/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Sand drops at a rate of \(2000 \... [FREE SOLUTION] | 91Ó°ÊÓ

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Sand drops at a rate of \(2000 \mathrm{~kg} / \mathrm{min}\) from the bottom of a stationary hopper onto a belt conveyer moving horizontally at \(250 \mathrm{~m} / \mathrm{min}\). Determine the force needed to drive the conveyer, neglecting friction. [Hint: How much momentum must be imparted to the sand each second?]

Short Answer

Expert verified
The force needed to drive the conveyer is approximately 139.0 N.

Step by step solution

01

Calculate the Sand's Mass Contribution per Second

First, convert the rate of sand dropping from kg/min to kg/s. The rate is given as 2000 kg/min. To find kg/s, divide by 60 seconds:\[\text{Mass rate (kg/s)} = \frac{2000}{60} \approx 33.33 \text{ kg/s}\]
02

Determine the Velocity of the Belt Conveyor

The belt conveyor moves at a speed of 250 m/min. Convert this speed to m/s by dividing by 60 (since there are 60 seconds in a minute):\[\text{Velocity (m/s)} = \frac{250}{60} \approx 4.167 \text{ m/s}\]
03

Calculate the Change in Momentum per Second (Force)

The momentum imparted to the sand per second is the product of the mass rate and the velocity of the conveyor belt. According to Newton's second law, the change in momentum per unit time is the force:\[F = \dot{m} \times v = 33.33 \text{ kg/s} \times 4.167 \text{ m/s} \approx 139.0 \text{ N}\]
04

State the Required Force to Drive the Conveyor

The force needed to drive the conveyor, neglecting friction, is equal to the change in momentum per second. Thus, the required force is 139.0 N.

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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 cornerstone in the world of physics. It explains how the velocity of an object changes when it is subjected to an external force. The law is encapsulated in the equation: \[ F = m imes a \] where \( F \) is the total force exerted on the object, \( m \) is the object's mass, and \( a \) is its acceleration. This relationship is fundamental because it shows a direct connection between force, mass, and acceleration. In everyday situations, this means that a greater force is needed to accelerate a larger mass. Conversely, a smaller mass will accelerate more easily with the same force. Think about pushing a small box versus pushing a car; the principles of Newton's Second Law apply to both situations. When applied to the conveyor belt problem, this law helps determine the force that needs to be imparted to the mass (sand) to maintain a constant speed, reflecting how vital an understanding of force dynamics is in practical applications.
Momentum in Physics
Momentum is a key concept in physics that refers to the quantity of motion an object has. It is defined as the product of an object's mass and its velocity, expressed mathematically as:\[ p = m imes v \]where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. Momentum measures how difficult it is to stop an object in motion. The greater the momentum, the harder it is to halt. There are two crucial points about momentum:
  • Momentum is a vector quantity, meaning it has both magnitude and direction.
  • The conservation of momentum states that in a closed system, absent of external forces, the total momentum before an event is equal to the total momentum after the event.
In the context of the exercise, momentum is imparted to the sand each second as it falls onto the moving belt. The problem requires calculating how much momentum needs to be added to keep the system at a steady state, essential for keeping the conveyor running smoothly.
Force Calculation
Calculating force is a basic but crucial task in physics, especially when dealing with motion and momentum. Within the conveyor belt problem, we find the force required to maintain the belt's constant speed using the change in momentum over time. According to Newton's Second Law, we know that the force can be expressed as:\[ F = \dot{m} \times v \]where \( \dot{m} \) is the mass flow rate (rate at which mass is added, in kg/s) and \( v \) is the velocity (m/s) at which the object must be accelerated. The calculated result gives the net force needed to keep the conveyor moving at a constant speed. In our exercise, the sand contributes \( 33.33 \) kg/s to the system, while the belt maintains a speed of \( 4.167 \) m/s. By multiplying these values, we obtain \( 139.0 \) N, which is the precise force necessary to keep the system in balance without any external friction. Understanding these calculations allows you to troubleshoot and balance systems in practical engineering tasks efficiently.

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Most popular questions from this chapter

Two identical railroad cars sit on a horizontal track, with a distance \(D\) between their two centers of mass. By means of a cable between them, a winch on one is used to pull the two together. (a) Describe their relative motion. ( \(b\) ) Repeat the analysis if the mass of one car is now three times that of the other. Keep in mind that the velocity of the center of mass of a system can only be changed by an external force. Here the forces due to the cable acting on the two cars are internal to the two-car system. The net external force on the system is zero, and so its center of mass does not move, even though each car travels toward the other. Taking the origin of coordinates at the center of mass, $$ x_{\mathrm{cm}}=0=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} $$ where \(x_{1}\) and \(x_{2}\) are the positions of the centers of mass of the two cars. (a) If \(m_{1}=m_{2}\), this equation becomes $$ 0=\frac{x_{1}+x_{2}}{2} \quad \text { or } \quad x_{1}=-x_{2} $$ The two cars approach the center of mass, which is originally midway between the two cars (that is, \(D / 2\) from each), in such a way that their centers of mass are always equidistant from it. (b) If \(m_{1}=3 m_{2}\), then we have $$ 0=\frac{3 m_{2} x_{1}+m_{2} x_{2}}{3 m_{2}+m_{2}}=\frac{3 x_{1}+x_{2}}{4} $$ from which \(x_{1}=-x_{2} / 3\). Since \(m_{1}>m_{2}\), it must be that \(x_{1}

Four masses are positioned in the \(x y\) -plane as follows: \(300 \mathrm{~g}\) at \((x=0, y=2.0 \mathrm{~m}), 500 \mathrm{~g}\) at \((-20 \mathrm{~m},-3.0 \mathrm{~m})\), \(700 \mathrm{~g}\) at \((50 \mathrm{~cm}, 30 \mathrm{~cm})\), and \(900 \mathrm{~g}\) at \((-80 \mathrm{~cm}, 150 \mathrm{~cm})\). Find their center of mass.

A system consists of the following masses in the \(x y\) -plane: \(4.0 \mathrm{~kg}\) at coordinates \((x=0, y=5.0 \mathrm{~m})\), \(7.0 \mathrm{~kg}\) at \((3.0 \mathrm{~m}, 8.0 \mathrm{~m})\), and \(5.0 \mathrm{~kg}\) at \((-3.0 \mathrm{~m},-6.0 \mathrm{~m})\). Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\sum m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$ and \(z_{\mathrm{cm}}=0\). These distances are, of course, measured from the origin \((0,0,0,\), .

Two balls of equal mass, moving with speeds of \(3 \mathrm{~m} / \mathrm{s}\), collide head-on. Find the speed of each after impact if (a) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(1 / 3\).

An empty \(15000-\mathrm{kg}\) coal car is coasting on a level track at \(5.00 \mathrm{~m} / \mathrm{s}\). Suddenly \(5000 \mathrm{~kg}\) of coal is dumped into it from directly above it. The coal initially has zero horizontal velocity with respect to the ground. Find the final speed of the car.

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