Problem 71 - College Physics

Question

You are trying to raise a bicycle wheel of mass \(m\) and radius \(R\) up over a curb of height \(h .\) To do this, you apply a horizontal force \(\vec{F}\) (Fig. \(10.81 ) .\) What is the smallest magnitude of the force \(\vec{F}\) that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel, and (b) at the top of the wheel? (c) In which case is less force required?

Step-by-Step Solution

Answer

(b) requires less force if \( R > 2h \).

Step 1: Analyze the geometry of the problem

When lifting the wheel over the curb, the pivot point is the upper edge of the curb. The wheel should rotate about this point when the force is applied.

Step 2: Establish conditions for equilibrium

For the wheel to begin moving over the curb, the torque due to the applied force needs to balance out the torque due to the weight of the wheel. Apply the condition for rotational equilibrium: \( F \times R_f = mg \times R_w \), where \( R_f \) and \( R_w \) are perpendicular distances from the pivot to the line of force and weight, respectively.

Step 3: Calculate the torque when force is applied at the center of the wheel

Here, \( R_f = \sqrt{R^2 - h^2} \) and \( R_w = h \). So, the equilibrium condition is \( F \times \sqrt{R^2 - h^2} = mg \times h \). Solving for \( F \), we get \( F = \frac{mg \cdot h}{\sqrt{R^2 - h^2}} \).

Step 4: Calculate the torque when force is applied at the top of the wheel

For this scenario, \( R_f = 2R - h \) and \( R_w = h \). This gives us the condition \( F \times (2R - h) = mg \times h \). Solving for \( F \), we have \( F = \frac{mg \cdot h}{2R - h} \).

Step 5: Compare forces from different scenarios

We need to determine when \( F = \frac{mg \cdot h}{\sqrt{R^2 - h^2}} \) and when \( F = \frac{mg \cdot h}{2R - h} \). Generally, \( F \) needed when force is applied at the top (part b) is less if \( R > 2h \).

Key Concepts

Torque

In physics, torque is a measure of how much a force acting on an object causes that object to rotate. Imagine trying to open a door; it's easier to push from the edge than near the hinge. That's because the torque is greater when the force is applied further from the pivot point. The formula for torque (τ) is:\[ \tau = F \times r \times \sin(\theta) \]where:

• \(F\) is the force applied,
• \(r\) is the distance from the pivot point,
• \(\theta\) is the angle between the force and the lever arm.

Torque plays a crucial role in this bicycle wheel exercise. To understand the problem, the torque created by the applied force and the gravitational force must be compared. At equilibrium, these torques must be balanced for the wheel to overcome the curb.

Rotational Equilibrium

An object is in rotational equilibrium when the sum of all torques acting on it is zero. This means it's not rotating, or it rotates at a constant angular velocity. For lifting the wheel over the curb, we apply this concept by setting the torque created by the gravitational force equal to the torque of the applied force. This is essential to predict how a force can rotate the wheel up over the curb. The condition for rotational equilibrium in this exercise can be formulated as:\[ F \times R_f = mg \times R_w \]where:

• \(F\) is the force applied,
• \(R_f\) is the distance from the pivot to where the force is applied,
• \(mg\) is the weight of the wheel,
• \(R_w\) is the vertical distance from the pivot point to the wheel's center of mass.

Understanding this balance helps determine the minimum force needed to lift the wheel over the curb.

Forces and Motion

In this context, forces refer to either the applied force trying to lift the wheel or the gravitational force pulling the wheel down. Motion occurs when forces cause a change in the wheel's position or rotation. In mechanics, it's crucial to analyze how forces interact with objects, especially when dealing with rotational situations. The elevating wheel over the curb requires us to balance these forces carefully.- **Applied Force (\(F\))**: Attempts to create upward motion and overcome gravity.- **Gravitational Force (\(mg\))**: Acts downward and resists the motion.By analyzing the forces and resultant motion, one can determine the necessary conditions to begin lifting the wheel over the curb — ensuring that the applied force results in sufficient torque.

Mechanics

Mechanics is a branch of physics focusing on the behavior of physical bodies when subjected to forces or displacements. It includes the study of objects at rest (statics) or in motion (dynamics). The exercise of lifting a bicycle wheel over a curb encompasses these fundamental principles: - **Statics**: Involves analyzing forces without causing motion. The setup at the curb requires static analysis to ensure forces and torques are balanced initially. - **Dynamics**: Once forces create sufficient torque, dynamics explains the subsequent motion over the curb. Understanding these principles helps solve complex problems like raising the bicycle wheel, offering a deeper appreciation for how forces influence movement and stability.