Chapter 9, Problem 118P

(a)

To determine

To Calculate: The initial angular acceleration of the structure.

Answer to Problem 118P

  $0.3{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given information:

Height of each vertical beam $=12.0\text{ft}$

Width of each vertical beam $=2.0\text{ft}$

Length horizontal cross-member $=6.0\text{ft}$

The mass of the vertical beam $=350\text{ kg}$

The mass of the horizontal beam $=280\text{ kg}$

Formula Used:

From Newton’s second law of motion

  $F=ma$

Where, F is the net force, m is the mass and a is the acceleration.

Calculation:

  

Here, one of the structures initially in its upright position. But later, it is about to strike the floor. The moments about the axis of rotation (a line through point P) can be considered.

Use the parallel-axis theorem to find the moments of inertia of the two parts of this composite structure. Let the numeral 1 denote the vertical member and the numeral 2 the horizontal member.

Apply Newton’s second law of motion in rotational form to the structure to express its angular acceleration in terms of the net torque causing it to fall and its moment of inertia with respect to point P.

Taking clockwise rotation to be positive, use ${\displaystyle \sum \tau =I_P}\alpha$ .

  $m_{2}g\left(\frac{l_2}{2}\right)-m_{1}g\left(\frac{w}{2}\right)=I_p\alpha$

Here,

The mass of the vertical beam = $m_1$

mass of the horizontal beam = $m_2$

Length of vertical beam = $l_2$

Width of the beam = $w$

Acceleration of the beam = $\alpha$

Gravitational acceleration = $g$

  $\alpha =\frac{m_{2}g{l}_2-m_{1}gw}{2I_p}$

Because $I_P=I_{1P}+I_{2P}$

  $\alpha =\frac{g\left(m_2{l}_2-m_{1}w\right)}{2\left(I_{1p}+I_{2p}\right)}$

  $\begin{array}{l}{l}_1=10.0\text{ft}\times \frac{1\text{m}}{3.281\text{f}t}\\ l_1=3.05\text{m}\end{array}$

  $\begin{array}{l}{l}_1=6.0\text{ft}\times \frac{1\text{m}}{3.281\text{f}t}\\ l_1=1.83\text{m}\end{array}$

  $\begin{array}{l}w=2.0\text{ft}\times \frac{1\text{m}}{3.281\text{f}t}\\ l_1=0.61\text{m}\end{array}$

  $\begin{array}{l}{I}_{1p}=\frac{1}{3}{m}_1{l}_1^2+m_1{\left(\frac{w}{2}\right)}^2\\ I_{1p}=m_1\left(\frac{1}{3}{l}_1^2+\frac{1}{4}{w}^2\right)\end{array}$

  $\begin{array}{l}{I}_{1p}=\left(350\text{kg}\right)\left[\frac{1}{3}{\left(3.05\text{m}\right)}^2+\frac{1}{4}{\left(0.610\text{m}\right)}^2\right]\\ I_{1p}=1.12\times {10}^3{\text{kgm}}^{\text{2}}\end{array}$

  $I_{2p}=I_{2,cm}+m_2{d}^2$

  $\begin{array}{c}{d}^2={\left(l_1+\frac{1}{2}w\right)}^2+{\left(\frac{1}{2}{l}_2-w\right)}^2\\ d=\sqrt{{\left(l_1+\frac{1}{2}w\right)}^2+{\left(\frac{1}{2}{l}_2-w\right)}^2}\end{array}$

  $d=\sqrt{{\left[3.05\text{m}+\left(0.610\text{m}\right)\right]}^2+{\left[\frac{1}{2}\left(1.83\text{m}\right)-0.610\text{m}\right]}^2}=3.67\text{m}$

  $I_{2,cm}=\frac{1}{12}{m}_2{l}_2^2$

  $\begin{array}{c}{I}_{2p}=\frac{1}{12}{m}_2{l}_2^2+m_2{d}^2\\ I_{2p}=m_2\left(\frac{1}{12}{l}_2^2+d^2\right)\end{array}$

  $\begin{array}{l}{I}_{2p}=\left(280\text{kg}\right)\left[\frac{1}{12}{\left(1.83\text{m}\right)}^2+{\left(3.67\text{m}\right)}^2\right]\\ I_{2p}=3.85\times {10}^3\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

  $\begin{array}{l}\alpha =\frac{\left(9.81\text{m}\cdot {\text{s}}^{\text{-2}}\right)\left[\left(280\text{kg}\right)\left(1.83\text{m}\right)-\left(350\text{kg}\right)\left(0.61\text{m}\right)\right]}{2\left(1.12+3.85\right)\times {10}^3\text{kg}\cdot {\text{m}}^{\text{2}}}\\ \alpha =0.3{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

The initial angular acceleration of the structure is $0.3{\text{rad/s}}^{\text{2}}$ .

