Chapter 12.1, Problem 12.70P

Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the fixed circular slot DE of radius $b=20$ in. Neglecting friction and assuming that $\stackrel{\text{'}}{\theta }=15$ rad/s and $\stackrel{\text{'}\text{'}}{\theta }=250$ rad/s² for the position $\theta =20°$ , determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE.

  

To determine

(a)

The radial and transverse component of the resultant force at given condition.

Answer to Problem 12.70P

The required value of the radial component is $-13.16lb$ and transverse component is $2.10lb.$

Explanation of Solution

Given information:

$\begin{array}{l}{w}_B=4oz\\ b=20in\\ \dot{\theta }=15rad/s\\ \ddot{\theta }=250rad/s^2\\ \theta ={20}^o\end{array}$

Formula used:

$a_r=\ddot{r}-r{\dot{\theta }}^2$

$a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }$

$F=ma$

Calculation:

According to the geometry of the system we have.

$\Rightarrow r=2b\cos \theta$.

$\Rightarrow \dot{r}=-\left(2b\sin \theta \right)\dot{\theta }$.

$\Rightarrow \ddot{r}=-2b\left(\ddot{\theta }\sin \theta +{\dot{\theta }}^2\cos \theta \right)$.

And, $a_r=\ddot{r}-r{\dot{\theta }}^2$.

$\Rightarrow -2b\left(\ddot{\theta }\sin \theta +{\dot{\theta }}^2\cos \theta \right)-\left(2b\cos \theta \right){\dot{\theta }}^2$.

$\Rightarrow -2b\left(\ddot{\theta }\sin \theta +2{\dot{\theta }}^2\cos \theta \right)$.

$\Rightarrow -2\left(\frac{20}{12}ft\right)\left[\left(250rad/s^2\right)\sin {20}^o+2{\left(15rad/s\right)}^2\cos {20}^o\right]$.

$\Rightarrow -1694.56ft/s^2$.

We know that, $a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }$.

$\Rightarrow \left(2b\cos \theta \right)\ddot{\theta }+2\left(-2b\dot{\theta }\sin \theta \right)\dot{\theta }$.

$\Rightarrow 2b\left(\ddot{\theta }\cos \theta -2{\dot{\theta }}^2\sin \theta \right)$.

$\Rightarrow 2\left(\frac{20}{12}ft\right)\left[\left(250rad/s^2\right)\cos {20}^o-2\left(15rad/s^2\right)\sin {20}^o\right]$.

$\Rightarrow 270.05ft/s^2$.

We know that, $F_r=m{a}_r$.

$\Rightarrow \frac{1/4lb}{32.2ft/s^2}\times \left(-1694.56ft/s^2\right)$

$\Rightarrow -13.1565lb$.

$F_r=-13.16lb$

And also, $F_{\theta }=m{a}_{\theta }$.

$\Rightarrow \frac{1/4lb}{32.2ft/s^2}\times \left(270.05ft/s^2\right)$.

$\Rightarrow 2.0967lb$.

$\Rightarrow F_{\theta }=2.10lb$

To determine

(b)

The forces P and Q exerted on pin B at given condition.

Answer to Problem 12.70P

The required value of the force is $P=6.89lb$ and $Q=14lb$.

Explanation of Solution

Given information:

$\begin{array}{l}{w}_B=4oz\\ b=20in\\ \dot{\theta }=15rad/s\\ \ddot{\theta }=250rad/s^2\\ \theta ={20}^o\end{array}$

Formula used:

$F=ma$

Calculation:

The FBD is depicted below.

We know that, $\sum F_r=-F_r$.

$\Rightarrow -Q\cos {20}^o$.

$\Rightarrow Q=\frac{1}{\cos {20}^o}\left(13.1565lb\right)$.

$\Rightarrow 14.0009lb$.

Now the sum of the angular forces.

$\Rightarrow \sum F_{\theta }=P-Q\sin {20}^o$.

$\Rightarrow P=\left(2.0967+14.0009\sin {20}^o\right)$.

$\Rightarrow P=6.89lb$.

$\Rightarrow Q=14lb$