Chapter 16.2, Problem 16.135P

To determine

(a)

To find:

the value the couple M applied at disk A.

Answer to Problem 16.135P

At disk A, couple applied magnitude is $M=-36.3N\cdot m$

Explanation of Solution

Given information:

Rod BC mass, m = 6kg

Disk mass, m = 10kg

Rod CD mass, m = 5kg

Disk AB velocity

$v_{AB}=\left(r_{AB}{\omega }_{AB}\searrow {30}^{\circ }\right)$

Disk radius, $r_{AB}=200mm$

Disk angular velocity, ${\omega }_{AB}=36rad/s$

$v_{AB}=\left(200mm\frac{1m}{1000mm}\right)\left(36rad/s\right)$

$=\left(7.2m/s\searrow {30}^{\circ }\right)$

Since point C velocity is parallel to point B velocity, the point C velocity magnitude and direction is same as point B.

Rod CD angular velocity

${\omega }_{CD}=\frac{v_C}{L_{CD}}$

$v_C=7.2m/s$

$L_{CD}=250mm$

${\omega }_{CD}=\frac{7.2m/s}{250mm\frac{1m}{1000mm}}$

${\omega }_{CD}=28.8rad/s$

Disk B acceleration,

$a_B=0-\left(36rad/s^2\right)\left(200mm\frac{1m}{1000mm}\right)$

$=\left(259.2m/s^2\swarrow {60}^{\circ }\right)$

Rod BC acceleration tangential component,

${\left(a_{BC}\right)}_t=L_{BC}{\alpha }_{BC}$

$L_{BC}=400mm$

${\left(a_{BC}\right)}_t=\left(400mm\frac{1m}{1000mm}\right){\alpha }_{BC}$

$=\left(0.4{\alpha }_{BC}\right)\uparrow$

Rod BC acceleration,

$a_C=a_B+{\left(a_{BC}\right)}_t-{\omega }_{BC}^2{L}_{BC}$

${\omega }_{BC}=0$

$a_C=\left(259.2m/s^2\swarrow {60}^{\circ }\right)+\left(0.4{\alpha }_{BC}\right)\uparrow -0$

$=\left(259.2m/s^2\swarrow {60}^{\circ }\right)+\left(0.4{\alpha }_{BC}\right)\uparrow$

$=\left\{\left[\left(259.2m/s^2\right)\left(\cos {60}^{\circ }\right)\right]\leftarrow +\left[\left(259.2m/s^2\right)\left(\sin {60}^{\circ }\right)\right]\downarrow +\left(0.4{\alpha }_{BC}\right)\uparrow \right\}$ ......(Equation A)

$=\left[129.6m/s^2\right]\leftarrow +\left[224.4737m/s^2\right]\downarrow +\left(0.4{\alpha }_{BC}\right)\uparrow$

Rod CD acceleration tangential component,

${\left(a_{CD}\right)}_t=L_{CD}{\alpha }_{CD}$

$L_{CD}=250mm$

${\left(a_{CD}\right)}_t=\left(250mm\frac{1m}{1000mm}\right){\alpha }_{CD}$

$=\left(0.25{\alpha }_{CD}\swarrow {30}^{\circ }\right)$

Rod CD acceleration,

$a_C={\left(a_{CD}\right)}_t-{\omega }_{CD}^2{L}_{CD}$

${\omega }_{CD}=28.8rad/s^2$

$L_{CD}=0.25m$

$a_C=\left(0.25{\alpha }_{CD}\swarrow {30}^{\circ }\right)-\left(28.8rad/s^2\right)\left(0.25\right)$

$=\left(0.25{\alpha }_{CD}\swarrow {30}^{\circ }\right)-\left(207.36m/s^2\swarrow {60}^{\circ }\right)$

$=\left\{\left(0.25{\alpha }_{CD}\cos {30}^{\circ }\right)\leftarrow +\left(0.25{\alpha }_{CD}\sin {30}^{\circ }\right)\downarrow -\left(\left(207.36m/s^2\right)\cos {60}^{\circ }\right)\leftarrow -\left(\left(207.36m/s^2\right)\sin {60}^{\circ }\right)\downarrow \right\}$ .....(Equation B)

