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Chapter 1: Q5E (page 357)

Raising a Ladder. A ladder carried by a fire truck is 20.0mlong. The ladder weighs 3400Nand 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 8.0mfrom A, and the force F→exerted by the piston makes an angle of 40°with the ladder. What magnitude F→must have to just lift the ladder off the support bracket at B? Start with a free-body diagram of the ladder.

Short Answer

Expert verified

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

Step by step solution

01

Step-by-Step Solution Step 1: Equilibrium

The condition for translational equilibrium is: ∑Fext=0.

And that for rotational equilibrium is:∑τext=0

02

Find the Force

The pivoted ladder has its center of gravity at its center, and the piston exerts the force F→which is 40∘apart from the ladder.

Let the whole set up illustrated as a free body diagram for force F→shown in the figure as:

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

∑τext=0Fsin40°·8.00-3400×10.00=0Fsin40°×8.00=3400×10.00F=6610N

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

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