(b)

To determine

ToCalculate: The magnitude of the initial linear acceleration of the right end of the horizontal beam.

Answer to Problem 118P

  $1.16\text{m}\cdot {\text{s}}^{\text{-2}}$

Explanation of Solution

Given information:

Height of each vertical beam $=12.0\text{ft}$

Width of each vertical beam $=2.0\text{ft}$

Length horizontal cross-member $=6.0\text{ft}$

The mass of the vertical beam $=350\text{ kg}$

The mass of the horizontal beam $=280\text{ kg}$

The initial angular acceleration of the structure is $0.3{\text{rad/s}}^{\text{2}}$ .

Formula used:

From Newton’s second law of motion

  $F=ma$

Where, F is the net force, m is the mass and a is the acceleration.

Linear acceleration (a) in terms of angular acceleration ( $\alpha$ ) and radius (R) is:

  $a=\alpha R$

Calculation:

  

$R^2=\left(l_1+w\right)+\left(l_2-w\right)$

  $\begin{array}{l}R=\sqrt{{\left(l_1+w\right)}^2+{\left(l_2-w\right)}^2}\\ R=\sqrt{{\left(3.05\text{m}+0.61\text{m}\right)}^2+{\left(1.83\text{m-}0.61\text{m}\right)}^2}\\ R=3.86\text{m}\end{array}$

  $\begin{array}{l}a=\left(0.3{\text{rad/s}}^{\text{2}}\right)\left(3.86\text{m}\right)\\ a=1.16\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

Conclusion:

The magnitude of the initial linear acceleration of the right end of the horizontal beam is $1.16\text{m}\cdot {\text{s}}^{\text{-2}}$ .

(c)

To determine

To explain: The horizontal component of the initial linear acceleration be at this same location.

Answer to Problem 118P

  $1.1\text{m}\cdot {\text{s}}^{\text{-2}}$

Explanation of Solution

Given information:

Height of each vertical beam $=12.0\text{ft}$

Width of each vertical beam $=2.0\text{ft}$

Length horizontal cross-member $=6.0\text{ft}$

The mass of the vertical beam $=350\text{ kg}$

The mass of the horizontal beam $=280\text{ kg}$

The magnitude of the initial linear acceleration of the right end of the horizontal beam is $1.16\text{m}\cdot {\text{s}}^{\text{-2}}$ .

Formula used:

Horizontal component of acceleration is $a_x=a\cos \theta$ .

Calculation:

  $\begin{array}{c}{a}_x=a\cos \theta \\ a_x=a\frac{l_1+w}{R}\end{array}$

  $\begin{array}{l}{a}_x=\left(1.16\text{m}\cdot {\text{s}}^{\text{-2}}\right)\frac{3.05\text{m}+0.610\text{m}}{3.86\text{m}}\\ a_x=1.1\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

Conclusion:

The horizontal component of the initial linear acceleration be at this same location is $1.1\text{m}\cdot {\text{s}}^{\text{-2}}$ .

(d)

To determine

To Calculate: The beam’s rotational speed when they caught it.

Answer to Problem 118P

  $1.05\text{rad/s}$

Explanation of Solution

Given information:

Height of each vertical beam $=12.0\text{ft}$

Width of each vertical beam $=2.0\text{ft}$

Length horizontal cross-member $=6.0\text{ft}$

The mass of the vertical beam $=350\text{ kg}$

The mass of the horizontal beam $=280\text{ kg}$

Formula used:

Rotational kinetic energy:

  $K.E_r=\frac{1}{2}I{\omega }^2$

Where, I is the moment of inertia and $\omega$ is the angular speed.

Potential energy, $U=mgh$

Where, m is the mass, g is the acceleration due to gravity and h is the height.

Calculation:

By applying the conservation of mechanical energy to the beam:

  $\Delta K+\Delta U_s=K_f-K_i+U_f-U_i=0$

  $U_g=0$

  $K_f-U_i=0$

  $\frac{1}{2}{I}_p{\omega }_f^2-mg\left(\Delta h\right)=0$

  $\Delta h\approx \frac{1}{4}{l}_1$

  $\begin{array}{c}\frac{1}{2}{I}_p{\omega }_f^2-\frac{1}{4}mg{l}_1=0\\ {\omega }_f=\sqrt{\frac{mg{l}_1}{2I_p}}\end{array}$

  $\begin{array}{l}{\omega }_f=\sqrt{\frac{\left(350\text{kg}\right)\left(9.81{\text{ms}}^{\text{-2}}\right)\left(3.05\text{m}\right)}{2\left(1.12+3.85\right)\times {10}^3{\text{kgm}}^{\text{-2}}}}\\ {\omega }_f=1.05\text{rad/s}\end{array}$

Conclusion:

The beam’s rotational speed when they caught it is $1.05\text{rad/s}$ .