$=\left\{\left(0.2165{\alpha }_{CD}\right)\leftarrow +\left(0.2165{\alpha }_{CD}\right)\downarrow -\left(103.68m/s^2\right)\leftarrow -\left(179.579m/s^2\right)\downarrow \right\}$

Equation forces horizontal component from equations A and B,

$\left[129.6m/s^2\right]\leftarrow =\left(0.2165{\alpha }_{CD}\right)\leftarrow +\left(103.68m/s^2\right)\leftarrow$

$0.2165{\alpha }_{CD}=129.6m/s^2-103.68m/s^2$

${\alpha }_{CD}=\frac{25.92m/s^2}{0.2165}$

$=119.719rad/s^2$

Equation forces vertical component from equations A and B,

$\left[224.4737m/s^2\right]\downarrow +\left(0.4{\alpha }_{BC}\right)\uparrow =\left\{-\left(0.25{\alpha }_{CD}\sin {30}^{\circ }\right)\downarrow +\left(\left(207.36m/s^2\right)\sin {60}^{\circ }\right)\downarrow \right\}$

${\alpha }_{CD}=119.719rad/s^2$

$\left[224.4737m/s^2\right]\downarrow -\left(0.4{\alpha }_{BC}\right)\downarrow =\left\{-\left(0.25\left(119.719rad/s^2\right)\sin {30}^{\circ }\right)\downarrow +\left(\left(207.36m/s^2\right)\sin {60}^{\circ }\right)\downarrow \right\}$

$0.4{\alpha }_{BC}=224.4737m/s^2+14.9648m/s^2-179.579m/s^2$

${\alpha }_{BC}=\frac{59.8595}{0.4}$

$=149.649rad/s^2$

$a_P=a_B+{\alpha }_{BC}{r}_{PB}-{\omega }_{BC}^2{r}_{PB}$

$r_{PB}=0.2m$

Acceleration of point A is zero since it is pivoted

Rod BC acceleration of mass centre P,

$a_P=\left(259.2m/s^2\swarrow {60}^{\circ }\right)+\left[\left(149.649rad/s^2\right)\left(0.2m\right)\right]\uparrow -0$

$=\left(259.2m/s^2\swarrow {60}^{\circ }\right)+\left(29.9298rad/s^2\right)\uparrow$

Rod CD acceleration of mass centre Q,

$a_Q={\alpha }_{CD}{r}_{QD}-{\omega }_{CD}^2{r}_{QD}$

$r_{QD}=0.125m$

$a_Q=\left(119.719rad/s^2\right)\left(0.125m\right)-{\left(28.8rad/s\right)}^2\left(0.125m\right)$

$=\left(14.9648m/s^2\nwarrow {30}^{\circ }\right)-\left(103.6m/s^2\swarrow {60}^{\circ }\right)$

Disk AB effective force at mass centre,

${\left(F_{eff}\right)}_{AB}=m_{AB}{a}_A$

$m_{AB}=10kg$

$a_A=0$

${\left(F_{eff}\right)}_{AB}=\left(10kg\right)\left(0\right)$

$=0$

Disk AB moment of inertia,

$I_{AB}=\frac{m_{AB}{r}_{AB}^2}{2}$

$I_{AB}=\frac{\left(10kg\right){\left(0.2m\right)}^2}{2}$

$=0.2kg\cdot m^2$

Rod BC effective force at mass centre,

${\left(F_{eff}\right)}_{BC}=m_{BC}{a}_P$

$m_{BC}=6kg$

$\begin{array}{l}{\left(F_{eff}\right)}_{BC}=\left(6kg\right)\left[\left(259.2m/s^2\swarrow {60}^{\circ }\right)+\left(29.9298rad/s^2\right)\uparrow \right]\\ {\left(F_{eff}\right)}_{BC}=\left(1555.2N\swarrow {60}^{\circ }\right)+\left(179.58N\right)\uparrow \end{array}$

