Chapter 3, Problem 3.39P

To determine

(a)

The equation of motion of the scale.

Answer to Problem 3.39P

$\left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)$.

Explanation of Solution

Given:

Friction of the pivot and mass of the scale arm are neglected.

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: $I_o\dot{\omega }=M_o$

Where $I_o$ = Mass moment of Inertia of the body about point O.

$\omega$ = Angular velocity of the mass about an axis through a point O.

$M_o$ = Sum of the moments applied to the body about the point O.

Let the angular displacement be $\theta$.

The angular velocity, $\omega =\frac{d\theta }{dt}=\dot{\theta }$

$\Rightarrow \dot{\omega }=\frac{d\omega }{dt}=\frac{d^2\theta }{d{t}^2}=\ddot{\theta }$

Hence, the equation of motion of this object can be rewritten by substituting, $\dot{\omega }=\ddot{\theta }$:

$I_o\ddot{\theta }=M_o$

Derivation of Equation of motion:

Free body diagram:

To find the equation of motion, the required unknowns are $I_A$ (Inertia about point A ) and $M_A$ (Sum of moments about pointA ).

The mass moment of Inertia, I about a specified reference axis is given as:

$I={\displaystyle \int r^{2}dm}$

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = ${\displaystyle \int r^{2}dm=m{r}^2}$

Inertia of mass, $m_c$, is $m_c{L}_1^2$

Inertia of mass, $m$, is $m{L}_2^2$

Total Inertia, $I_A=m_c{L}_1^2+m{L}_2^2$.

Sum of moments about the point, A:

$M_A=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)$

$\begin{array}{l}{I}_A\ddot{\theta }=M_A\\ I_A=m_c{L}_1^2+m{L}_2^2\\ M_A=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\\ \left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\end{array}$.

Conclusion:

The equation of motion of the scale is $\left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)$.

To determine

(b)

The equilibrium relation between weight and angle.

Answer to Problem 3.39P

$mg=\frac{m_{c}g{L}_1\sin \theta }{L_2\cos \left(\beta -\theta \right)}$.

Explanation of Solution

Given:

The system is in equilibrium.

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: $I_o\dot{\omega }=M_o$

Where $I_o$ = Mass moment of Inertia of the body about point O.

$\omega$ = Angular velocity of the mass about an axis through a point O.

$M_o$ = Sum of the moments applied to the body about the point O.

Let the angular displacement be $\theta$.

The angular velocity, $\omega =\frac{d\theta }{dt}=\dot{\theta }$

$\Rightarrow \dot{\omega }=\frac{d\omega }{dt}=\frac{d^2\theta }{d{t}^2}=\ddot{\theta }$

Hence, the equation of motion of this object can be rewritten by substituting, $\dot{\omega }=\ddot{\theta }$:

$I_o\ddot{\theta }=M_o$

Derivation of mg at equilibrium:

Free body diagram:

To find the equation of motion, the required unknowns are $I_A$ (Inertia about point A ) and $M_A$ (Sum of moments about point A ).

The mass moment of Inertia, I about a specified reference axis is given as:

$I={\displaystyle \int r^{2}dm}$

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = ${\displaystyle \int r^{2}dm=m{r}^2}$

Inertia of mass, $m_c$, is $m_c{L}_1^2$

Inertia of mass, $m$, is $m{L}_2^2$

Total Inertia, $I_A=m_c{L}_1^2+m{L}_2^2$.

Sum of moments about the point, A:

$M_A=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)$

$\begin{array}{l}{I}_A\ddot{\theta }=M_A\\ I_A=m_c{L}_1^2+m{L}_2^2\\ M_A=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\\ \left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\to \left(1\right)\end{array}$

At equilibrium, the angular acceleration, $\ddot{\theta }$ is zero.

