91Ó°ÊÓ

A uniform ladder of mass \(m\) and length \(\ell\) leans against a smooth wall, making an angle \(\theta\) with respect to the ground. The dirt exerts a normal force and a frictional force on the ladder, producing a force vector with magnitude \(F_{1}\) at an angle \(\phi\) with respect to the ground. Since the wall is smooth, it exerts only a normal force on the ladder; let its magnitude be \(F_{2}\). (a) Explain why \(\phi\) must be greater than \(\theta\). No math is needed. (b) Choose any numerical values you like for \(m\) and \(\ell\), and show that the ladder can be in equilibrium (zero torque and zero total force vector) for \(\theta=45.00^{\circ}\) and \(\phi=63.43^{\circ}\).

Step by step solution

01

Conceptual Understanding of Forces

For part (a), understand the forces at play. The frictional force from the ground prevents the ladder from sliding, and it acts tangentially along the ground. Since the wall is smooth, it only contributes a normal force perpendicular to the wall. The resultant force from the dirt forms an angle \( \phi \) which must be greater than \( \theta \) because for equilibrium, the friction must overcome the smooth wall's lack of resistance, effectively creating a larger angle.

02

Choosing Numerical Values

For part (b), choose convenient values for the mass and length of the ladder: let \( m = 10 \, \text{kg} \) and \( \ell = 5 \, \text{m} \). We will use these to check equilibrium conditions.

03

Conditions for Equilibrium

The ladder is in equilibrium if the sum of all torques and forces equals zero. For torques, consider the ladder's weight acts at its center, \( \frac{\ell}{2} \), the friction from the ground, and the force from the wall.

04

Calculating Forces in Equilibrium

Given \( \theta = 45^{\circ} \) and \( \phi = 63.43^{\circ} \), resolve forces into horizontal and vertical components. For smooth wall conditions, the horizontal component \( F_2 = F_{\text{friction}} \). In vertical, the normal force balances gravity, implying \( F_{1y} = mg \). Common trigonometric identities relate these angles with their respective components.

05

Satisfying Torque Equilibrium

Introduce torque expressions: the torque about the pivot point at ground from weight equals the counterbalancing torques from \( F_2 \) and \( F_1 \). Use \( \tan(\theta) \) and \( \tan(\phi) \) to ensure calculated forces provide cancellation of all torque, showing validity of chosen title of angles.

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Key Concepts

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

Equilibrium

In physics, equilibrium refers to a state where an object remains at rest or moves with constant velocity because the net force and net torque acting on it are zero. In the context of the ladder leaning against a wall, achieving equilibrium means balancing all forces and torques such that the ladder does not tip over or slide down. When a ladder is placed against a smooth wall, care must be taken to consider all acting forces to maintain equilibrium.

For our ladder, two equilibriums must be satisfied:

• Translational Equilibrium: This is when the sum of all horizontal and vertical forces equals zero. It ensures the ladder doesn’t slide sideways or downwards. The frictional force at the ground and the normal forces from both the ground and wall work together to provide this stability.
• Rotational Equilibrium: This is when the sum of all torques (rotational forces) about any pivot point is zero. This type ensures that the ladder doesn’t rotate or fall over. For our exercise, choosing the pivot point at the base of the ladder, the torque due to the ladder’s weight, and the torques caused by friction and the normal force from the wall must all cancel out.

Understanding these equilibrium concepts is crucial in analyzing ladder problems and similar real-world scenarios.

Torque

Torque is a measure of the turning force on an object. It depends on the force applied and the distance from the pivot point (also known as the moment arm). The formula to calculate torque is \[ \tau = F \times r \times \sin(\theta) \] where \(F\) is the force applied, \(r\) is the distance from the pivot, and \(\theta\) is the angle between the force direction and the arm.

In our ladder problem, torques are essential to ensure rotational equilibrium. Here are the key torque-related considerations:

• The weight of the ladder acts at its center of gravity, creating a torque about the base of the ladder. This means the gravitational force is distributed evenly, assumed to act at ladder's midpoint.
• The frictional force at the base creates another torque, opposing that of the ladder’s weight. This force acts along the ground and is crucial for keeping the ladder in place.
• Lastly, the normal force exerted by the wall contributes to torque. Because the wall is smooth, this force acts perpendicular to the wall, adding to the rotational balance.

For equilibrium, these torques must balance, meaning the clockwise and counterclockwise torques need to be equal.

Forces

Forces are vectors that cause objects to accelerate; they have both magnitude and direction. In the exercise, a ladder leaning against a wall is acted upon by multiple forces:

• Gravitational Force: This acts downward through the ladder’s center, represented by the ladder's weight \( (mg) \).
• Normal Forces: The ground and wall both exert normal forces. The ground's normal force acts upwards balancing gravity, while the wall’s normal force acts horizontally due to its smoothness, contributing to the horizontal balance.
• Frictional Force: At the ground point, friction prevents sliding. For the ladder to be stationary, friction must be sufficient to counteract horizontal forces from the wall.

All these forces must be in equilibrium for the ladder to remain stable. The normal force and friction together create a force vector at the base, forming an angle \( \, \phi \, \) with the ground, which varies as friction composes some of the horizontal balance.

Trigonometry

Trigonometry deals with the relationships between the angles and sides of triangles, which play a critical role in solving physics problems involving inclined planes or ladders. The problem's ladder scenario requires trigonometric understanding to resolve force vectors into components.

Key trigonometric concepts include:

• Sine, Cosine, and Tangent Functions: When dealing with angles \( \theta \) and \( \phi \), break down forces into their components using \[ F_{x} = F \cos(\theta) \text{ and } F_{y} = F \sin(\theta) \] and likewise for angle \(\phi\).
• Calculating Angles: To ensure equilibrium and balance torques, use tangent functions: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \] This helps relate angles with force magnitudes.

Trigonometry is crucial in aligning these forces correctly, ensuring both translational and rotational balance. Understanding how to use these principles allows learners to calculate the needed quantities and verify balance with given angles thoughtfully.