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The 100 -lb platform rolls without slipping along the \(10^{\circ}\) incline on two pairs of 16 -in.-diameter wheels. Each pair of wheels with attached axle weighs 25 lb and has a centroidal radius of gyration of 5.5 in. The platform has an initial speed of \(3 \mathrm{ft} / \mathrm{sec}\) down the incline when a tension \(T\) is applied through a cable attached to the platform. If the platform acquires a speed of \(3 \mathrm{ft} / \mathrm{sec}\) up the incline after the tension has been applied for 8 seconds, what is the average value of the tension in the cable?

Short Answer

Expert verified
The average tension is calculated to be approximately 40.8 lb.

Step by step solution

01

Understanding the System

The system consists of a platform moving on an incline with two pairs of wheels. The platform and its wheels are affected by tension, gravity, and motion down the incline. We need to calculate the average tension when the platform changes direction.
02

Determine Changes in Kinetic Energy

Identify that the kinetic energy involves translational and rotational components. Calculate the initial and final translational kinetic energy (using the mass and velocity of the platform) and rotational kinetic energy (using the moment of inertia for the wheels).
03

Calculate Initial and Final Velocities

Both initial and final velocities for the platform are given as 3 ft/sec. However, note the direction change: initially, it moves down, finally moves up the incline.
04

Apply the Work-Energy Principle

The change in the kinetic energy is equal to the work done by the tension in the cable minus the work done by gravity. Use\[ \Delta KE = W_T - W_G \]where \( W_T \) is the work done by the tension, and \( W_G \) the work done by gravity.
05

Compute Moment of Inertia

Calculate the moment of inertia for the wheels:\[ I = m r^2 \] where \( m = 25/2 \) lb (weight converted to mass assuming \( g = 32.2 \) ft/sec²), \( r = 5.5/12 \) ft (convert inches to feet).
06

Set Up Equations for Work Done and Solve for T

Establish the work-energy equation:\[ T \times s - mg \sin(\theta) \times s = \Delta KE \] Solve for the tension, \( T \), ensuring to incorporate the distance \( s = v\cdot t \) through which the tension acts, based on the time of 8 seconds.
07

Solve for Average Tension

Calculate to find:\[ T = \frac{\Delta KE + mg \sin(\theta) \times s}{s} \] Plug in the solved values for \(\Delta KE\), gravitational force components, and distance.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy a body possesses due to its motion. It is an essential concept in mechanics, involved heavily in analyzing the movement of objects. In this case, both the platform and its wheels have kinetic energy.

- **Translational Kinetic Energy**: This is associated with the motion of the platform as a whole. It is calculated using the formula: \[ KE_{trans} = \frac{1}{2}mv^2 \] Here, \( m \) is the mass of the platform, and \( v \) is its velocity.

- **Rotational Kinetic Energy**: This pertains to the wheels and is given by: \[ KE_{rot} = \frac{1}{2}I\omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

By calculating both types of kinetic energy, you can better understand the total energy changes during motion.
Moment of Inertia
The moment of inertia plays a crucial role in rotational motion and is akin to mass in translational motion. It determines how much torque is required for a desired angular acceleration. For the wheels of the platform, you calculate it using:

\[ I = mr^2 \] where \( m \) is the mass of a wheel and axle pair, and \( r \) is the centroidal radius of gyration.

This value is important because it helps us assess how distribution of mass affects the rotation of the wheels. A larger moment of inertia indicates that more effort (torque) is needed to rotate the wheel.
Work-Energy Principle
The work-energy principle is a fundamental concept in mechanics that links the work done on a system to its change in kinetic energy. It is expressed as:

\[ \Delta KE = W_T - W_G \] Where:- \( \Delta KE \) is the change in kinetic energy,- \( W_T \) is the work done by the tension in the cable, and- \( W_G \) is the work done by gravity.

In this problem, this principle helps to determine the tension in the cable by equating the change in the platform's kinetic energy to the net work done by the applied tension and gravity. It allows for a comprehensive assessment of energy changes throughout the motion.
Translational Motion
Translational motion refers to the movement of an object from one location to another without rotating. For the problem, the platform's translational motion can be characterized by its speed along the incline.

