. While pulley $\text{P}_1$ only rotates with its center remaining fixed, $\text{P}_2$ rotates as well as moves. Assume both pulleys to be ideal (massless and frictionless) so that tension on the two sides of each pulley have the same magnitude. Do not assume tensions in the two strings are equal but you can assume that the length of strings do not change during the motion. Find the tensions in the two strings and accelerations of the three blocks.

Answer Hint: when $m_1=m_2=m_3\text{,}$ $|a_1| = \frac{1}{3}g\text{.}$ Try this case first before you solve the general problem.
Solution

No solution provided.