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Chapter 11: Problem 10

A uniform ladder 5.0 \(\mathrm{m}\) long rests against a frictionless, vertical wall with its lower end 3.0 \(\mathrm{m}\) from the wall. The ladder weighs 160 \(\mathrm{N}\) . The coefficient of static friction between the foot of the ladder and the ground is \(0.40 .\) A man weighing 740 \(\mathrm{N}\) climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum frictional force that the ground can exert on the ladder at its lower end? (b) What is the actual frictional force when the man has climbed 1.0 \(\mathrm{m}\) along the ladder? (c) How far along the ladder can the man climb before the ladder starts to slip?

Short Answer

Expert verified
(a) 360 N (b) 80 N (c) 4.05 m

Step by step solution

01

Understanding the Problem

A uniform ladder leans against a frictionless wall. The length of the ladder is 5.0 m, and it is placed 3.0 m away from the wall, making a right triangle with the ground and the wall. It has a weight of 160 N. The coefficient of static friction between the ladder's foot and the ground is 0.40. A man weighing 740 N climbs up the ladder. We need to determine the maximum frictional force, the actual frictional force when the man is 1.0 m up, and how far the man can climb before slipping occurs.
02

Calculate Maximum Frictional Force

The maximum frictional force that the ground can exert at the ladder's lower end is given by the equation: \( f_{max} = \mu_s N \). Since the ladder is in equilibrium, the normal force \( N \) is equal to the total weight (man + ladder), which is 900 N (740 N + 160 N). The static friction coefficient \( \mu_s \) is 0.40. Therefore, \[ f_{max} = 0.40 \times 900 = 360 \, \text{N}. \]
03

Calculating the Actual Frictional Force

To calculate the actual frictional force when the man climbs 1.0 m, first determine the torques about the base of the ladder. Let's consider torques and assume counterclockwise torque as positive. The torques are calculated as follows:- Torque due to the ladder's weight \( = 160 \times \left( \frac{3.0}{2} \right); \) this is the perpendicular distance from the base to the ladder's center of mass.- Torque due to the man \( = 740 \times 1.0 \times \left( \frac{4.0}{5.0} \right); \) as the perpendicular portion of the ladder that the man has climbed.Set up the rotational equilibrium condition about the base for the torques:\[ \text{Net Torque} = 0 = 160 \times 1.5 - 740 \times 0.8 + f \times 5.0 \]Solving for \( f \):\[ f = \frac{160 \times 1.5 - 740 \times 0.8}{5.0} = 80 \, \text{N}. \]
04

Determine Maximum Climbing Distance

For the ladder not to slip, frictional force must not exceed 360 N. Solving for the maximum distance the man can climb:Using torques, the distance \( x \) he can climb satisfies:\[ 160 \times 1.5 = 740 \times x \times \left( \frac{4.0}{5.0} \right) + 360 \times 5.0 \]Rearranging this equation:\[ 240 = 592x + 1800 \]\[ 592x = 2400 \ x \approx 4.05 \, \text{m}. \]
05

Conclusion Based on Calculations

The maximum frictional force that the ground can exert is 360 N. When the man is 1.0 m up the ladder, the actual frictional force is 80 N. The man can climb up to approximately 4.05 m before the ladder starts to slip.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free-body Diagrams
Free-body diagrams (FBDs) are essential tools in physics to simplify and solve problems involving forces. By visualizing all the forces acting on a single object, we can better understand how these forces affect the object's motion or stability.

When drawing a free-body diagram for our ladder problem, consider the following components:
  • Gravity: The ladder and the man are both subject to gravitational force, pulling them downwards toward the ground.
  • Normal Force: The ground pushes upward on the ladder's base, counteracting gravity.
  • Frictional Force: Since the wall is frictionless, the friction only acts at the base of the ladder, resisting any sliding.
  • Reaction at the Wall: The wall provides a horizontal reaction force acting perpendicularly to its surface, pushing the ladder outwardly away.
Visualizing these forces helps determine their relative magnitudes and angles. By arranging them graphically on paper, we gain a clearer perspective of how they contribute to the ladder's equilibrium, guiding us to the right calculations and solutions.
Equilibrium of Forces
Understanding the equilibrium of forces is crucial to solving problems like the ladder scenario. When an object, such as a ladder, is in equilibrium, all net forces and net torques acting on it must be zero. This ensures the ladder is neither moving nor rotating.

