Chapter 12.1, Problem 12.60P

(a)

To determine

Find whether the block will slide in the slot if it is released in the position corresponding to $\theta =80°$.

Find the magnitude and the direction of the friction force exerted on the block immediately after it is released.

Explanation of Solution

Given information:

The radius of semicircular slot is $10\text{in}\text{.}$.

The angular velocity $\left({\varphi }_{ABCD}\right)$ of flat plate about the vertical AD is 14 rad/s.

The weight $\left(W\right)$ of block E is 0.8 lb.

The coefficient of static friction $\left({\mu }_s\right)$ is 0.35.

The coefficient of kinetic friction $\left({\mu }_k\right)$ is 0.25.

Calculation:

Write the general equation of mass of block B (m).

$m=\frac{W}{g}$

Here, g is the acceleration due to gravity.

Write the general equation (a) of acceleration in curved path.

$a=\frac{v^2}{\rho }$

Here, v is the velocity in the curved path.

Find the horizontal distance (d) between line of origin to block E using the equation:

$\sin \theta =\frac{d}{r}$

Substitute 10 in. for r.

$\begin{array}{l}\sin \theta =\frac{d}{10}\\ d=10\sin \theta \end{array}$

Find the distance (D) between edge AD and block E using the equation:

$D=26-d$

Substitute $10\sin \theta$ for d.

$D=26-10\sin \theta$

Find the radius $\left(\rho \right)$ of rotation of flat plate using the equation:

$\rho =\frac{1}{12}D$

Substitute $26-10\sin \theta$ for D.

$\rho =\frac{1}{12}\left(26-10\sin \theta \right)$

Write the equation velocity of block E

$v_E=\rho {\varphi }_{ABCD}$

Find the equation of acceleration of block E.

$a_n=\frac{v_E^2}{\rho }$

Substitute $\rho {\varphi }_{ABCD}$ for $v_E$.

$\begin{array}{c}{a}_n=\frac{{\left(\rho {\varphi }_{ABCD}\right)}^2}{\rho }\\ =\rho {\varphi }_{ABCD}^2\end{array}$

Substitute $\frac{1}{12}\left(26-10\sin \theta \right)$ for $\rho$ and 14 rad/s for ${\varphi }_{ABCD}$.

$\begin{array}{c}{a}_n=\frac{1}{12}\left(26-10\sin \theta \right){\left(14\right)}^2\\ =\frac{196}{12}\left(2\right)\left(13-5\sin \theta \right)\\ =\frac{98}{3}\left(13-5\sin \theta \right)\end{array}$

Assume that the block is at rest with respect to the plate.

Sketch the free body diagram of block E as shown in Figure 1.

Here, F is the friction force and N is the normal force.

Refer Figure 1.

Apply Newton’s law of equation along x-axis (Consider forces along x-axis is negative).

$\begin{array}{l}\Sigma F_x=m{a}_x\\ N+W\cos \theta =m{a}_x\end{array}$

Substitute $a_n\sin \theta$ for $a_x$ and $\frac{W}{g}$ for m.

$\begin{array}{c}N+W\cos \theta =\frac{W}{g}{a}_n\sin \theta \\ N=\frac{W}{g}{a}_n\sin \theta -W\cos \theta \\ =W\left(\frac{a_n}{g}\sin \theta -\cos \theta \right)\end{array}$ (1)

Apply Newton’s law of equation along y-axis (Consider forces along x-axis is negative).

$\begin{array}{l}\Sigma F_y=m{a}_y\\ -F+W\sin \theta =-m{a}_y\end{array}$

Substitute $a_n\cos \theta$ for $a_y$ and $\frac{W}{g}$ for m.

$\begin{array}{c}-F+W\sin \theta =-\left(\frac{W}{g}\right)a_n\cos \theta \\ F=W\sin \theta +\left(\frac{W}{g}\right)a_n\cos \theta \\ =W\left(\sin \theta +\frac{a_n}{g}\cos \theta \right)\end{array}$ (2)

Find the normal force (N) when $\theta =80°$ using the equation (2).

$N=W\left(\frac{a_n}{g}\sin \theta -\cos \theta \right)$

Substitute $\frac{98}{3}\left(13-5\sin \theta \right)$ for $a_n$.

