Chapter 15.1, Problem 15.23P

Three belts move over tow pulleys without slipping in the speed reduction system shown. At the instant shown, the velocity of point A on the input belt is 2 ft/s to the right, decreasing at the rate of 6 ft/s². Determine, at the instant, (a) the velocity and acceleration of point C on the output belt, (b) the acceleration of point B on the output pulley.

To determine

(a)

The velocity and acceleration of point C.

Answer to Problem 15.23P

The velocity of point C is $0.\text{5}\text{ft}/\text{s}$.

The acceleration of point C is $1.5\text{ft}/\text{s}$.

Explanation of Solution

Given Information:

The velocity of point A on the input belt is $2\text{ft}/\text{s}$, the acceleration of the point A is $6\text{ft}/{\text{s}}^{\text{2}}$.

Write the expression for the angular velocity of the left side pulley.

${\omega }_L=\frac{2v_A}{d_2}$ ...... (I)

Here, the velocity of point A is $v_A$ and the outer diameter of the left side pulley is $d_2$.

Write the expression for the angular acceleration of the left side pulley.

${\alpha }_1=\frac{2{\left(a_A\right)}_t}{d_2}$ ...... (II)

Here, the tangential acceleration of point A is ${\left(a_A\right)}_t$.

Write the expression for the velocity of belt.

$v_1=d_1{\omega }_1$ ...... (III)

Here, the inner diameter of the left side pulley is $d_1$.

Write the expression for acceleration of intermediate belt.

${\left(a_1\right)}_t=\frac{d_1{\alpha }_1}{2}$ ...... (IV)

Write the expression for the angular velocity of right side pulley.

${\omega }_2=\frac{2v_1}{d_4}$ ...... (V)

Here, the outer diameter of the right-side pulley is $d_4$.

Write the expression to calculate the angular acceleration of the right side pulley.

${\alpha }_2=\frac{2{\left(a_1\right)}_t}{d_4}$ ...... (VI)

Write the expression to calculate the velocity of point C.

$v_C=\frac{{\omega }_2{d}_3}{2}$ ...... (VII)

Here, the inner diameter of the right-side pulley is $d_3$.

Write the expression to calculate the acceleration of point C.

${\left(a_C\right)}_t=r_3{\alpha }_2$ ...... (VIII)

Calculation:

Substitute $\text{2}\text{ft}/\text{s}$ for $v_A$ in Equation (I).

$\begin{array}{c}{\omega }_1=\frac{2\left(2\text{ft}/\text{s}\right)}{8\text{in}}\\ =\frac{4\text{ft}/\text{s}}{8\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}\\ =\frac{4\text{ft}/\text{s}}{0.666\text{ft}}\\ =6\text{rad}/\text{s}\end{array}$

Substitute $6\text{ft}/{\text{s}}^2$ for ${\left(a_A\right)}_t$ and $8\text{in}$ for $d_2$ in Equation (II).

$\begin{array}{c}{\alpha }_1=\frac{2\left(6\text{ft}/{\text{s}}^2\right)}{8\text{in}}\\ =\frac{2\left(6\text{ft}/{\text{s}}^2\right)}{8\text{in}\left(\frac{1\text{m}}{39.37\text{in}}\right)}\\ =18\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Substitute $\text{4}\text{in}$ for $d_1$ and $6\text{rad}/\text{s}$ for ${\omega }_1$ in Equation (III).

$\begin{array}{c}{v}_1=\frac{4\text{in}\times \text{6}\text{rad}/\text{s}}{2}\\ =\frac{24\text{in}/\text{s}\left(\frac{1\text{ft}}{12\text{in}}\right)}{2}\\ =1\text{ft}/\text{s}\end{array}$

Substitute $18\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_1$ and $4\text{in}$ for $d_1$ in Equation (IV).