Rod BC moment of inertia,

$I_{BC}=\frac{\left(6kg\right){\left(0.4m\right)}^2}{12}$

$=0.08kg\cdot m^2$

Rod CD effective force at mass centre,

${\left(F_{eff}\right)}_{CD}=m_{CD}{a}_Q$

$m_{CD}=5kg$

${\left(F_{eff}\right)}_{CD}=\left(5kg\right)\left[\left(14.9648m/s^2\nwarrow {30}^{\circ }\right)-\left(103.6m/s^2\swarrow {60}^{\circ }\right)\right]$

$=\left(74.824N\nwarrow {30}^{\circ }\right)-\left(518N\swarrow {60}^{\circ }\right)$

Rod CD moment of inertia,

$I_{CD}=\frac{m_{CD}{L}_{CD}^2}{12}$

$L_{CD}=0.25m$

$I_{CD}=\frac{\left(5kg\right){\left(0.25m\right)}^2}{12}$

$=0.02604kg\cdot m^2$

Rod BC free body diagram

Figure A

Moment at point B from above figure,

$L_{BC}{c}_y-\frac{L_{BC}}{2}{m}_{BC}g=\left\{I_{BC}{\alpha }_{BC}+\left(0.2m\right)\left(179.58N\right)-\left(0.2m\right)\left(1555.2N\right)\sin {60}^{\circ }\right\}$

$g=9.81m/s^2$

$\left(0.4m\right)C_y-\frac{0.4m}{2}\left(6kg\right)\left(9.81m/s^2\right)=\left\{\left(0.08kg\cdot m^2\right)\left(149.649rad/s^2\right)+\left(0.2m\right)\left(179.58N\right)-\left(0.2m\right)\left(1555.2N\right)\sin {60}^{\circ }\right\}$

$0.4C_y-11.772N\cdot m=11.9719N\cdot m+35.916N\cdot m-269.3685N\cdot m$

$C_y=-524N$

Rod CD free body diagram

Figure B

Moment at point D from above figure,

$\left\{C_x{L}_{CD}\cos {30}^{\circ }+\left(524N\right)\left(0.125m\right)-m_{CD}g\left(0.0625m\right)\right\}=I_{CD}{\alpha }_{CD}+\left(74.824N\right)\left(0.125m\right)$

$\left\{C_x\left(0.25m\right)\cos {30}^{\circ }+\left(524N\right)\left(0.125m\right)-\left(5kg\right)\left(9.81m/s^2\right)\left(0.0625m\right)\right\}=\left\{\left(0.02604kg\cdot m^2\right)\left(119.719rad/s^2\right)+\left(74.824N\right)\left(0.125m\right)\right\}$

$0.2165C_x+65.5339N\cdot m-3.0656N\cdot m=3.11754N\cdot m+9.353N\cdot m$

$C_x=\frac{49.9979}{0.2165}$

$=-231N$

Combined disk AB and rod BC free body diagram

Figure C

From above figure, take moment at point A,

$\left\{-M-\left(524.47N\right)\left(0.4m+0.2\sin {30}^{\circ }\right)+\left(230.93\right)\left(0.2\cos {30}^{\circ }\right)-\left(6kg\right)\left(9.81m/s^2\right)\left(0.2+0.2\sin {30}^{\circ }\right)\right\}$

$=\left\{I_{BC}{\alpha }_{BC}+\left(179.58N\right)\left(0.2+0.2\sin {30}^{\circ }\right)-\left(1555.2\cos {30}^{\circ }\right)\left(0.2\right)\right\}$

M is couple applied at point A

$\left\{-M-\left(524.47N\right)\left(0.4m+0.2\sin {30}^{\circ }\right)+\left(230.93\right)\left(0.2\cos {30}^{\circ }\right)-\left(6kg\right)\left(9.81m/s^2\right)\left(0.2+0.2\sin {30}^{\circ }\right)\right\}$

$=\left\{\left(0.08kg\cdot m^2\right)\left(149.649rad/s^2\right)+\left(179.58N\right)\left(0.2+0.2\sin {30}^{\circ }\right)-\left(1555.2\cos {30}^{\circ }\right)\left(0.2\right)\right\}$

$\left\{-M-262.235N\cdot m+39.9982N\cdot m-17.658N\cdot m\right\}=\left\{11.9719N\cdot m+53.874N\cdot m-269.3685N\cdot m\right\}$

