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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? (b) What is the actual friction force when the man has climbed 1.0 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) 326.67 N (c) 1.14 m.

Step by step solution

01

Understanding the Problem

We have a ladder leaning against a frictionless wall and a person climbing it. The objective is to analyze forces acting on the ladder with different climbing positions and determine maximum allowable forces due to static friction before slipping occurs.
02

Free-Body Diagram

Identify all the forces acting on the ladder: the weight of the ladder acting downward at its center (160 N), the weight of the man acting downward at his climbing position, the normal force from the ground upward at the base, a horizontal force exerted by the wall (since it's frictionless), and finally, the frictional force at the base opposing the ladder from slipping. The ladder forms a right triangle with the wall and ground.
03

Calculate Maximum Friction Force

The maximum static friction force is given by the formula \( f_{\text{max}} = \mu_s N \), where \( \mu_s = 0.40 \) is the coefficient of static friction and \( N \) is the normal force (equal to the total weight on the ladder in this case). With the ladder weight 160 N and man weighing 740 N, maximum friction force is \( f_{\text{max}} = 0.40 \times (160 + 740) = 360 \text{ N} \).
04

Calculate Actual Friction Force at 1.0 m Climb

At 1.0 m (from the base), calculate the torque about the base. Torque is contributed by the ladder's own weight, man's weight, and normal force from the wall. Balance these torques assuming no rotational motion: \( 160 \times \frac{3}{2} + 740 \times 1 = F_N \times 3 \). Solve for \( F_N \) (normal force from the wall), then use \( F_{\text{friction}} = F_N \) since ladder is not accelerating horizontally. Here \( 240 + 740 = 3 F_N \) gives \( F_N = 326.67 \text{ N} \), thus \( F_{\text{friction}} = 326.67 \text{ N} \).
05

Maximum Climb before Slip

To determine how far the man can climb, calculate critical climbing distance to reach maximum friction; balance torque as before, now with man's weight positioned at unknown distance \( x \): \[ 160 \times \frac{3}{2} + 740 \times x = 360 \times 3 \]Solving, \[ 240 + 740x = 1080 \]yields \( x = 1.14 \text{ m} \).
06

Conclusion

We used equilibrium of forces and static friction principles to determine max climb before slippage. Calculated maximum friction and applied equilibrium conditions allowed determining man's climb limit.

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Key Concepts

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

Free-Body Diagram
When solving physics problems involving forces, a free-body diagram is an essential tool. It helps visualize the different forces acting on an object. For the ladder resting against a frictionless wall, several forces need to be considered.
  • The ladder's weight acts downward from its center, which is at 2.5 meters along the ladder (since it's uniform).
  • The man’s weight acts downward at the position he climbs to.
  • A normal force from the ground acts upward at the base of the ladder, countering the downward forces.
  • A horizontal force from the wall, which is frictionless, pushes against the ladder. This force keeps the ladder steady against the wall.
  • Finally, static friction at the bottom surface keeps the ladder from slipping. This friction acts in the opposite direction of any tendency to slip.
By labeling these forces on a diagram, you organize the problem visually and can better apply physics principles like Newton's laws, making the later calculations more approachable.
Static Friction
Static friction is the force that prevents surfaces from sliding past each other. In this context, it's the force stopping the ladder from slipping on the ground. The maximum static friction force is crucial because it determines how much force is needed to cause slipping. We can calculate this maximum using the formula:
\[ f_{\text{max}} = \mu_s N \]
* \( \mu_s \) is the coefficient of static friction (in this case 0.40).
* \( N \) represents the normal force, which equals the total downward forces acting at the base. Here, it’s the combined weight of the ladder and the person climbing.
With this information, multiply the total weight by the coefficient of static friction to find the maximum friction the ground can exert. This tells us how close the system is to slipping when the man climbs the ladder.
Stay under this maximum friction to keep the ladder stable.
Torque
Torque involves rotational force, pivotal for understanding how forces cause rotation. With the ladder leaning against the wall, torque calculations help determine stability and predict slipping. Torque is the turning effect around a pivot point, in this case, the ladder's base. The formula for torque (\(\tau\)) is:
\[ \tau = F \times d \]
- \( F \) is the force applied.
- \( d \) is the perpendicular distance from the pivot point to the line of action of the force.
For the ladder, consider:
  • The ladder’s own weight producing torque at its midpoint (2.5 meters entire length but from its base only 1.5 meters horizontally).
  • The man's weight adding torque, calculated at whatever distance he has covered while climbing.
  • The horizontal force at the wall counteracts some of this torque.
Balancing these torques involves setting the clockwise torques equal to counterclockwise torques. This is crucial for equilibrium and ensuring stability. Understanding these torque relationships helps explain what will cause the ladder to start slipping and how the weight's position directly affects stability.

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Most popular questions from this chapter

A steel cable with cross-sectional area 3.00 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.

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

You are a summer intern for an architectural firm. An 8.00-m-long uniform steel rod is to be attached to a wall by a frictionless hinge at one end. The rod is to be held at 22.0\(^\circ\) below the horizontal by a light cable that is attached to the end of the rod opposite the hinge. The cable makes an angle of 30.0\(^\circ\) with the rod and is attached to the wall at a point above the hinge. The cable will break if its tension exceeds 650 N. (a) For what mass of the rod will the cable break? (b) If the rod has a mass that is 10.0 kg less than the value calculated in part (a), what are the magnitude and direction of the force that the hinge exerts on the rod?

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