$\begin{array}{c}{\left(a_1\right)}_t=\frac{18\text{rad}/{\text{s}}^{\text{2}}\times 4\text{in}}{2}\\ =36\text{in}/{\text{s}}^{\text{2}}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =3\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Substitute $1\text{ft}/\text{s}$ for $v_1$ and $\text{8}\text{in}$ for $d_4$ in Equation (V).

$\begin{array}{c}{\omega }_2=\frac{2\left(1\text{ft}/\text{s}\right)}{8\text{in}}\\ =\frac{2\text{ft}/\text{s}}{8\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}\\ =3\text{rad}/\text{s}\end{array}$

Substitute $3\text{ft}/{\text{s}}^2$ for ${\left(a_1\right)}_t$ and $\text{8}\text{in}$ for $d_4$ in equation (V).

$\begin{array}{c}{\alpha }_2=\frac{2\left(3\text{ft}/{\text{s}}^{\text{2}}\right)}{8\text{in}}\\ =\frac{6\text{ft}/{\text{s}}^{\text{2}}}{8\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}\\ =9\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Substitute $\text{3}\text{rad}/\text{s}$ for ${\omega }_2$ and $\text{4}\text{in}$ for $d_3$ in Equation (VI).

$\begin{array}{c}{v}_C=\frac{\left(\text{3}\text{rad}/\text{s}\right)\left(4\text{in}\right)}{2}\\ =6\text{in}/\text{s}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =0.5\text{ft}/\text{s}\end{array}$

Substitute $\text{4}\text{in}$ for $d_3$ and $9\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_2$ in Equation (VII).

$\begin{array}{c}{\left(a_C\right)}_t=\frac{\left(4\text{in}\right)\left(9\text{rad}/{\text{s}}^{\text{2}}\right)}{2}\\ =18\text{in}/{\text{s}}^{\text{2}}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =1.5\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

The velocity of point C is $0.\text{5}\text{ft}/\text{s}$.

The acceleration of point C is $1.5\text{ft}/\text{s}$.

To determine

(b)

The acceleration of point B.

Answer to Problem 15.23P

The acceleration of point B is $4.24\text{ft}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression for the normal acceleration of point B.

${\left(a_B\right)}_n=\frac{{\omega }_2^2{d}_4}{2}$ ...... (VIII)

Write the expression for tangential acceleration of point B.

${\left(a_B\right)}_t=\frac{{\alpha }_2{d}_4}{2}$ ...... (IX)

Write the expression for the resultant acceleration.

$a_B=\sqrt{{\left(a_B\right)}_n^2+{\left(a_B\right)}_t^2}$ ...... (X)

Calculation:

Substitute $8\text{in}$ for $d_4$ and $3\text{rad}/\text{s}$ for ${\omega }_2$ in Equation (VIII).

$\begin{array}{c}{\left(a_B\right)}_n=\frac{{\left(3\text{rad}/\text{s}\right)}^2\left(8\text{in}\right)}{2}\\ =36\text{in}/{\text{s}}^{\text{2}}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =3\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Substitute $\text{9}\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_2$ and $8\text{in}$ for $d_4$ in Equation (IX).

$\begin{array}{c}{\left(a_B\right)}_t=\frac{\left(9\text{rad}/{\text{s}}^{\text{2}}\right)\left(8\text{in}\right)}{2}\\ =36\text{in}/{\text{s}}^{\text{2}}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =3\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Substitute $3\text{ft}/{\text{s}}^{\text{2}}$ for ${\left(a_B\right)}_n$ and $3\text{ft}/{\text{s}}^{\text{2}}$ for in Equation (X).

$\begin{array}{c}{a}_B=\sqrt{{\left(3\text{ft}/{\text{s}}^{\text{2}}\right)}^2+{\left(3\text{ft}/{\text{s}}^{\text{2}}\right)}^2}\\ =\sqrt{18{\text{ft}}^2/{\text{s}}^4}\\ =4.24\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

The acceleration of point B is $4.24\text{ft}/{\text{s}}^{\text{2}}$.