Let's break it down:
  • Translational Equilibrium: The sum of all horizontal and vertical forces must be zero. Since the ladder is not sliding down the wall or lifting off the ground, the upward normal force balances the downward gravitational force of the ladder and the man. Similarly, friction balances the horizontal force exerted by the wall.
  • Rotational Equilibrium: The sum of all torques about any pivot point must be zero. By choosing the ladder's base as our pivot point, we apply this principle to tackle scenarios like the maximum distance the man can climb. The clockwise and counterclockwise torques generated by the ladder's and man's weights must be countered by the torque due to static friction.
Ensuring both conditions of equilibrium are satisfied allows the ladder to remain stable and secure. Thus, analyzing these forces and torques enables us to predict the maximum safe point the man can reach before the ladder begins to slip.
Torque Calculations
Torque is the measure of the rotational force applied around a pivot point. It's crucial in understanding how ladders and similar structures maintain balance and avoid rotation. To gain insights into our ladder problem, we perform torque calculations about its base, acknowledging how different forces contribute to rotation:

  • Torque due to Ladder's Weight: The gravitational force acts at the ladder's midpoint, creating a counterclockwise torque.
  • Torque due to Man's Weight: The man's position on the ladder contributes another counterclockwise torque, influenced by his distance from the ladder's base and the ladder's angle.
  • Frictional Torque: Friction at the ladder's base creates a clockwise torque to balance the system, preventing rotation.
From our calculations, we determine the effective torques, ensuring they net to zero for equilibrium. This helps us find out how far the man can climb while maintaining balance, as exceeding maximum torque thresholds would induce slipping.

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Most popular questions from this chapter

A uniform, horizontal flagpole 5.00 \(\mathrm{m}\) long with a weight of 200 \(\mathrm{N}\) is hinged to a vertical wall at one end. \(\mathrm{A} 600-\mathrm{N}\) stuntwoman hangs from its other end. The flagpole is supported by a guy wire running from its outer end to a point on the wall directly above the pole. (a) If the tension in this wire is not to exceed \(1000 \mathrm{N},\) what is the minimum height above the pole at which it may be fastened to the wall? (b) If the flagpole remains horizontal, by how many newtons would the tension be increased if the wire were fastened 0.50 m below this point?

A Truck on a Drawbridge. A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity three-quarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge (Fig. P11.56). The drawbridge is 40.0 \(\mathrm{m}\) long and has a mass of \(18,000 \mathrm{kg} ;\) its center of gravity is at its midpoint. The cement mixer, with driver, has mass \(30,000\) kg. When the drawbridge has been raised to an angle of \(30^{\circ}\) above the horizontal, the cable makes an angle of \(70^{\circ}\) with the surface of the bridge. (a) What is the tension \(T\) in the cable when the drawbridge is held in this position? (b) What are the horizontal and vertical components of the force the hinge exerts on the span?

BIO Leg Raises. In a simplified version of the musculature action in leg raises, the abdominal muscles pull on the femur (thigh bone) to raise the leg by pivoting it about one end (Fig. P11.57. When you are lying. horizontally, these muscles make an angle of approximately \(5^{\circ}\) with the femur, and if you raise your legs, the muscles remain approximately horizontal, so the angle \(\theta\) increases. We shall assume for simplicity that these muscles attach to the femur in only one place, 10 \(\mathrm{cm}\) from the hip joint (although, in reality, the situation is more complicated). For a certain \(80-\mathrm{kg}\) person having a leg 90 \(\mathrm{cm}\) long, the mass of the leg is 15 \(\mathrm{kg}\) and its center of mass is 44 \(\mathrm{cm}\) from his hip joint as measured along the leg. If the person raises his leg to \(60^{\circ}\) above the horizontal, the angle between the abdominal muscles and his femur would also be about \(60^{\circ} .\) (a) With his leg raised to \(60^{\circ},\) find the tension in the abdominal muscle on each leg. As usual, begin your solution with a free-body diagram. (b) When is the tension in this muscle greater: when the leg is raised to \(60^{\circ}\) or when the person just starts to raise it off the ground? Why? (Try this yourself to check your answer.) (c) If the abdominal muscles attached to the femur were perfectly horizontal when a person was lying down, could the person raise his leg? Why or why not?

Cp A steel cable with cross-sectional area 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum upward acceleration that can be given a 1200 -kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

BIO Forearm. In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 \(\mathrm{cm}\) from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 \(\mathrm{N}\) and that their center of gravity is 15.0 \(\mathrm{cm}\) from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds a 80.0 -N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 \(\mathrm{cm}\) from the elbow. Construct a free-body diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part (b), find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the \(80.0-\mathrm{N}\) weight, the person raises his forearm until it is at an angle of \(53.0^{\circ}\) above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearm is in this position? Has the force increased or decreased from its value in part (b)? Explain why this is so, and test your answer by actually doing this with your own arm.
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