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

Given information:

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

The angle is calculated as,

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

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

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

The formula to calculate normal reaction is,

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

• $N$ is the normal reaction.
• $m$ is the mass of block.
• $g$ 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 is,

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

• $\mu$ is the coefficient of kinetic friction.
• $a$ is the acceleration.

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 $a$.

$\begin{array}{c}\left(\text{10 N}\right)\cos \left(14.03°\right)-\left(0.400\right)\left(19.13\text{ N}\right)=\left(\text{2}\text{.20 kg}\right)a\\ 2.04\text{ N}=\left(\text{2}\text{.20 kg}\right)a\\ a=\frac{2.04}{\left(\text{2}\text{.20 kg}\right)}\\ =0.93\text{m}/{\text{s}}^2\end{array}$

Conclusion:

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

To determine

To describe: 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 is decreased and increased simultaneously as the value of $\theta$ varies.

Explanation of Solution

Given information:

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 $0.400$.

Substitute $mg-T\sin \theta$ for $N$ in above equation (I).

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

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.

$\begin{array}{c}a=\frac{\left(\text{10 N}\right)}{\left(\text{2}\text{.20 kg}\right)}\left(\cos \theta +\left(0.400\right)\sin \theta \right)-\left(0.400\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =\left[\left(4.54\right)\left(\cos \theta +\left(0.400\right)\sin \theta \right)-3.92\right]\text{m}/{\text{s}}^2\end{array}$         (III)

As the value of $\theta$ varies, the acceleration is decreased, then increase and increased then decreased. So there is no change in the acceleration.

Conclusion:

Therefore, the acceleration is decreased and increased simultaneously as the value of $\theta$ varies.

To determine

The maximum value of the acceleration and position $x$ for which it is occur.

Answer to Problem 51CP

The maximum value of the acceleration is $0.76\text{m}/{\text{s}}^2$ occurs at position $1.14\text{ m}$.

Explanation of Solution

Given information:

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 $0.400$.

Differentiate the equation (III) with respect to $\theta$ to obtain maximum value.

$\begin{array}{c}\frac{da}{d\theta }=0\\ \left[\left(4.54\right)\left(\cos \theta +\left(0.400\right)\sin \theta \right)-3.92\right]\text{m}/{\text{s}}^2=0\\ \left(4.54\right)\sin \theta =\left(0.400\right)\cos \theta \\ \tan \theta =0.088\end{array}$

Further solve for $\theta$.

$\begin{array}{c}\tan \theta =0.088\\ \theta ={\tan }^{-1}\left(0.088\right)\\ =5.02°\end{array}$

Substitute $5.02°$ for $\theta$ to in equation (III) to find maximum value of acceleration.

$\begin{array}{c}a=\left[\left(4.54\right)\left(\cos \left(5.02°\right)+\left(0.400\right)\sin \left(5.02°\right)\right)-3.92\right]\text{m}/{\text{s}}^2\\ =0.76\text{m}/{\text{s}}^2\end{array}$

The angle is calculated as,

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

Substitute $5.02°$ for $\theta$, $0.400\text{ m}$ for $x$ and $0.100\text{ m}$ for $H$ to in above equation to find maximum value of $x$.

$\begin{array}{c}\tan \left(5.02°\right)=\left(\frac{0.100\text{ m}}{x}\right)\\ 0.087=\left(\frac{0.100\text{ m}}{x}\right)\\ x=\frac{\left(0.100\text{ m}\right)}{\left(0.087\right)}\\ =1.14\text{ m}\end{array}$

Conclusion:

Therefore, the maximum value of the acceleration is $0.76\text{m}/{\text{s}}^2$ occurs at position $1.14\text{ m}$.

To determine

The value of $x$ for which acceleration is zero.

Answer to Problem 51CP

The position $x$ for which acceleration is zero is $1.6\times {10}^{-3}\text{ m}$.

Explanation of Solution

Given information:

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 $0.400$.

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

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

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

$\begin{array}{c}\left(\text{10 N}\right)\left(\cos \theta +\left(0.400\right)\sin \theta \right)-\left(0.400\right)\left(9.8\text{m}/{\text{s}}^2\right)=0\\ \left(\text{10 N}\right)\cos \theta +4\sin \theta -3.92\text{m}/{\text{s}}^2=0\end{array}$

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 $x$.

$\begin{array}{c}\left(\text{10}\right)\left(\frac{x}{\sqrt{x^2+{\left(0.100\text{ m}\right)}^2}}\right)+4\left(\frac{0.100\text{ m}}{\sqrt{x^2+{\left(0.100\text{ m}\right)}^2}}\right)=3.92\\ \left(\text{10}\right)x+\left(0.4\right)=3.92\sqrt{x^2+{\left(0.100\text{ m}\right)}^2}\\ {\left(\text{10}x+0.4\right)}^2=15.36\left(x^2+0.01\right)\\ 100x^2+0.16+8x=15.36x^2+0.1536\end{array}$

Simplify the above expression.

$\begin{array}{c}100x^2+0.16+8x=15.36x^2+0.1536\\ 84.64x^2+8x-0.0136=0\\ x=1.6\times {10}^{-3}\text{ m}\end{array}$

Conclusion:

Therefore, the position $x$ for which acceleration is zero is $1.6\times {10}^{-3}\text{ m}$.