Substitute the value of $\ddot{\theta }$ in equation, $\left(1\right)$:

$\begin{array}{l}\left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\to \left(1\right)\\ \ddot{\theta }=0\\ \left(m_c{L}_1^2+m{L}_2^2\right)\times 0=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\\ 0=mg{L}_2\cos \left(\beta -\theta \right)-m_{c}g{L}_1\sin \theta \\ mg{L}_2\cos \left(\beta -\theta \right)=m_{c}g{L}_1\sin \theta \\ mg=\frac{m_{c}g{L}_1\sin \theta }{L_2\cos \left(\beta -\theta \right)}\end{array}$.

Conclusion:

When system is at equilibrium, the relation between mg and $\theta$ is $mg=\frac{m_{c}g{L}_1\sin \theta }{L_2\cos \left(\beta -\theta \right)}$.

To determine

(c)

The value of weightof the object.

Answer to Problem 3.39P

22. 713N.

Explanation of Solution

Given:

The mass of counterweight, $m_c=5kg$

$L_1=0.2m$

$L_2=0.15m$

$\beta =30°$

$\theta =20°$

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: $I_o\dot{\omega }=M_o$

Where $I_o$ = Mass moment of Inertia of the body about point O.

$\omega$ = Angular velocity of the mass about an axis through a point O.

$M_o$ = Sum of the moments applied to the body about the point O.

Let the angular displacement be $\theta$.

The angular velocity, $\omega =\frac{d\theta }{dt}=\dot{\theta }$

$\Rightarrow \dot{\omega }=\frac{d\omega }{dt}=\frac{d^2\theta }{d{t}^2}=\ddot{\theta }$

Hence, the equation of motion of this object can be rewritten by substituting, $\dot{\omega }=\ddot{\theta }$:

$I_o\ddot{\theta }=M_o$

Derivation of mg at equilibrium:

Free body diagram:

To find the equation of motion, the required unknowns are $I_A$ (Inertia about point A ) and $M_A$ (Sum of moments about point A ).

The mass moment of Inertia, I about a specified reference axis is given as:

$I={\displaystyle \int r^{2}dm}$

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = ${\displaystyle \int r^{2}dm=m{r}^2}$

Inertia of mass, $m_c$, is $m_c{L}_1^2$

Inertia of mass, $m$, is $m{L}_2^2$

Total Inertia, $I_A=m_c{L}_1^2+m{L}_2^2$.

Sum of moments about the point, A:

$M_A=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)$

$\begin{array}{l}{I}_A\ddot{\theta }=M_A\\ I_A=m_c{L}_1^2+m{L}_2^2\\ M_A=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\\ \left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\to \left(1\right)\end{array}$

At equilibrium, the angular acceleration, $\ddot{\theta }$ is zero.

Substitute the value of $\ddot{\theta }$ in equation, $\left(1\right)$:

$\begin{array}{l}\left(m_c{L}_1^2+m{L}_2^2\right)\ddot{\theta }=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\to \left(1\right)\\ \ddot{\theta }=0\\ \left(m_c{L}_1^2+m{L}_2^2\right)\times 0=-m_{c}g{L}_1\sin \theta +mg{L}_2\cos \left(\beta -\theta \right)\\ 0=mg{L}_2\cos \left(\beta -\theta \right)-m_{c}g{L}_1\sin \theta \\ mg{L}_2\cos \left(\beta -\theta \right)=m_{c}g{L}_1\sin \theta \\ mg=\frac{m_{c}g{L}_1\sin \theta }{L_2\cos \left(\beta -\theta \right)}\to \left(2\right)\end{array}$

Substitute the given values in equation $\left(2\right)$ to find the weight, mg.

$\begin{array}{l}mg=\frac{m_{c}g{L}_1\sin \theta }{L_2\cos \left(\beta -\theta \right)}\to \left(2\right)\\ m_c=5kg,L_1=0.2m,L_2=0.15m,\beta =30°,\theta =20°\\ mg=\frac{5g\times 0.2\sin 20°}{0.15\cos \left(30°-20°\right)}\\ mg=\frac{g\sin 20°}{0.15\cos \left(10°\right)}\\ mg=22.713N\end{array}$.

Conclusion:

The weight of the object is 22. 713N.