Q5 E - University Physics with Modern Physics

Question

Raising a Ladder. A ladder carried by a fire truck is$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}$ long. The ladder weighs $\mathbf{3400}\mathbf{ }\mathit{N}$and its center of gravity is at its center. The ladder is pivoted at one end (A) about a pin (Fig. E11.5); ignore the friction torque at the pin. The ladder is raised into position by a force applied by a hydraulic piston at C. Point C is $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}$from A, and the force $\stackrel{\mathbf{u}}{\mathbf{F}}$exerted by the piston makes an angle of $40°$with the ladder. What magnitude must $\stackrel{\mathbf{u}}{\mathbf{F}}$have to just lift the ladder off the support bracket at B? Start with a free-body diagram of the ladder.

Step-by-Step Solution

Answer

A force of magnitude 6610 N is required to lift the ladder.

Step 1: Equilibrium

The condition for translational equilibrium is:$\underset{}{\sum F_{ext }= 0}$.

And that for rotational equilibrium is: $\underset{}{\sum {\mathrm{\pi }}_{ext}=0}$.

Step 2: Find the Force

Given,

The pivoted ladder has its center of gravity at its center, and the piston exerts the force $\stackrel{\mathit{u}}{F}$which is $40°$apart from the ladder.

Let the whole set up illustrated as a free body diagram for force $\stackrel{\mathit{u}}{F}$shown in the figure as:

Considering anticlockwise rotation as positive and applying the condition for rotational equilibrium at the A, we have:

$$\begin{gathered} \underset{}{ \sum {\mathrm{\pi }}_{ext}=0} \\ \mathrm{F} \sin 40°.\left(8.00\right)-3400\times \left(10.00\right)=0 \\ \mathrm{F} \sin 40°\times \left(8.00\right)=3400\times \left(10.00\right) \\ \mathrm{F} = 6610 \mathrm{N} \end{gathered}$$

Thus, a force of magnitude 6610 N is required to lift the ladder.