/*! 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 125 Water is discharged from a nozzl... [FREE SOLUTION] | 91Ó°ÊÓ

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Water is discharged from a nozzle with a velocity of \(12 \mathrm{m} / \mathrm{s}\) and strikes the blade mounted on the 20 -kg cart. Determine the tension developed in the cord, needed to hold the cart stationary, and the normal reaction of the wheels on the cart. The nozzle has a diameter of \(50 \mathrm{mm}\) and the density of water is \(\rho_{w}=1000 \mathrm{kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
Tension developed in the cord is equal to the force due to the discharge of water and the normal reaction of the wheels on the cart is equal to the weight of the cart.

Step by step solution

01

Calculate the Force Due to the Discharge of Water

To find the fluid force, first calculate the volume flow rate then the mass flow rate, which is used to find the fluid force. The volume flow rate \(Q\) is given by \(Q = \pi D^2 v / 4\) where \(D = 50 \times 10^{-3}\) m is the diameter of the nozzle and \(v = 12\) m/s is the velocity of the water. The mass flow rate \(\dot{m}\) is given by \(\dot{m} = \rho Q\) where \(\rho = 1000\) kg/m³ is the density of water. Finally, the fluid force \(F_{w}\) is given by \(F_{w} = \dot{m} v\).
02

Calculate the Tension in the Cord

The cart is stationary, so the tension \(T\) in the cord holding the cart should balance with the fluid force \(F_{w}\) acting on the cart due to the discharge of water. So tension \(T = F_{w}\).
03

Calculate the Normal Reaction of the Wheels on the Cart

The normal reaction \(N\) is the force exerted by the surface that opposes the weight of an object. Here, the normal reaction on the wheels of the cart is equal to the total weight of the cart. The normal reaction \(N = mg\) where \(m = 20\) kg is the mass of the cart and \(g = 9.8\) m/s² is the acceleration due to gravity.

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Key Concepts

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

Volume Flow Rate
Volume flow rate is a measure of the volume of fluid that passes through a given surface per unit time. It's an important concept in fluid dynamics because it helps us understand how much fluid is moving and at what speed. To visualize it, imagine water flowing through a hose; the volume flow rate would tell us how much water comes out of the hose every second.

In the exercise, we calculate the volume flow rate using the formula
\[Q = \frac{\pi D^2 v}{4}\]
where
\(D\) is the diameter of the nozzle, and \(v\) is the velocity of the water. By plugging in the nozzle diameter (\(50 mm\) converted to meters) and the water velocity (\(12 m/s\)), we determine how much water flows through the nozzle each second. Once we know the volume flow rate, we can move on to calculate the mass flow rate, which is crucial for understanding the forces involved in the exercise.
Mass Flow Rate
Mass flow rate is closely related to volume flow rate but focuses on the mass of fluid passing through a surface per unit time, rather than its volume. It is calculated by multiplying the volume flow rate by the fluid’s density (\(\rho\)).

Here’s the formula used for the exercise:
\[\dot{m} = \rho Q\]
It tells us the mass of water coming out of the nozzle every second. The density of water (\(\rho_w\)) is typically \(1000 kg/m^3\), which we use in our calculation. The mass flow rate is a key quantity for determining the fluid force exerted by the water on the cart’s blade, as a higher mass flow rate would result in a larger force.
Normal Reaction Force
The normal reaction force is the support force exerted by a surface when an object is in contact with it. It's always perpendicular to the surface, hence the term 'normal', which in geometry means perpendicular.

In our exercise, the normal reaction force is the force exerted by the ground on the wheels of the cart. It is denoted as \(N\) and it counteracts the cart's weight. The formula used is:
\[N = mg\]
where \(m\) is the mass of the cart and \(g\) is the acceleration due to gravity (approximately \(9.8 m/s^2\)). This force is critical as it ensures that the cart remains in contact with the ground and doesn't lift off due to the opposing force of the fluid.
Tension in a Cord
Tension is the force exerted along a cord, rope, or chain when it's used to convey a force. In physics, it's often treated as a pulling force because cords cannot push effectively.

In the problem at hand, the tension in the cord \(T\) must be equal to the fluid force \(F_w\) to keep the cart stationary. This is found through the relation:
\[T = F_w\]
Once we've calculated the fluid force resulting from the water's mass flow rate and its velocity, this same force is the amount of tension that the cord must withstand. Solving for tension helps us understand the stress on the cord and ensures the integrity of the system.

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Most popular questions from this chapter

Sand drops onto the 2 -Mg empty rail car at \(50 \mathrm{kg} / \mathrm{s}\) from a conveyor belt. If the car is initially coasting at \(4 \mathrm{m} / \mathrm{s}\) determine the speed of the car as a function of time.

As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the 10 -kg block which slides along the smooth surface and is subjected to a horizontal force of \(6 \mathrm{N}\). If observer \(A\) is in a fixed frame \(x\), determine the final speed of the block in 4 s if it has an initial speed of \(5 \mathrm{m} / \mathrm{s}\) measured from the fixed frame. Compare the result with that obtained by an observer \(B\), attached to the \(x^{\prime}\) axis that moves at a constant velocity of \(2 \mathrm{m} / \mathrm{s}\) relative to \(A\).

For a short period of time, the frictional driving force acting on the wheels of the \(2.5-\mathrm{Mg}\) van is \(F_{D}=\left(600 t^{2}\right) \mathrm{N}\) where \(t\) is in seconds. If the van has a speed of \(20 \mathrm{km} / \mathrm{h}\) when \(t=0,\) determine its speed when \(t=5 \mathrm{s}\).

The balloon has a total mass of \(400 \mathrm{kg}\) including the passengers and ballast. The balloon is rising at a constant velocity of \(18 \mathrm{km} / \mathrm{h}\) when \(h=10 \mathrm{m} .\) If the man drops the 40 -kg sand bag, determine the velocity of the balloon when the bag strikes the ground. Neglect air resistance.

A \(1-\) lb ball \(A\) is traveling horizontally at \(20 \mathrm{ft} / \mathrm{s}\) when it strikes a 10 -lb block \(B\) that is at rest. If the coefficient of restitution between \(A\) and \(B\) is \(e=0.6,\) and the coefficient of kinetic friction between the plane and the block is \(\mu_{k}=0.4,\) determine the time for the block \(B\) to stop sliding.

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