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Chapter 2: Problem 29

An open-topped tank, in the form of a cube of \(900 \mathrm{~mm}\) side, has a mass of \(340 \mathrm{~kg}\). It contains \(0.405 \mathrm{~m}^{3}\) of oil of relative density \(0.85\) and is accelerated uniformly up a long slope at arctan (1/3) to the horizontal. The base of the tank remains parallel to the slope, and the side faces are parallel to the direction of motion. Neglecting the thickness of the walls of the tank, estimate the net force (parallel of the slope) accelerating the tank if the oil is just on the point of spilling.

Short Answer

Expert verified
The net force accelerating the tank up the slope is approximately 2236.22 N.

Step by step solution

01

Convert Units

Convert the side length of the tank from millimeters to meters. Given side length is 900 mm, so: \[900 \text{ mm} = 0.9 \text{ m} \]
02

Calculate the Volume of the Tank

The volume of a cube is calculated using the formula: \[V = \text{side}^3\] Substitute the side length: \[V = 0.9^3 = 0.729 \text{ m}^3 \]
03

Oil's Volume and Density

Extract the given oil volume and determine the oil density using relative density: \[V_{\text{oil}} = 0.405 \text{ m}^3 \]\[ \text{Relative density} = 0.85 \]Water density is 1000 kg/m³, so oil density is: \[ \rho_{\text{oil}} = 0.85 \times 1000 = 850 \text{ kg/m}^3 \]
04

Calculate Oil Mass

Calculate the oil mass using the volume and density: \[ m_{\text{oil}} = V_{\text{oil}} \times \rho_{\text{oil}} \]\[ m_{\text{oil}} = 0.405 \times 850 = 344.25 \text{ kg} \]
05

System Total Mass

Calculate the total mass of the tank and oil combined: \[ m_{\text{total}} = m_{\text{tank}} + m_{\text{oil}} \]\[ m_{\text{total}} = 340 \text{ kg} + 344.25 \text{ kg} = 684.25 \text{ kg} \]
06

Calculate Angle of Slope

Determine the angle of the slope using the given arctan value: \[ \theta = \text{arctan}\frac{1}{3} \]Since \(\theta\) is the angle where \(\tan^{-1} (\frac{1}{3})\), it's calculated as: \[ \theta = 18.435^\text{o} \]
07

Calculate the Net Force

Net force is parallel to the slope, calculated using:\[ F = m_{\text{total}} \times g \times \tan(\theta) \]\( g\) (acceleration due to gravity) = 9.81 m/s², \[F = 684.25 \times 9.81 \times \frac{1}{3} = 2236.215 \text{ N}\]

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Key Concepts

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

Oil Density
In this problem, the density of oil is crucial in calculating the mass of the oil contained in the tank. Density is defined as mass per unit volume, typically given the symbol \( \rho \). To find the oil density, we use the relative density (or specific gravity) of the oil. Relative density is the ratio of the density of a substance to the density of water. Given that water has a density of \( 1000 \text{ kg/m}^3 \), and the oil has a relative density of 0.85, we calculate the oil density as follows: \[ \rho_{\text{oil}} = 0.85 \times 1000 = 850 \text{ kg/m}^3 \]
Relative Density
Relative density, also known as specific gravity, is the ratio of the density of a substance to the density of a reference substance, commonly water for liquids. It is dimensionless, meaning it has no units. In context, the relative density of the oil is given as 0.85. This means the oil is 85% as dense as water. Calculating the actual density using this ratio simplifies many other calculations, particularly when dealing with different substances.
Net Force Calculation
To find the net force acting on the tank and its contents, we need to calculate the force parallel to the inclined plane. The formula for the net force in this scenario is: \[ F = m_{\text{total}} \times g \times \tan( \theta ) \] where:
- \( F \) is the net force,
- \( m_{\text{total}} \) is the total mass (tank + oil),
- \( g \) is the acceleration due to gravity (9.81 \text{ m/s}^2),
- \( \theta \) is the angle of the incline.
The angle is found using arctan: \( \theta = \tan^{-1} ( \frac{1}{3} ) \). By plugging in the values, we get: \[ F = 684.25 \times 9.81 \times \frac{1}{3} = 2236.215 \text{ N} \]
Inclined Plane
An inclined plane is a flat surface tilted at an angle, with one end higher than the other. It is used to study how objects behave under gravity when they are not on a horizontal surface. In this problem, the incline is described by \( \theta = \tan^{-1} ( \frac{1}{3} ) \), which simplifies many force calculations. When an object is on the inclined plane, the component of gravitational force acting along the plane is crucial for understanding how the object will move.
Mass Calculation
The mass of the system, consisting of both the tank and its oil content, needs to be calculated to determine the net force. Starting with the empty tank mass of 340 kg and adding the mass of the oil, found by multiplying the oil’s volume by its density: \[ m_{\text{oil}} = 0.405 \text{ m}^3 \times 850 \text{ kg/m}^3 = 344.25 \text{ kg} \] Therefore, the total mass \( m_{\text{total}} \) is: \[ m_{\text{total}} = 340 + 344.25 = 684.25 \text{ kg} \]

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Most popular questions from this chapter

Two small vessels are connected to a U-tube manometer containing mercury (relative density 13.56) and the connecting tubes are filled with alcohol (relative density \(0.82\) ). The vessel at the higher pressure is \(2 \mathrm{~m}\) lower in elevation than the other. What is the pressure difference between the vessels when the steady difference in level of the mercury menisci is \(225 \mathrm{~mm}\) ? What is the difference of piezometric head? If an inverted U-tube manometer containing a liquid of relative density \(0.74\) were used instead, what would be the manometer reading for the same pressure difference?

A vessel of water of total mass \(5 \mathrm{~kg}\) stands on a parcel balance. An iron block of mass \(2.7 \mathrm{~kg}\) and relative density \(7.5\) is suspended by a fine wire from a spring balance and is lowered into the water until it is completely immersed. What are the readings on the two balances?

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A solid uniform cylinder of length \(150 \mathrm{~mm}\) and diameter \(75 \mathrm{~mm}\) is to float upright in water. Between what limits must its mass be?

A circular opening \(1.2 \mathrm{~m}\) in diameter in the vertical side of a reservoir is closed by a disc which just fits the opening and is pivoted on a shaft along its horizontal diameter. Show that, if the water level in the reservoir is above the top of the disc, the turning moment on the shaft required to hold the disc vertical is independent of the head of water. Calculate the amount of this moment.
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