Chapter 5, Problem 51CP

A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.51. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the acceleration of the block when x = 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x = 0. (c) Find the maximum value of the acceleration and the position x for which it occurs. (d) Find the value of x for which the acceleration is zero.

Figure P5.51

To determine

The acceleration of the block when $x=0.400\text{ m}$.

Answer to Problem 51CP

The acceleration of the block when $x=0.400\text{ m}$ is $0.93\text{m}/{\text{s}}^2$.

Explanation of Solution

The mass of block is $\text{2}\text{.20 kg}$, the tension $T$ is $\text{10 N}$, the height of pulley above the top of block is $\text{0}\text{.100 m}$, and the coefficient of kinetic friction is $0.400$.

Consider the free body diagram give below.

Figure I

Write the expression for the angle the block makes with the pulley

    $\theta ={\tan }^{-1}\left(\frac{H}{x}\right)$                                                              (I)

Here, $\theta$ is the angle that the pulley makes with the block, $H$ is the vertical distance between the pulley and the block and $x$ is the horizontal distance between the block and the pulley.

Substitute $0.400\text{ m}$ for $x$ and $0.100\text{ m}$ for $H$ to find $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0.100\text{ m}}{0.400\text{ m}}\right)\\ =14.03°\end{array}$

Write the formula to calculate normal reaction

    $\begin{array}{c}N+T\sin \theta =mg\\ N=mg-T\sin \theta \end{array}$

Here, $N$ is the normal reaction, $T$ is the tension, $m$ is the mass of block and $g$ is the acceleration due to gravity.

Substitute $\text{2}\text{.20 kg}$ for $m$, $\text{10 N}$ for $T$, $14.03°$ for $\theta$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $N$.

    $\begin{array}{c}N=\left(\text{2}\text{.20 kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)-\left(\text{10 N}\right)\sin \left(14.03°\right)\\ =19.13\text{ N}\end{array}$

The equation for the force in x direction

    $T\cos \theta -\mu N=ma$                                                               (II)

Here, $\mu$ is the coefficient of kinetic friction and $a$ is the acceleration.

Conclusion:

Substitute $\text{2}\text{.20 kg}$ for $m$, $\text{10 N}$ for $T$, $14.03°$ for $\theta$, $0.400$ for $\mu$ and $19.13\text{ N}$ for $N$ in above equation to find $a$.

    $\begin{array}{c}\left(\text{10 N}\right)\cos \left(14.03°\right)-\left(0.400\right)\left(19.13\text{ N}\right)=\left(\text{2}\text{.20 kg}\right)a\\ 2.05\text{ N}=\left(\text{2}\text{.20 kg}\right)a\\ a=\frac{2.05}{\left(\text{2}\text{.20 kg}\right)}\\ =0.931\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of the block when $x=0.400\text{ m}$ is $0.931\text{m}/{\text{s}}^2$.

To determine

The general behavior of the acceleration as the block slides from a location where $x$ is larger to $x=0$.

Answer to Problem 51CP

The acceleration increases initially then decreases and becomes negative as the value of $x$ decreases.

Explanation of Solution

Substitute $mg-T\sin \theta$ for $N$ in above equation (II) and rearrange for $a$

    $\begin{array}{c}T\cos \theta -\mu \left(mg-T\sin \theta \right)=ma\\ a=\frac{T\cos \theta -\mu \left(mg-T\sin \theta \right)}{m}\\ =\frac{T}{m}\left(\cos \theta +\mu \sin \theta \right)-\mu g\end{array}$               (III)

Substitute $\text{2}\text{.20 kg}$ for $m$, $\text{10 N}$ for $T$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $0.400$ for $\mu$ in above equation.

    $\begin{array}{c}a=\frac{\left(\text{10 N}\right)}{\left(\text{2}\text{.20 kg}\right)}\left(\cos \theta +\left(0.400\right)\sin \theta \right)-\left(0.400\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =\left[\left(4.55\right)\left(\cos \theta +\left(0.400\right)\sin \theta \right)-3.92\right]\text{m}/{\text{s}}^2\end{array}$         (IV)

Substitute $0°$ for $\theta$ in the above equation

   $\begin{array}{c}a=\left[\left(4.545\right)\left(\cos 0°+\left(0.400\right)\sin 0°\right)-3.92\right]\text{m}/{\text{s}}^2\\ =0.625\text{m}/{\text{s}}^2\end{array}$

Substitute $90°$ for $\theta$ in the above equation

   $\begin{array}{c}a=\left[\left(4.545\right)\left(\cos 90°+\left(0.400\right)\sin 90°\right)-3.92\right]\text{m}/{\text{s}}^2\\ =-2.10\text{m}/{\text{s}}^2\end{array}$

When $x$ is very large, $\theta$ approaches zero and as $x$ decreases, $\theta$ increases. The value of acceleration varies as $\left(\cos \theta +\left(0.400\right)\sin \theta \right)-3.92$. Thus, as $x$ decreases, the acceleration increases initially, then drops quickly and becomes negative.

