/*! 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 29 An open-topped tank, in the form... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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/m鲁, 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/s虏, \[F = 684.25 \times 9.81 \times \frac{1}{3} = 2236.215 \text{ N}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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鈥檚 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: \[ m_{\text{total}} = 340 + 344.25 = 684.25 \text{ kg} \]

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A tank with vertical sides contains water to a depth of \(1.2 \mathrm{~m}\) and a layer of oil \(800 \mathrm{~mm}\) deep which rests on top of the water. The relative density of the oil is \(0.85\) and above the oil is air at atmospheric pressure. In one side of the tank, extending through its full height, is a protrusion in the form of a segment of a vertical circular cylinder. This is of radius \(700 \mathrm{~mm}\) and is riveted across an opening \(500 \mathrm{~mm}\) wide in the plane wall of the tank. Calculate the total horizontal thrust tending to force the protrusion away from the rest of the tank and the height of the line of action of this thrust above the base of the tank.

A canal lock is \(6 \mathrm{~m}\) wide and has two vertical gates which make an angle of \(120^{\circ}\) with each other. The depths of water on the two sides of the gates are \(9 \mathrm{~m}\) and \(2.7 \mathrm{~m}\) respectively. Each gate is supported on two hinges, the lower one being \(600 \mathrm{~mm}\) above the bottom of the lock. Neglecting the weight of the gates themselves, calculate the thrust between the gates and the height of the upper hinges if the forces on them are to be half those on the lower hinges.

An open channel has a cross-section in the form of an equilateral triangle with \(2.5 \mathrm{~m}\) sides and a vertical axis of symmetry. Its end is closed by a triangular vertical gate, also with \(2.5 \mathrm{~m}\) sides, supported at each corner. Calculate the horizontal thrust on each support when the channel is brim-full of water.

A rectangular pontoon \(6 \mathrm{~m}\) by \(3 \mathrm{~m}\) in plan, floating in water, has a uniform depth of immersion of \(900 \mathrm{~mm}\) and is subjected to a torque of \(7600 \mathrm{~N} \cdot \mathrm{m}\) about the longitudinal axis. If the centre of gravity is \(700 \mathrm{~mm}\) up from the bottom, estimate the angle of heel.

A hollow cylinder with closed ends is \(300 \mathrm{~mm}\) diameter and \(450 \mathrm{~mm}\) high, has a mass of \(27 \mathrm{~kg}\) and has a small hole in the base. It is lowered into water so that its axis remains vertical. Calculate the depth to which it will sink, the height to which the water will rise inside it and the air pressure inside it. Disregard the effect of the thickness of the walls but assume that it is uniform and that the compression of the air is isothermal. (Atmospheric pressure \(=101.3 \mathrm{kPa}\) )
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Physics of Motion

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.