Chapter 14, Problem 32PQ

A 215-kg robotic arm at an assembly plant is extended horizontally (Fig. P14.32). The massless support rope attached at point B makes an angle of 15.0° with the horizontal, and the center of mass of the arm is at point C. a. What is the tension in the support rope? b. What are the magnitude and direction of the force exerted by the hinge A on the robotic arm to keep the arm in the horizontal position?

FIGURE P14.32

To determine

Tension in the support rope.

Answer to Problem 32PQ

The tension in the support rope is $\underset{\_}{1.20\times {10}^4\text{N}}$.

Explanation of Solution

The figure given below is the free body diagram of the robotic arm.

In the equilibrium state of the system, the net torque experienced by the robotic arm is zero. Take shoulder joint at point A as the pivot point. The forces acting on the body are the force of tension on the rope and the force at robotic arm due to gravitational field. Torque acts on the arm due to these forces.

Write the equation to find the torque produced by the tension on the rope.

${\tau }_r=r{F}_T\sin \theta$ (I)

Here, ${\tau }_r$ is the torque due to tension on rope, $r$ is the perpendicular distance between the point of application of force and the point of rotation, $F$ is the force of tension, and $\theta$ is the angle made by the rope with the horizontal.

Write the equation to find the torque acting on the arm due to earth’s gravity.

${\tau }_g=r_g{F}_g$ (II)

Here, ${\tau }_g$ is the torque acting on the arm due to force of gravity, $r_g$ is the perpendicular distance between the point of application of force and the axis of rotation, and $F_g$ is the gravitational force.

Write the equation to find the gravitational force.

  $F_g=mg$

Substitute the above equation in (II) to get ${\tau }_g$.

${\tau }_g=-r_{g}mg$ (III)

The force of gravity is in down ward direction, therefore the torque due to gravitational force has a negative sign.

Add equations (III) and (I) to find the sum of torques at the arm.

${\displaystyle \sum \tau }=\left(F_T\sin \theta \right)r-mg{r}_g$ (IV)

Since the sum of torque is zero in equilibrium condition, equate equation (IV) to zero and solve for $F_T$.

$\begin{array}{c}\left(F_T\sin \theta \right)r-mg{r}_g=0\\ \left(F_T\sin \theta \right)r=mg{r}_g\\ F_T=\frac{mg{r}_g}{r\sin \theta }\end{array}$ (V)

Conclusion:

Substitute $15.0°$ for $\theta$, $1.50\text{m}$ for $r$, $215\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $2.20\text{m}$ for $r_g$ in equation (V) to get $F_T$.

  $\begin{array}{c}{F}_T=\frac{\left(215\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2.20\text{m}\right)}{\left(1.50\text{m}\right)\sin 15.0°}\\ =1.20\times {10}^4\text{N}\end{array}$

Therefore, the tension in the support rope is $\underset{\_}{1.20\times {10}^4\text{N}}$.

To determine

The magnitude and direction of force exerted by hinge $A$ on arm of the robot.

Answer to Problem 32PQ

The magnitude of the force exerted by hinge $A$ on arm is $\underset{\_}{1.16\times {10}^4\text{N}}$ and the force makes an angle of $\underset{\_}{4.89°}$ with the horizontal.

Explanation of Solution

At equilibrium the sum of both x and y components of all the force is equal to zero.

The forces in the y direction are the force exerted by the hinge, force due to the tension on rope, and the force due to gravity.

Write the equation to find the y component of force due tension on the rope.

$F_{Ty}=F_T\sin \theta$ (VI)

Here, $F_{Ty}$ is the force of tension on the rope.

Write the equation to find the force due to gravity.

$F_g=-mg$ (VII)

Add equation (VI) and (VII) to get the sum of forces.

${\displaystyle \sum F_y}=F_T\sin \theta -mg$

In addition to these two forces there is the force exerted by the hinge on the arm of the robot. Let this force be $F_{hy}$ acting in the negative direction.

Add $F_{hy}$ to the above equation to find ${\displaystyle \sum F_y}$.

${\displaystyle \sum F_y=F_T\sin \theta -mg-F_{hy}}$ (VIII)

Here, $F_{hy}$ is the force exerted by the hinge on the robotic arm.