Conclusion:

Therefore, the acceleration increases initially then decreases and becomes negative as the value of $x$ decreases.

To determine

The maximum value of the acceleration and its corresponding position.

Answer to Problem 51CP

The maximum value of the acceleration is $0.975\text{m}/{\text{s}}^2$ which occurs at the position $0.250\text{ m}$.

Explanation of Solution

Differentiate the equation (IV) with respect to $\theta$ in order to find the maximum value.

    $\begin{array}{c}\frac{da}{d\theta }=0\\ \left[\left(4.545\right)\left(-\sin \theta +\left(0.400\right)\cos \theta \right)-0\right]\text{m}/{\text{s}}^2=0\\ \sin \theta =\left(0.400\right)\cos \theta \\ \tan \theta =0.400\end{array}$

Further solve for $\theta$.

    $\begin{array}{c}\tan \theta =0.400\\ \theta ={\tan }^{-1}\left(0.400\right)\\ =21.8°\end{array}$

Substitute $21.8°$ for $\theta$ to in equation (IV) to find maximum value of acceleration.

    $\begin{array}{c}a=\left[\left(4.545\right)\left(\cos \left(21.8°\right)+\left(0.400\right)\sin \left(21.8°\right)\right)-3.92\right]\text{m}/{\text{s}}^2\\ =0.975\text{m}/{\text{s}}^2\end{array}$

Rearrange (I) for $x$

    $x=\left(\frac{H}{\tan \theta }\right)$

Substitute $0.400$ for $\tan \theta$ and $0.100\text{ m}$ for $H$ in above equation to find value of $x$ which has maximum acceleration

    $\begin{array}{c}x=\left(\frac{0.100\text{ m}}{0.400}\right)\\ =0.250\text{ m}\end{array}$

Conclusion:

Therefore, the maximum value of the acceleration is $0.975\text{m}/{\text{s}}^2$ which occurs at position $0.250\text{ m}$.

To determine

The position where acceleration is zero.

Answer to Problem 51CP

The acceleration is zero at the position $0.0610\text{ m}$.

Explanation of Solution

Substitute $0$ for $a$ in equation (III).

    $\begin{array}{c}0=\frac{T}{m}\left(\cos \theta +\mu \sin \theta \right)-\mu g\\ \frac{T}{m}\left(\cos \theta +\mu \sin \theta \right)-\mu g=0\end{array}$

Substitute $\text{10 N}$ for $T$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $0.400$ for $\mu$ in above equation.

    $\begin{array}{c}\frac{\left(\text{10 N}\right)}{\left(2.20\text{kg}\right)}\left(\cos \theta +\left(0.400\right)\sin \theta \right)-\left(0.400\right)\left(9.8\text{m}/{\text{s}}^2\right)=0\\ \left(\text{4}{\text{.545 Nkg}}^{-1}\right)\cos \theta +1.82\sin \theta -3.92\text{m}/{\text{s}}^2=0\end{array}$

Conclusion:

Substitute $\frac{x}{\sqrt{x^2+{\left(0.100\text{ m}\right)}^2}}$ for $\cos \theta$ and $\frac{0.100\text{ m}}{\sqrt{x^2+{\left(0.100\text{ m}\right)}^2}}$ for $\sin \theta$ in the above equation to calculate $x$.

    $\begin{array}{c}\left(\text{4}\text{.545}\right)\left(\frac{x}{\sqrt{x^2+{\left(0.100\right)}^2}}\right)+1.82\left(\frac{0.100\text{ m}}{\sqrt{x^2+{\left(0.100\right)}^2}}\right)=3.92\\ \left(\text{4}\text{.545}\right)x+\left(0.182\right)=3.92\sqrt{x^2+{\left(0.100\right)}^2}\\ 20.7x^2+0.0331+0.165x=15.4\left(x^2+0.010\right)\\ 5.29x^2+1.65x-0.121=0\end{array}$

Solving the above expression for $x$ and consider only the positive root

    $x=0.0610\text{ m}$

Therefore, the position $x$ for which acceleration is zero is $0.0610\text{ m}$.