Chapter 12, Problem 18P

A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole as shown in Figure P12.10. A cable at an angle of θ = 30.0° with the beam helps support the light. (a) Draw a force diagram for the beam. By computing torques about an axis at the hinge at the left-hand end of the beam, find (b) the tension in the cable, (c) the horizontal component of the force exerted by the pole on the beam, and (d) the vertical component of this force. Now solve the same problem from the force diagram from part (a) by computing torques around the junction between the cable and the beam at the right-hand end of the beam. Find (e) the vertical component of the force exerted by the pole on the beam, (f) the tension in the cable, and (g) the horizontal component of the force exerted by the pole on the beam. (h) Compare the solution to parts (b) through (d) with the solution to parts (c) through (g). Is either solution more accurate?

Figure P12.10

To determine

To draw: A force diagram for the beam.

Answer to Problem 18P

A force diagram for the beam is,

Explanation of Solution

Force diagram contains all the forces acting on the body. It contains the direction of each force acting on the body represent by vectors.

The hinged support have two reactions, one is vertical reaction and another is horizontal reaction. The tension force is acting at an angle $\theta$ as shown in the force diagram. The floodlight in the park is supported at the end of a horizontal beam of the negligible mass is hinged at a pole. So, the weight of the floodlight is acting downward as shown in force diagram.

A force diagram for the beam is,

To determine

The tension in the cable.

Answer to Problem 18P

The tension in the cable is $392\text{N}$.

Explanation of Solution

Given Information: The mass of a floodlight supported at the end of a horizontal beam is $20.0\text{kg}$. The angle of cable with beam is $30°$

Write the expression for the weight of the floodlight is acting downward.

$T_2=mg$

Here,

$T_2$ is the weight of the floodlight is acting downward.

$m$ is the mass of floodlight.

$g$ is the acceleration due to gravity.

Write the expression for the moment at the hinged point is,

${\displaystyle \sum T}=0$

Calculate moment at hinge point.

${\displaystyle \sum T}=mgL-T_1\sin \theta L$ (1)

Here,

$T_1$ is the tension in the cable.

Substitute $0$ for ${\displaystyle \sum T}$ in equation (1) to find the tension,

$\begin{array}{c}0=mgL-T_1\sin \theta L\\ T_1\sin \theta L=mgL\\ T_1\sin \theta =mg\\ T_1=\frac{mg}{\sin \theta }\end{array}$

Substitute $20\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $30°$ for $\theta$ in the above equation to find the tension,

$\begin{array}{r}{T}_1=\frac{20\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}}{\sin 30°}\\ =\frac{196\text{kgm}/{\text{s}}^{\text{2}}}{0.5}\\ =392\text{N}\end{array}$

Conclusion:

Therefore, the tension in the cable is $392\text{N}$.

To determine

The horizontal component of the force exerted by the pole on the beam.

Answer to Problem 18P

The horizontal component of the force exerted by the pole on the beam is $339.48\text{N}$.

Explanation of Solution

Given Information: The tension in the cable is $392\text{N}$. The angle of cable with beam is $30°$.

Write the expression for the horizontal component of the force exerted by the pole on the beam.

$F_x=T_1\cos \theta$

Here,

$F_x$ is the horizontal component.

$T$ is the tension.

Substitute $392\text{N}$ for $T_1$ and $30°$ for $\theta$ in above equation to find $F_x$,

$\begin{array}{c}{F}_x=392\text{N}\times \cos 30°\\ =339.48\text{N}\end{array}$

Conclusion:

Therefore, the horizontal component of the force exerted by the pole on the beam is $339.48\text{N}$.

To determine

The vertical component of the force exerted by the pole on the beam.

Answer to Problem 18P

The vertical component of the force exerted by the pole on the beam is $0\text{N}$.

Explanation of Solution

Given Information: The mass of a floodlight supported at the end of a horizontal beam is $20.0\text{kg}$, the tension in the cable is $392\text{N}$, acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$ and the angle of cable with beam is $30°$.

Write the expression for the vertical component of the force exerted by the pole on the beam.

$F_y=mg-T_1\sin \theta$

Here,

$F_y$ is the vertical component.

$T$ is the tension.

$m$ is the mass of floodlight.

$g$ is the acceleration due to gravity.

Substitute $392\text{N}$ for $T_1$, $20\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$, and $30°$ for $\theta$ in above equation to find $F_y$.

$\begin{array}{c}{F}_y=20\text{kg}\times \text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}-392\text{N}\times \text{sin30}°\\ \text{=196}\text{kgm}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kgm}/{\text{s}}^{\text{2}}}\right)-196\text{N}\\ \text{=196}\text{N}-196\text{N}\\ =\text{0}\text{N}\end{array}$

Thus, the vertical component of the force exerted by the pole on the beam is $0\text{N}$.

Conclusion:

Therefore, the vertical component of the force exerted by the pole on the beam is $0\text{N}$.