At equilibrium, the sum of y component of forces is equal to zero. Equate equation (VIII) to zero and solve for $F_{hy}$.

$\begin{array}{c}\\ F_T\sin \theta -mg-F_{hy}=0\\ F_{hy}=F_T\sin \theta -mg\end{array}$ (IX)

Substitute equation (V) in (IX) and solve for $F_{hy}$.

$\begin{array}{c}{F}_{hy}=F_T\sin \theta -mg\\ =\frac{mg{r}_g}{r\sin \theta }\sin \theta -mg\end{array}$ (X)

The forces acting in the x direction are the force due to tension on rope and the horizontal component of force exerted by the hinge.

Write the equation to find the x component of the tension force on the rope.

$F_{Tx}=F_T\cos \theta$ (XI)

Here, $F_{Tx}$ is the force of tension in the x direction.

Substitute equation (V) in (XI).

$F_{Tx}=\frac{mg{r}_g}{r\sin \theta }\cos \theta$ (XII)

Write the equation to find the sum of forces in the x direction.

${\displaystyle \sum F_x}=F_{hx}+F_{Tx}$

Here, $F_{hx}$ is the horizontal component of force exerted by the hinge.

Substitute equation (XII) to the above equation to get ${\displaystyle \sum F_x}$.

${\displaystyle \sum F_x}=F_{hx}+\frac{mg{r}_g}{r\sin \theta }\cos \theta$ (XIII)

In equilibrium, sum of forces in the x direction is zero. Equate equation (XIII) to zero.

$F_{hx}+\frac{mg{r}_g}{r\sin \theta }\cos \theta =0$ (XIV)

Solve equation (XIV) to get $F_{hx}$.

$F_{hx}=\frac{mg{r}_g}{r\sin \theta }\cos \theta$ (XV)

Write the equation to find the resultant force exerted by hinge.

$F_h=\sqrt{F_{hx}{}^2+F_{hy}}$ (XVI)

Here, $F_h$ is the resultant force exerted by the hinge.

Write the equation to find the angle made by the resultant force with the horizontal.

$\theta ={\tan }^{-1}\left|\frac{F_{hy}}{F_{hx}}\right|$ (XVII)

Here, $\theta$ is the angle made by the resultant force exerted by the hinge with the horizontal.

Conclusion:

Substitute $215\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.20\text{m}$ for $r_g$, $1.50\text{m}$ for $r$, and $15.0°$ for $\theta$ in equation (X) to get $F_{hy}$.

  $\begin{array}{c}{F}_{hy}=\frac{\left(215\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2.20\text{m}\right)}{\left(1.50\text{m}\right)\left(\sin 15.0°\right)}\sin \left(15.0°\right)-\left(215\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =984\text{N}\end{array}$

Substitute $215\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.20\text{m}$ for $r_g$, $1.50\text{m}$ for $r$, and $15.0°$ for $\theta$ in equation (XV) to get $F_{hx}$.

  $\begin{array}{c}{F}_{hx}= \frac{\left(215\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(2.20\text{m}\right)}{\left(1.50\text{m}\right)\sin 15.0°} \cos 15.0°\\ =1.15\times {10}^4\text{N}\end{array}$

Substitute $984\text{N}$ for $F_{hy}$ and $1.15\times {10}^4\text{N}$ for $F_{hx}$ in equation (XVI) to get $F_h$.

  $\begin{array}{c}{F}_h=\sqrt{{\left(984\text{N}\right)}^2+{\left(1.15\times {10}^4\text{N}\right)}^2}\\ =1.16\times {10}^4\text{N}\end{array}$

Substitute $984\text{N}$ for $F_{hy}$ and $1.15\times {10}^4\text{N}$ for $F_{hx}$ in equation (XVII) to get $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left|\frac{984\text{N}}{1.15\times {10}^4\text{N}}\right|\\ =4.89°\end{array}$

Therefore, the magnitude of the force exerted by hinge on arm is $\underset{\_}{1.16\times {10}^4\text{N}}$ and the force makes an angle of $\underset{\_}{4.89°}$ with the horizontal.