$N=W\left[\frac{\frac{98}{3}\left(13-5\sin \theta \right)}{g}\sin \theta -\cos \theta \right]$

Substitute 0.8 lb for W, $80°$ for $\theta$, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}N=0.8\left[\frac{\frac{98}{3}\left(13-5\sin 80°\right)}{32.2}\sin 80°-\cos 80°\right]\\ =6.3159\text{lb}\end{array}$

Find the maximum friction force $\left(F_{\max }\right)$ using the equation:

$F_{\max }={\mu }_{s}N$

Substitute 0.35 for ${\mu }_s$ and 6.3159 lb for N.

$\begin{array}{c}{F}_{\max }=0.35\left(6.3159\right)\\ =2.2106\text{lb}\end{array}$

Find the friction force (F) using Equation (1):

$F=W\left(\sin \theta +\frac{a_n}{g}\cos \theta \right)$

Substitute $\frac{98}{3}\left(13-5\sin \theta \right)$ for $a_n$.

$F=W\left\{\sin \theta +\frac{\left[\frac{98}{3}\left(13-5\sin \theta \right)\right]}{g}\cos \theta \right\}$

Substitute 0.8 lb for W, $80°$ for $\theta$, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}F=0.8\left[\sin 80°+\frac{\left[\frac{98}{3}\left(13-5\sin 80°\right)\right]}{32.2}\cos 80°\right]\\ =1.92601\text{lb}\end{array}$

The friction force on the system is less than the maximum friction force. Therefore, the block does not slide in the slot.

Thus, the friction force exerted on the block immediately after it is released is $\underset{\_}{1.92601\text{lb}}$ and the direction is along y-axis.

(b)

To determine

Find whether the block will slide in the slot if it is released in the position corresponding to $\theta =40°$.

Find the magnitude and the direction of the friction force exerted on the block immediately after it is released.

Explanation of Solution

Calculation:

Find the normal force (N) when $\theta =80°$ using the equation (2).

$N=W\left(\frac{a_n}{g}\sin \theta -\cos \theta \right)$

Substitute $\frac{98}{3}\left(13-5\sin \theta \right)$ for $a_n$.

$N=W\left[\frac{\frac{98}{3}\left(13-5\sin \theta \right)}{g}\sin \theta -\cos \theta \right]$

Substitute 0.8 lb for W, $40°$ for $\theta$, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}N=0.8\left[\frac{\frac{98}{3}\left(13-5\sin 40°\right)}{32.2}\sin 40°-\cos 40°\right]\\ =4.4924\text{lb}\end{array}$

Find the maximum friction force $\left(F_{\max }\right)$ using the equation:

$F_{\max }={\mu }_{s}N$

Substitute 0.35 for ${\mu }_s$ and 6.3159 lb for N.

$\begin{array}{c}{F}_{\max }=0.35\left(6.3159\right)\\ =2.2106\text{lb}\end{array}$

Find the friction force (F) using Equation (1):

$F=W\left(\sin \theta +\frac{a_n}{g}\cos \theta \right)$

Substitute $\frac{98}{3}\left(13-5\sin \theta \right)$ for $a_n$.

$F=W\left\{\sin \theta +\frac{\left[\frac{98}{3}\left(13-5\sin \theta \right)\right]}{g}\cos \theta \right\}$

Substitute 0.8 lb for W, $40°$ for $\theta$, and $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}F=0.8\left[\sin 40°+\frac{\left[\frac{98}{3}\left(13-5\sin 40°\right)\right]}{32.2}\cos 40°\right]\\ =6.5984\text{lb}\end{array}$

The friction force in the system is greater than the maximum friction force. Therefore the block will slide downward in the slot.

The sliding leads to kinetic equation.

Find the magnitude and the direction of the friction force (F) exerted on the block immediately after it is released using the equation:

$F={\mu }_{k}N$

Substitute 0.25 for ${\mu }_k$ and $4.4924\text{lb}$ for N.

$\begin{array}{l}F=\left(0.25\right)\left(4.4924\right)\\ =1.123\text{lb}\end{array}$

Thus, the friction force exerted on the block immediately after it is released is $\underset{\_}{1.123\text{lb}}$ and the direction is along y-axis