To determine

The vertical component of the force exerted by the pole on the beam from the force diagram.

Answer to Problem 18P

The vertical component of the force exerted by the pole on the beam from the force diagram is $0\text{N}$.

Explanation of Solution

Given Information: The mass of a floodlight supported at the end of a horizontal beam is $20.0\text{kg}$, the tension in the cable is $392\text{N}$, acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$ and the angle of cable with beam is $30°$.

Write the expression for the torque around the junction between the cable and the beam at the right hand end of the beam.

$\tau =rF$

Here,

$\tau$ is the torque.

$r$ is the distance of the point at which the torque is determined.

$F$ is the force.

Substitute $0\text{m}$ for $r$  in above equation to find $\tau$,

$\begin{array}{c}\tau =0\text{m}\times F\\ =0\end{array}$

Write the expression for the torque at the right hand end of the beam.

$\tau =F_y\text{N}\times l\text{m}$

Here,

$\tau$ is the torque.

$F_y$ is the vertical component.

$l$ is the distance of the beam.

Substitute $0$ for $\tau$ in above equation to find $F_y$,

$\begin{array}{c}0=F_y\text{N}\times l\text{m}\\ F_y=0\text{N}\end{array}$

Conclusion:

Therefore, the vertical component of the force exerted by the pole on the beam from the force diagram is $0\text{N}$.

To determine

The tension in the cable from the force diagram.

Answer to Problem 18P

The tension in the cable from the force diagram is $0\text{N}$.

Explanation of Solution

Given Information: The mass of a floodlight supported at the end of a horizontal beam is $20.0\text{kg}$, the tension in the cable is $392\text{N}$, acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$ and the angle of cable with beam is $30°$.

Write the expression for the torque around the junction between the cable and the beam at the right hand end of the beam.

$\tau =rF$

Here,

$\tau$ is the torque.

$r$ is the distance of the point at which the torque is determined.

$F$ is the force.

Substitute $0\text{m}$ for $r$ in above equation to find $\tau$,

$\begin{array}{c}\tau =0\text{m}\times F\\ =0\text{N}-\text{m}\end{array}$

Since the torque is $0\text{N}-\text{m}$, therefore the tension in the cable is also $0\text{N}$.

Conclusion:

Therefore, the tension in the cable from the force diagram is $0\text{N}$.

To determine

The horizontal component of the force exerted by the pole on the beam from the force diagram.

Answer to Problem 18P

The horizontal component of the force exerted by the pole on the beam from the force diagram is $0\text{N}$.

Explanation of Solution

Given Information: The mass of a floodlight supported at the end of a horizontal beam is $20.0\text{kg}$, the tension in the cable is $392\text{N}$, acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$ and the angle of cable with beam is $30°$.

Write the expression for the torque around the junction between the cable and the beam at the right hand end of the beam.

$\tau =rF$

Here,

$\tau$ is the torque.

$r$ is the distance of the point at which the torque is determined.

$F$ is the force.

Substitute $0\text{m}$ for $r$ in above equation to find $\tau$,

$\begin{array}{c}\tau =0\text{m}\times F\\ =0\end{array}$

Write the expression for the torque at the right hand end of the beam.

$\tau =F_y\text{N}\times l\text{m}$

Here,

$\tau$ is the torque.

$F_y$ is the vertical component.

$l$ is the distance of the beam.

Since the line of action is same for both horizontal component and torque at the right hand end of the beam, the horizontal component of the force exerted by the pole on the beam from the force diagram is $0\text{N}$.

Conclusion:

Therefore, the horizontal component of the force exerted by the pole on the beam from the force diagram is $0\text{N}$.

To determine

The accuracy of solution.

Answer to Problem 18P

The solution is not accurate because the value of tension is different in part (b) and part (f) and the value of horizontal component is also different in part (c) and in part (g) but the value of vertical component is same in part (d) and part (e)

Explanation of Solution

Given Information: The mass of a floodlight supported at the end of a horizontal beam is $20.0\text{kg}$, the tension in the cable is $392\text{N}$, acceleration due to gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$ and the angle of cable with beam is $30°$.

The tension in the cable from part (b) is $392\text{N}$ and tension in the cable from part (f) is $0\text{N}$.

The vertical component of the force exerted by the pole on the beam from part (d) is $0\text{N}$ and the vertical component of the force exerted by the pole on the beam from part (e) is $0\text{N}$.

The horizontal component of the force exerted by the pole on the beam from part (c) is $339.48\text{N}$ and the horizontal component of the force exerted by the pole on the beam from part (g) is $0\text{N}$.

Since the value of tension is different in part (b) and part (f) and the value of horizontal component is also different in part (c) and in part (g) but the value of vertical component is same in part (d) and part (e), therefore the solution is not accurate.

Conclusion:

Thus, the solution is not accurate because the value of tension is different in part (b) and part (f) and the value of horizontal component is also different in part (c) and in part (g) but the value of vertical component is same in part (d) and part (e)