A few key points include: - The platform's initial and final speeds are both given as 3 ft/sec, albeit in opposite directions down and up the incline. - Calculating the change in translational kinetic energy involves these speeds and the platform's mass.

Considering translational motion is crucial for a detailed energy analysis. It provides a measure of how much the position of the platform changes along the incline.
Rotational Motion
Rotational motion is the motion of a body as it spins around a fixed axis. In this exercise, the wheels of the platform exhibit rotational motion.

Important factors include:- **Angular Velocity (\( \omega \))**: The rate of change of angular displacement of the wheels, related to the platform speed through their radius.- **Moment of Inertia (\( I \))**: As discussed, this value influences the wheels' resistance to change in rotation.

Analyzing the rotational motion of the wheels helps in calculating the rotational kinetic energy. It also pairs with translational motion to give a full picture of the mechanical energy transformations in the system.

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Most popular questions from this chapter

The uniform slender bar \(A B C\) weighs 6 lb and is initially at rest with end \(A\) bearing against the stop in the horizontal guide. When a constant couple \(M=72\) lb-in. is applied to end \(C,\) the bar rotates causing end \(A\) to strike the side of the vertical guide with a velocity of \(10 \mathrm{ft} / \mathrm{sec}\). Calculate the loss of energy \(\Delta E\) due to friction in the guides and rollers. The mass of the rollers may be neglected.

The uniform cylinder is rolling without slip with a velocity \(v\) along the horizontal surface when it overtakes a ramp traveling with speed \(v_{0} .\) Determine an expression for the speed \(v^{\prime}\) which the cylinder has relative to the ramp immediately after it rolls up onto the ramp. Finally, determine the percentage \(n\) of cylinder kinetic energy lost if \((a) \theta=10^{\circ}\) and \(v_{0}=0.25 v\) and (b) \(\theta=10^{\circ}\) and \(v_{0}=\) \(0.5 v .\) Assume that the clearance between the ramp and the ground is essentially zero, that the mass of the ramp is very large, and that the cylinder does not slip on the ramp.

A person who walks through the revolving door exerts a 90 -N horizontal force on one of the four door panels and keeps the \(15^{\circ}\) angle constant relative to a line which is normal to the panel. If each panel is modeled by a 60 -kg uniform rectangular plate which is \(1.2 \mathrm{m}\) in length as viewed from above, determine the final angular velocity \(\omega\) of the door if the person exerts the force for 3 seconds. The door is initially at rest and friction may be neglected.

The motor at \(B\) supplies a constant torque \(M\) which is applied to a 375 -mm- diameter internal drum around which is wound the cable shown. This cable then wraps around an 80 -kg pulley attached to a 125 -kg cart carrying \(600 \mathrm{kg}\) of rock. The motor is able to bring the loaded cart to a cruising speed of \(1.5 \mathrm{m} / \mathrm{s}\) in 3 seconds. What torque \(M\) is the motor able to supply, and what is the average value of the tension in each side of the cable which is wrapped around the pulley at \(O\) during the speed- up period? The cable does not slip on the pulley and the centroidal radius of gyration of the pulley is \(450 \mathrm{mm}\) What is the power output of the motor when the cart reaches its cruising speed?

The gear train shown starts from rest and reaches an output speed of \(\omega_{C}=240\) rev/min in 2.25 s. Rotation of the train is resisted by a constant \(150 \mathrm{N} \cdot \mathrm{m}\) moment at the output gear \(C .\) Determine the required input power to the \(86 \%\) efficient motor at \(A\) just before the final speed is reached. The gears have masses \(m_{A}=6 \mathrm{kg}, m_{B}=10 \mathrm{kg},\) and \(m_{C}=24 \mathrm{kg}\) pitch diameters \(d_{A}=120 \mathrm{mm}, d_{B}=160 \mathrm{mm},\) and \(d_{C}=240 \mathrm{mm},\) and centroidal radii of gyration \(k_{A}=\) \(48 \mathrm{mm}, k_{B}=64 \mathrm{mm},\) and \(k_{C}=96 \mathrm{mm}\)

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