Problem 72 - University Physics with Modern Physics

Question

You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is 6.00 m long and inclined at 37.0\(^\circ\) above the horizontal. Each cylinder is fitted with a light, frictionless yoke through its center, and a light (but strong) rope is attached to the yoke. Each cylinder is uniform and has mass 460 kg and radius 0.300 m. The cylinders are pulled up the ramp by applying a constant force \(\overrightarrow{F}\) to the free end of the rope. \(\overrightarrow{F}\) is parallel to the surface of the ramp and exerts no torque on the cylinder. The coefficient of static friction between the ramp surface and the cylinder is 0.120. (a) What is the largest magnitude \(\overrightarrow{F}\) can have so that the cylinder still rolls without slipping as it moves up the ramp? (b) If the cylinder starts from rest at the bottom of the ramp and rolls without slipping as it moves up the ramp, what is the shortest time it can take the cylinder to reach the top of the ramp?

Step-by-Step Solution

Answer

(a) Maximum \( \overrightarrow{F} \) is 3171.3 N. (b) Solve \( t = \sqrt{\frac{2d}{a}} \) for time.

Step 1: Understand the Problem and Identify Forces

We need to ensure the cylinder rolls without slipping as it moves up the ramp. The force \( \overrightarrow{F} \), frictional forces, and gravitational forces are involved. The force \( \overrightarrow{F} \) should not exceed the maximum frictional force that prevents slipping.

Step 2: Calculate Maximum Static Frictional Force

The maximum static frictional force is given by \( f_s = \mu_s N \), where \( \mu_s = 0.120 \) and \( N \) is the normal force. The normal force can be calculated by \( N = mg \cos\theta \), where \( m = 460\,\text{kg} \), \( g = 9.81\,\text{m/s}^2 \), and \( \theta = 37.0^\circ \). Thus, \( N = 460 \times 9.81 \times \cos(37.0^\circ) \).

Step 3: Solve for Normal Force

\( N = 460 \times 9.81 \times \cos(37.0^\circ) = 460 \times 9.81 \times 0.7986 = 3598.3 \text{N} \).

Step 4: Solve for Maximum Frictional Force

\( f_s = 0.120 \times 3598.3 = 431.8 \text{N} \). This is the maximum static friction force available to prevent slipping.

Step 5: Set Up the Force Balance Equation Along the Ramp

Apply the force balance equation along the ramp, factoring the gravitational component along the ramp: \( F - mg\sin\theta = f_s \). Substitute \( g \) and \( \theta \) for the inclined force values.

Step 6: Rearrange and Solve for \( F \)

Solve for \( F \): \( F = mg\sin\theta + f_s = 460 \times 9.81 \times \sin(37.0^\circ) + 431.8 = 2739.5 \text{N} + 431.8 \text{N} = 3171.3 \text{N} \).

Step 7: Calculate Minimum Time to Reach the Top of the Ramp

Use the energy conservation approach for the rolling object. The gravitational potential energy at the top is \( mgh \), where \( h = 6\sin(37^\circ) = 3.6 \) meters. The kinetic energy at the bottom is zero since it starts from rest.

Step 8: Solve Energy Equation for Acceleration

Apply the work-energy principle. The net work is \( W = Fd - mgd\sin\theta \). Equate work done by force \( F \) and change in kinetic energy to find acceleration.

Step 9: Determine Acceleration from Energy Relations

Using \( F = ma \), solve for acceleration: \( a = \frac{F}{m} \), with \( F \) being the net force excluding friction.

Step 10: Determine Time Using Kinematics

Use the kinematic equation \( d = V_0 t + \frac{1}{2} a t^2 \), where \( V_0 = 0 \), \( d = 6 \) meters, and \( a \) from previous calculations, solve for \( t \): \( t = \sqrt{\frac{2d}{a}} \).

Step 11: Solve the Equation and Provide the Result

Substitute the known parameters into \( t = \sqrt{\frac{2 \times 6}{a}} \) after solving for \( a \) to find \( t \).

Key Concepts

Static Friction in Rotational Motion

Static friction is a crucial concept in the realm of rotational motion, especially when dealing with objects like cylinders rolling on surfaces. In this problem, static friction is what keeps the cylinder from slipping as it moves up the inclined ramp.

Static friction is described by the equation \( f_s = \mu_s N \), where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force. For the cylinder to roll without slipping, the static frictional force must be strong enough to oppose any external forces trying to cause slipping.

• The maximum static frictional force is what determines the largest force \( \overrightarrow{F} \) that can be applied without causing slipping.
• It is crucial that \( \overrightarrow{F} \) does not exceed this value to maintain rolling motion without slip.
• The role of static friction here is not to cause the cylinder to move, but to ensure that it rolls smoothly.

Inclined Plane Dynamics

An inclined plane refers to a flat supporting surface tilted at an angle, used in this scenario to lift a cylinder up to a loading dock. The physics of inclined planes helps us understand how forces act on objects when the surface is not horizontal.

When analyzing the movement on an inclined plane, it is important to break down the gravitational force into components. One component acts perpendicular to the plane, while the other acts parallel.

• The normal force (\( N \)) keeps the cylinder from falling through the plane and is calculated by \( N = mg \cos\theta \).
• The gravitational force parallel to the ramp is \( mg \sin\theta \), pulling the cylinder back down.
• Inclination angle \( \theta \) affects these components, determining how steep the path is for rolling.

The balance of these forces contributes significantly to how the cylinder moves, influencing the frictional force needed to prevent slipping.

Energy Conservation for Moving Objects

Energy conservation is a fundamental principle that helps us calculate the movement and time taken by the cylinder on the inclined plane. When the cylinder moves up, its mechanical energy is conserved if there's no energy loss.

Initially, when the cylinder is at rest, all energy is stored as potential energy. As it moves up:

• Potential energy increases with elevation, calculated using \( mgh \), where \( h \) is the height gained.
• Kinetic energy is zero at the start since the initial velocity \( V_0 = 0 \).
• As the force \( \overrightarrow{F} \) acts on the cylinder, it does work, translating into kinetic and potential energy.
• By applying the work-energy principle, the energy change relates to the distance displaced by force \( \overrightarrow{F} \).

Exact energy calculations allow determination of the acceleration, which is critical to finding the shortest time required for the cylinder to reach the top of the ramp.

Force Analysis in Mechanics

Force analysis is essential to predict the motion of objects under various force influences. In this exercise, understanding the forces acting on the cylinder on the inclined plane assists in deriving the correct force balance equations.

Key elements in force analysis include:

• Identifying all forces: gravitational force, normal force, static friction, and the applied force \( \overrightarrow{F} \).
• Breaking forces into components: notably, how gravity's effect splits between pulling down the slope and pressing into the slope.
• Formulating equations: using \( F - mg\sin\theta = f_s \) to maintain equilibrium and prevent slipping.
• Solving for the unknowns: rearranging equations to find \( \overrightarrow{F} \), ensuring all forces are correctly accounted.

This systematic approach allows precise computation of necessary forces to keep the cylinder rolling smoothly up the ramp.