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? (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

(a) 360 N (b) 326.67 N (c) 1.14 m.

Step-by-step solution

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.

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.

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 \( f_{\text{max}} = 0.40 \times (160 + 740) = 360 \text{ N} \).

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 \( F_{\text{friction}} = 326.67 \text{ N} \).

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 \( x = 1.14 \text{ m} \).

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.

Key concepts

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.