Chapter 16.1, Problem 16.20P

The coefficients of friction between the 30-lb block and the 5-lb platform BD are μ_s = 0.50 and μ_k = 0.40. Determine the accelerations of the block and of the platform immediately after wire AB has been cut.

Fig. P16.20

To determine

Calculate the accelerations of the block and of the platform immediately after wire AB has been cut $\left(a_b\text{and}{a}_p\right)$.

Answer to Problem 16.20P

The accelerations of the block and of the platform immediately after wire AB has been cut $\left(a_b\text{and}{a}_p\right)$ are $\underset{\_}{17.01\text{ft}/{\text{s}}^{\text{2}}}$ at angle $\underset{\_}{67.1°}$ and $\underset{\_}{31.3\text{ft}/{\text{s}}^{\text{2}}}$ at angle $\underset{\_}{30°}$.

Explanation of Solution

Given information:

The weight of the block $\left(W_b\right)$ is $30\text{lb}$.

The weight of the platform BD $\left(W_p\right)$ is $5\text{lb}$.

The coefficients of static friction $\left({\mu }_s\right)$ is $0.50$.

The coefficients of kinetic friction $\left({\mu }_k\right)$ is $0.40$.

The length of the platform BD (L) is $18\text{in}\text{.}$.

Calculation:

Consider the acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Consider the block does not slide relative to the platform.

Show the free body and kinetic diagram of the platform and block as in Figure 1.

Here, $T_{BC}$ is the tension of the wire BC, $T_{DE}$ is the tension of the wire DE, $m$ is the mass of the block and of the platform, and $a$ is the acceleration of the block and of the platform.

Refer to Figure 1.

Calculate the forces at the tangent:

$\begin{array}{l}{\displaystyle \sum F_t}=m{a}_t\\ \left(W\times \sin 30°\right)=ma\\ W\sin 30°=ma\end{array}$

Substitute $mg$ for W.

$\begin{array}{l}mg\sin 30°=ma\\ g\sin 30°=a\\ a=g\sin 30°\end{array}$ (1)

Calculate the acceleration of the block and of the platform (a):

Substitute and $32.2{\text{ft/s}}^2$ for g in Equation (1).

$\begin{array}{c}a=32.2\times \sin 30°\\ =16.1\text{ft}/{\text{s}}^{\text{2}}\text{at an angle 30}°\end{array}$

Check whether or not the block will slide relative to the platform.

Show the free body and kinetic diagram of the block as in Figure 2.

Here, F is the horizontal force of the block and N is the vertical force of the block.

Refer to Figure 2

Calculate the horizontal forces by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum F_x}=ma\\ F=ma\cos 30°\end{array}$

Substitute $g\sin 30°$ for $a$

$\begin{array}{c}F=mg\sin 30°\cos 30°\\ =W_b\sin 30°\cos 30°\end{array}$

Substitute $30\text{lb}$ for $W_b$

$\begin{array}{c}F=30\times \sin 30°\cos 30°\\ =12.99\text{lb}\end{array}$

Calculate the vertical forces by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum F_y}=ma\\ W_b-N=ma\sin 30°\end{array}$

Substitute $g\sin 30°$ for $a$

$\begin{array}{l}{\displaystyle \sum F_y}=ma\\ W_b-N=mg\sin 30°\sin 30°\\ N=W_b-0.25W_b\\ N=W_b\left(1-0.25\right)\end{array}$

Substitute $30\text{lb}$ for $W_b$

$\begin{array}{c}N=30\times \left(1-0.25\right)\\ =22.5\text{lb}\end{array}$

Calculate the force of the block $\left(F_m\right)$:

$F_m={\mu }_{s}N$

Substitute 0.50 for ${\mu }_s$ and $22.5\text{lb}$ for N.

$\begin{array}{c}{F}_m=0.50\times 22.5\\ =11.25\text{lb}<F\end{array}$

Since, the block will slide.

Consider the block slides relative to the platform.

Calculate the Equations of motion for block:

Show the free body and kinetic diagram of the block as in Figure 3.

Here, ${\left(a_b\right)}_y$ is the acceleration of the block at y direction and ${\left(a_b\right)}_x$ is the acceleration of the block at x direction.

Refer to Figure 3

Calculate the vertical forces by applying the equation of equilibrium:

$\begin{array}{c}{\displaystyle \sum F_y}=ma\\ 30-N=\frac{30}{g}\times {\left(a_b\right)}_y\\ {\left(a_b\right)}_y=\left(\frac{30-N}{30}\right)g\\ =g\left(1-\frac{N}{30}\right)\end{array}$ (2)

Calculate the horizontal forces by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum F_x}=ma\\ F=\frac{30}{g}\times {\left(a_b\right)}_x\end{array}$

Substitute ${\mu }_{k}N$ for F.

