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Chapter 4: Problem 80

At a bulk loading station, gravel leaves the hopper at the rate of 220 lb/sec with a velocity of \(10 \mathrm{ft} / \mathrm{sec}\) in the direction shown and is deposited on the moving flatbed truck. The tractive force between the driving wheels and the road is \(380 \mathrm{lb}\), which overcomes the 200 lb of frictional road resistance. Determine the acceleration \(a\) of the truck 4 seconds after the hopper is opened over the truck bed, at which instant the truck has a forward speed of \(1.5 \mathrm{mi} / \mathrm{hr}\) The empty weight of the truck is \(12,000 \mathrm{lb}\).

Short Answer

Expert verified
The acceleration of the truck is approximately \(0.45 \; \text{ft/s}^2\).

Step by step solution

01

Convert Units

First, we need to convert the given speed of the truck from miles per hour to feet per second. Given speed: \(1.5\; \text{mi/hr}\)Convert miles per hour to feet per second:\[1.5 \; \text{mi/hr} \times \frac{5280 \; \text{ft/mi}}{3600 \; \text{s/hr}} = 2.2 \; \text{ft/sec}\]So, the initial speed of the truck is \(2.2 \text{ ft/sec}\).
02

Calculate Mass Flow Rate Increase

Determine how the flow of gravel increases the truck's mass. Gravel is leaving the hopper at the rate of \(220\; \text{lb/sec}\).In 4 seconds, the total mass of gravel added to the truck is:\[220 \; \text{lb/sec} \times 4 \; \text{sec} = 880 \; \text{lb}\].
03

Determine Forces Acting on the Truck

We calculate the total force available to accelerate the truck. The tractive force provided by the wheels is \(380 \; \text{lb}\), and the frictional resistance is \(200 \; \text{lb}\).The net force available for acceleration is:\[380 \; \text{lb} - 200 \; \text{lb} = 180 \; \text{lb}\].
04

Calculate the Total Mass of the Truck and Gravel

Find the total mass of the loaded truck at 4 seconds after loading begins.The initial weight of the truck is \(12,000 \; \text{lb}\).Adding the gravel mass:\[12,000 \; \text{lb} + 880 \; \text{lb} = 12,880 \; \text{lb}\].Convert the total weight to mass (in slugs) using \(g = 32.2 \; \text{ft/s}^2\):\[\text{mass} = \frac{12,880 \; \text{lb}}{32.2 \; \text{ft/s}^2} \approx 400 \; \text{slugs}\].
05

Determine the Acceleration of the Truck

Calculate the acceleration of the truck using Newton's second law:\[F = ma\]Where \(F\) is the net force and \(m\) is the mass.\[180 \; \text{lb} = 400 \; \text{slugs} \times a\]Solve for \(a\):\[a = \frac{180 \; \text{lb}}{400 \; \text{slugs}} \approx 0.45 \; \text{ft/s}^2\].

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Key Concepts

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

Newton's Second Law
Newton's second law is a fundamental principle in the field of dynamics. It describes how the motion of an object changes when it is acted upon by a force. In simple terms, the law states that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. This fundamental relationship is expressed mathematically as \( F = ma \), where \( F \) is the net force applied to an object, \( m \) is its mass, and \( a \) is the resulting acceleration.

In this exercise, Newton's second law is used to determine the truck's acceleration. By knowing the net force available (calculated after considering both the tractive force provided by the wheels and the opposing frictional force) and the total mass, we applied \( F = ma \) to solve for acceleration \( a \). Understanding this law helps in predicting how forces affect the motion of objects in many practical situations, from vehicles in motion to objects falling under gravity.
Frictional Force
Frictional force is a resistive force that opposes the motion of an object. It arises due to the interactions between the surfaces in contact, such as a vehicle's wheels on the road. This force is crucial in our context because it's part of what balances the other forces acting on the truck.