$M=-36.3N\cdot m$

At disk A, couple applied magnitude is $M=-36.3N\cdot m$

Conclusion:

At disk A, couple applied magnitude is $M=-36.3N\cdot m$

To determine

(b)

To find:

the force components exerted on rod BC

Answer to Problem 16.135P

The force horizontal component exerted at point C is $C_x=-231N$ and it acts in left direction and vertical component is $C_y=-524N$ and it also acts in left direction

Explanation of Solution

Given information:

Rod BC mass, m = 6kg

Disk mass, m = 10kg

Rod CD mass, m = 5kg

Rod BC free body diagram

Figure A

Moment at point B from above figure,

$L_{BC}{c}_y-\frac{L_{BC}}{2}{m}_{BC}g=\left\{I_{BC}{\alpha }_{BC}+\left(0.2m\right)\left(179.58N\right)-\left(0.2m\right)\left(1555.2N\right)\sin {60}^{\circ }\right\}$

$g=9.81m/s^2$

$\left(0.4m\right)C_y-\frac{0.4m}{2}\left(6kg\right)\left(9.81m/s^2\right)=\left\{\left(0.08kg\cdot m^2\right)\left(149.649rad/s^2\right)+\left(0.2m\right)\left(179.58N\right)-\left(0.2m\right)\left(1555.2N\right)\sin {60}^{\circ }\right\}$

$0.4C_y-11.772N\cdot m=11.9719N\cdot m+35.916N\cdot m-269.3685N\cdot m$

$C_y=-524N$

Rod CD free body diagram

Figure B

Moment at point D from above figure,

$\left\{C_x{L}_{CD}\cos {30}^{\circ }+\left(524N\right)\left(0.125m\right)-m_{CD}g\left(0.0625m\right)\right\}=I_{CD}{\alpha }_{CD}+\left(74.824N\right)\left(0.125m\right)$

$\left\{C_x\left(0.25m\right)\cos {30}^{\circ }+\left(524N\right)\left(0.125m\right)-\left(5kg\right)\left(9.81m/s^2\right)\left(0.0625m\right)\right\}=\left\{\left(0.02604kg\cdot m^2\right)\left(119.719rad/s^2\right)+\left(74.824N\right)\left(0.125m\right)\right\}$

$0.2165C_x+65.5339N\cdot m-3.0656N\cdot m=3.11754N\cdot m+9.353N\cdot m$

$C_x=\frac{49.9979}{0.2165}$

$=-231N$

Combined disk AB and rod BC free body diagram

Figure C

From above figure, take moment at point A,

$\left\{-M-\left(524.47N\right)\left(0.4m+0.2\sin {30}^{\circ }\right)+\left(230.93\right)\left(0.2\cos {30}^{\circ }\right)-\left(6kg\right)\left(9.81m/s^2\right)\left(0.2+0.2\sin {30}^{\circ }\right)\right\}$

$=\left\{I_{BC}{\alpha }_{BC}+\left(179.58N\right)\left(0.2+0.2\sin {30}^{\circ }\right)-\left(1555.2\cos {30}^{\circ }\right)\left(0.2\right)\right\}$

M is couple applied at point A

$\left\{-M-\left(524.47N\right)\left(0.4m+0.2\sin {30}^{\circ }\right)+\left(230.93\right)\left(0.2\cos {30}^{\circ }\right)-\left(6kg\right)\left(9.81m/s^2\right)\left(0.2+0.2\sin {30}^{\circ }\right)\right\}$

$=\left\{\left(0.08kg\cdot m^2\right)\left(149.649rad/s^2\right)+\left(179.58N\right)\left(0.2+0.2\sin {30}^{\circ }\right)-\left(1555.2\cos {30}^{\circ }\right)\left(0.2\right)\right\}$

$\left\{-M-262.235N\cdot m+39.9982N\cdot m-17.658N\cdot m\right\}=\left\{11.9719N\cdot m+53.874N\cdot m-269.3685N\cdot m\right\}$

$M=-36.3N\cdot m$

At disk A, couple applied magnitude is $M=-36.3N\cdot m$

Conclusion:

The force horizontal component exerted at point C is $C_x=-231N$ and it acts in left direction and vertical component is $C_y=-524N$ and it also acts in left direction.