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Chapter 3: Problem 121

A rectangular container, of base dimensions \(0.4 \mathrm{m} \times\) \(0.2 \mathrm{m}\) and height \(0.4 \mathrm{m},\) is filled with water to a depth of \(0.2 \mathrm{m} ;\) the mass of the empty container is \(10 \mathrm{kg} .\) The container is placed on a plane inclined at \(30^{\circ}\) to the horizontal. If the coefficient of sliding friction between the container and the plane is \(0.3,\) determine the angle of the water surface relative to the horizontal.

Short Answer

Expert verified
To find the angle of the water surface relative to the horizontal, the friction force, the normal force, and the component of the total weight along the plane need to be balanced appropriately. After setting up and simplifying the equation, the unknown angle can be solved by implementing appropriate trigonometrical methods.

Step by step solution

01

Set up the problem

Firstly, start by defining the relevant forces on the system. The weight of the container (\(W_c = m_c \cdot g\)) acts vertically downward, where \(m_c = 10\:kg\) is the mass of the container and \(g = 9.8\:m/s^2\) is the acceleration due to gravity. The weight of the water (\(W_w = m_w \cdot g\)) also acts vertically downward, where \(m_w\) is the mass of the water, which can be found using the volume of the water and the given depth, and \(g\) is again the acceleration due to gravity. The friction force (\(F\)) acts parallel to the plane and opposes the motion, and is given by \(F = \mu \cdot N\), where \(\mu = 0.3\) is the coefficient of sliding friction and \(N\) is the normal force.
02

Calculate the normal force

The normal force is equal to the component of the weight perpendicular to the plane. This can be calculated using trigonometry: \(N = (W_c + W_w) \cdot \cos(30^{\circ})\), where the total weight (\(W_c + W_w\)) is tilted at a \(30^{\circ}\) angle from the vertical direction.
03

Define the equilibrium condition

In the equilibrium condition, the sum of the forces along the plane should be zero. This can be written as: \(W_w \cdot \sin(\theta) + W_c \cdot \sin(30^{\circ}) = F\), where \(\theta\) is the angle of the water surface relative to the horizontal.
04

Solve for the unknown

Substituting the expression of the friction force from Step 1 into the equilibrium condition in Step 3, and knowing that \(W_w = m_w \cdot g\) and \(W_c = m_c \cdot g\), we get: \(m_w \cdot g \cdot \sin(\theta) + m_c \cdot g \cdot \sin(30^{\circ}) = \mu \cdot N = \mu \cdot (m_c \cdot g + m_w \cdot g) \cdot \cos(30^{\circ})\). After simplifying this equation, the unknown angle \(\theta\) can be solved by isolating it on one side of the equation. This should be achieved through trigonometrical methods.

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