$\begin{array}{l}{\mu }_{k}N=\frac{30}{g}\times {\left(a_b\right)}_x\\ {\left(a_b\right)}_x=\frac{g{\mu }_{k}N}{30}\end{array}$

Substitute 0.40 for ${\mu }_k$.

$\begin{array}{c}{\left(a_b\right)}_x=\frac{g\times 0.4\times N}{30}\\ =\frac{gN}{75}\end{array}$ (3)

Calculate the Equations of motion for platform:

Show the free body and kinetic diagram of the platform as in Figure 4.

Here, $a_p$ is the acceleration of the platform.

Refer to Figure 4

Calculate the forces at the tangent:

$\begin{array}{l}{\displaystyle \sum F_t}=m{a}_t\\ \left(N+5\right)\times \sin 30°-\left(F\times \cos 30°\right)=\frac{5}{g}{a}_p\\ 0.5N+2.5-0.866F=\frac{5}{g}{a}_p\end{array}$

Substitute $0.4N$ for F.

$\begin{array}{l}0.5N+2.5-\left(0.866\times 0.4N\right)=\frac{5}{g}{a}_p\\ a_p=\frac{\left(0.1536N+2.5\right)g}{5}\\ a_p=\left(0.030718N+0.5\right)g\end{array}$ (4)

If contact is maintained between block and platform.

Therefore,

$\begin{array}{c}{\left(a_b\right)}_y={\left(a_p\right)}_y\\ {\left(a_b\right)}_y=a_p\sin 30°\end{array}$ (5)

Calculate the vertical force of the block (N):

Substitute $g\left(1-\frac{N}{30}\right)$ for ${\left(a_b\right)}_y$ and $a_p$ for $\left(0.030718N+0.5\right)g$.

$\begin{array}{l}g\left(1-\frac{N}{30}\right)=\left(0.030718N+0.5\right)g\times \sin 30°\\ g-\frac{Ng}{30}=0.015359Ng+0.25g\\ 0.015359Ng+\frac{Ng}{30}=g-0.25g\\ 0.04869Ng=0.75g\end{array}$

$\begin{array}{l}0.04869N=0.75\\ N=\frac{0.75}{0.04869}\\ N=15.403\text{lb}\end{array}$

Calculate the acceleration of the block at x direction ${\left(a_b\right)}_x$:

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g and $15.403\text{lb}$ for N in Equation (3).

$\begin{array}{c}{\left(a_b\right)}_x=\frac{32.2\times 15.403}{75}\\ =6.61\text{ft}/{\text{s}}^{\text{2}}\to \end{array}$

Calculate the acceleration of the block at y direction ${\left(a_b\right)}_y$:

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g and $15.403\text{lb}$ for N in Equation (2).

$\begin{array}{c}{\left(a_b\right)}_y=32.2\times \left(1-\frac{15.403}{30}\right)\\ =15.67\text{ft}/{\text{s}}^{\text{2}}\downarrow \end{array}$

Calculate the accelerations of the block $\left(a_b\right)$:

$a_b=\sqrt{{\left[{\left(a_b\right)}_x\right]}^2+{\left[{\left(a_b\right)}_y\right]}^2}$

Substitute $6.61\text{lb}$ for ${\left(a_b\right)}_x$ and $15.67\text{lb}$ for ${\left(a_b\right)}_y$.

$\begin{array}{c}{a}_b=\sqrt{{\left(6.61\right)}^2+{\left(15.67\right)}^2}\\ =17.01\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Calculate the angle of the block $\left({\alpha }_b\right)$:

$\tan {\alpha }_b=\frac{{\left(a_b\right)}_y}{{\left(a_b\right)}_x}$

Substitute $6.61\text{lb}$ for ${\left(a_b\right)}_x$ and $15.67\text{lb}$ for ${\left(a_b\right)}_y$.

$\begin{array}{c}\tan {\alpha }_b=\frac{15.67}{6.61}\\ {\alpha }_b={\tan }^{-1}\left(\frac{15.67}{6.61}\right)\\ =67.1°\end{array}$

Calculate the accelerations of the platform $\left(a_p\right)$:

Substitute $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g and $15.403\text{lb}$ for N in Equation (4).

$\begin{array}{c}{a}_p=\left[\left(0.030718\times 15.403\right)+0.5\right]\times 32.2\\ =31.3\text{ft}/{\text{s}}^{\text{2}}\end{array}$

Calculate the angle of the platform $\left({\alpha }_p\right)$:

${\alpha }_p=90°-\theta$

Substitute $60°$ for $\theta$

$\begin{array}{c}{\alpha }_p=90°-60°\\ =30°\end{array}$

Hence, the accelerations of the block and of the platform immediately after wire AB has been cut $\left(a_b\text{and}{a}_p\right)$ are $\underset{\_}{17.01\text{ft}/{\text{s}}^{\text{2}}}$ at angle $\underset{\_}{67.1°}$ and $\underset{\_}{31.3\text{ft}/{\text{s}}^{\text{2}}}$ at angle $\underset{\_}{30°}$.