In this problem, friction acts against the truck's motion, quantified as road resistance of 200 pounds. To determine the net force for the truck's acceleration, the frictional force must be subtracted from the tractive force, which is the force propelling the truck forward. This yields a net force of 180 pounds, after overcoming the frictional opposition. Understanding and calculating friction is important in dynamics to ensure accurate predictions of movement in applied situations.
Mass Flow Rate
The mass flow rate is a measure of mass flowing through a point per unit time. It is an important concept when considering systems where mass changes over time, such as loading trucks in this scenario.

Here, gravel is added to the truck at a rate of 220 pounds per second. Over time, specifically after 4 seconds, this results in a total added mass of 880 pounds of gravel. This addition alters the truck's total mass, which directly affects its acceleration as calculated through Newton's second law. Recognizing how mass flow impacts dynamics allows us to calculate the changes in velocity and acceleration properly when dealing with complex systems.
Unit Conversion
Unit conversion is an essential skill in physics, allowing you to work with quantities in the units required by the problem. It ensures that all calculations are accurate and consistent.

In this exercise, we first needed to convert the truck's speed from miles per hour to feet per second for consistent units with other given values. By using the conversion factor \(1 ext{ mile} = 5280 ext{ feet}\) and \(1 ext{ hour} = 3600 ext{ seconds}\), the original speed of \(1.5 ext{ mi/hr}\) was converted to \(2.2 ext{ ft/sec}\). Proper unit conversion allows you to effectively apply formulas like \( F = ma \) correctly, ensuring all quantities are in compatible units, thus avoiding errors in problem-solving and ensuring precise calculations.

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Most popular questions from this chapter

The chain of length \(L\) and mass \(\rho\) per unit length is released from rest on the smooth horizontal surface with a negligibly small overhang \(x\) to initiate motion. Determine (a) the acceleration a as a function of \(x,(b)\) the tension \(T\) in the chain at the smooth corner as a function of \(x,\) and \((c)\) the velocity \(v\) of the last link \(A\) as it reaches the corner.

A railroad coal car weighs 54,600 lb empty and carries a total load of 180,000 lb of coal. The bins are equipped with bottom doors which permit discharging coal through an opening between the rails. If the car dumps coal at the rate of \(20,000 \mathrm{lb} / \mathrm{sec}\) in a downward direction relative to the car, and if frictional resistance to motion is 4 lb per ton of total remaining weight, determine the coupler force \(P\) required to give the car an acceleration of \(0.15 \mathrm{ft} / \mathrm{sec}^{2}\) in the direction of \(P\) at the instant when half the coal has been dumped.

The carriage of mass \(2 m\) is free to roll along the horizontal rails and carries the two spheres, each of mass \(m,\) mounted on rods of length \(l\) and negligible mass. The shaft to which the rods are secured is mounted in the carriage and is free to rotate. If the system is released from rest with the rods in the vertical position where \(\theta=0,\) determine the velocity \(v_{x}\) of the carriage and the angular velocity \(\dot{\theta}\) of the rods for the instant when \(\theta=180^{\circ} .\) Treat the carriage and the spheres as particles and neglect any friction.

The sprinkler is made to rotate at the constant angular velocity \(\omega\) and distributes water at the volume rate \(Q .\) Each of the four nozzles has an exit area \(A\). Water is ejected from each nozzle at an angle \(\phi\) that is measured in the horizontal plane as shown. Write an expression for the torque \(M\) on the shaft of the sprinkler necessary to maintain the given motion. For a given pressure and thus flow rate \(Q\), at what speed \(\omega_{0}\) will the sprinkler operate with no applied torque? Let \(\rho\) be the density of water.

When the rocket reaches the position in its trajectory shown, it has a mass of \(3 \mathrm{Mg}\) and is beyond the effect of the earth's atmosphere. Gravitational acceleration is \(9.60 \mathrm{m} / \mathrm{s}^{2}\). Fuel is being consumed at the rate of \(130 \mathrm{kg} / \mathrm{s},\) and the exhaust velocity relative to the nozzle is \(600 \mathrm{m} / \mathrm{s}\). Compute the \(n\) - and \(t\) -components of acceleration of the rocket.
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