Chapter 12.1, Problem 12.42P

To determine

Range of allowable values of $v$ so that the magnitudes of the forces in the links do not exceed $75\text{N}$.

Answer to Problem 12.42P

Range of allowable values of $v$ :

$0.73\text{m}/s<v<4.22\text{m}/s$

Explanation of Solution

Given:

$\rho =150\text{mm}=0.15\text{m}$

$m=0.5\text{kg}$

First of all,

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

Where,

$\rho =150\text{mm}=0.15\text{m}$

$\stackrel{+}{\leftarrow }\sum F_x=ma$ :

$T_{DA}\sin {20}^{\circ }+T_{DE}\sin {30}^{\circ }=\frac{W}{g}\frac{v^2}{\rho }$   ... (1)

$+\uparrow \sum F_y=0$ :

$T_{DA}\cos {20}^{\circ }+T_{DE}\cos {30}^{\circ }-W=0$   ... (2)

Note that Eq. (2) implies that

When

$T_{DE}={\left(T_{DE}\right)}_{\max }$ ,

$T_{DA}={\left(T_{DA}\right)}_{\max }$

When

$T_{DE}={\left(T_{DE}\right)}_{\min }$ ,

$T_{DA}={\left(T_{DA}\right)}_{\min }$

Case 1: $T_{DA}$ is maximum

Let

$T_{DA}=75\text{N}$

Eq. (2)

$\left(75\text{N}\right)\cos {20}^{\circ }-T_{DE}\cos {30}^{\circ }-\left(4.905\text{N}\right)$

$\left[\text{At }m=0.5\text{kg}\to W=4.905\text{N}\right]$

Or:

$T_{DE}=75.26\text{N}$, unacceptable $\left(>75\text{N}\right)$

Now let

$T_{DE}=75\text{N}$

Eq. (2)

$T_{DA}\cos {20}^{\circ }+\left(75\text{N}\right)\cos {30}^{\circ }-\left(4.905\text{N}\right)=0$

Or:

$T_{DA}=74.7\text{N}$ OK $\left(\le 75\text{N}\right)$

${\left(T_{DA}\right)}_{\max }=74.7\text{N}$ ,

${\left(T_{DE}\right)}_{\max }=75\text{N}$

Eq. (1)

${\left(v^2\right)}_{{\left(T_{DA}\right)}_{\max }}=\frac{\left(9.81\text{m}/s^2\right)\left(0.15\text{m}\right)}{4.905\text{N}}\left(74.7\sin {20}^{\circ }+75\sin {30}^{\circ }\right)\text{N}$

Or:

$v_{{\left(T_{DA}\right)}_{\max }}=4.22\text{m}/s$

Now form

$\left(\cos {30}^{\circ }\right)\times \left[\text{Eq}\text{.}\text{}\text{(1)}\right]+\left(\sin {30}^{\circ }\right)\times \left[\text{Eq}\text{.}\text{}\text{(2)}\right]$

$T_{DA}\sin {20}^{\circ }\cos {30}^{\circ }+T_{DA}\cos {20}^{\circ }\sin {30}^{\circ }=\frac{W}{g}\frac{v^2}{\rho }\cos {30}^{\circ }+W\sin {30}^{\circ }$

Or:

$T_{DA}\sin {50}^{\circ }=\frac{W}{g}\frac{v^2}{\rho }\cos {30}^{\circ }+W\sin {30}^{\circ }$   ... (3)

$v_{\max }$ occurs when $T_{DA}={\left(T_{DA}\right)}_{\max }$ ,

$v_{\max }=4.22\text{m}/s$

Case 2: $T_{DA}$ is minimum

Because ${\left(T_{DA}\right)}_{\min }$ occurs when $T_{DE}={\left(T_{DE}\right)}_{\min }$ ,

Let $T_{DE}=0$

Eq. (2)

$T_{DA}\cos {20}^{\circ }-4.905\text{N}=0$

Or:

$T_{DA}=5.6\text{N}$

Eq. (3) implies that when $T_{DA}={\left(T_{DA}\right)}_{\min }$ ,

$v=v_{\min }$

Then

Eq. (1)

${\left(v^2\right)}_{\min }=\frac{\left(9.81\text{m}/s^2\right)\left(0.15\text{m}\right)}{4.905\text{N}}\left(5.6\text{N}\sin {20}^{\circ }\right)$

Or:

$v_{\min }=0.73\text{m}/s$

$0<T_{AB},T_{BC},T_{AD},T_{DE}<75\text{N}$

When

$0.73\text{m}/s<v<2.42\text{m}/s$