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15. Rotational Equilibrium
Equilibrium in 2D - Ladder Problems
Problem 11.10a
Textbook Question
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weighs 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing 740 N climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum friction force that the ground can exert on the ladder at its lower end?

1
Start by drawing a free-body diagram of the ladder. Identify all the forces acting on the ladder: the weight of the ladder acting downward at its center of gravity, the weight of the man acting downward at his position on the ladder, the normal force exerted by the ground acting upward at the base of the ladder, the frictional force exerted by the ground acting horizontally at the base of the ladder, and the normal force exerted by the wall acting horizontally at the top of the ladder.
Calculate the maximum static friction force using the formula: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><sub>max</sub><mo>=</mo><mi>μ</mi><sub>s</sub><mi>N</mi></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi><sub>s</sub></math> is the coefficient of static friction and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math> is the normal force exerted by the ground.
Determine the normal force exerted by the ground. Since the ladder is in equilibrium, the sum of the vertical forces must be zero. Therefore, the normal force is equal to the sum of the weights of the ladder and the man: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi><mo>=</mo><mn>160</mn><mo>+</mo><mn>740</mn></math>.
Substitute the values into the formula for maximum static friction force: <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><sub>max</sub><mo>=</mo><mn>0.40</mn><mo>×</mo><mn>900</mn></math>. This will give you the maximum friction force that the ground can exert on the ladder at its lower end.
Verify that the ladder remains in equilibrium by checking the sum of the horizontal forces and the sum of the torques about any point. The frictional force must be sufficient to prevent slipping, and the torques must balance to ensure rotational equilibrium.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Free-Body Diagram
A free-body diagram is a graphical representation used to visualize the forces acting on an object. For the ladder problem, it involves identifying forces such as the weight of the ladder, the weight of the man, the normal force from the ground, and the frictional force. This diagram helps in setting up equations to solve for unknowns like the maximum friction force.
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Static Friction
Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It acts parallel to the surfaces and is proportional to the normal force, with the maximum static friction given by the product of the coefficient of static friction and the normal force. In this problem, it determines the maximum force the ground can exert before the ladder starts to slip.
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Torque and Equilibrium
Torque is the rotational equivalent of force, calculated as the product of force and the perpendicular distance from the pivot point. For the ladder in equilibrium, the sum of torques about any point must be zero. This concept helps in analyzing the rotational stability of the ladder and determining the conditions under which it remains stationary as the man climbs.
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