Problem 13 - Applied Mathematics

Question

A quad rilateral \(A B C D\) has opposite sides \(A B\) and DC parallel. Angle \(\mathrm{ABC}=150^{\circ}\) and angle \(\mathrm{BAD}=60^{\circ}\). Forces \(2 P, P, P, 2 P\) act along \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CD}}\), \(\overrightarrow{A D}\) respectively. Prove that the resultant has magnitude \(P(8+3 \sqrt{3})^{\frac{1}{2}}\) and find the tangent of the angle it makes with \(\mathrm{AB}\).

Step-by-Step Solution

Answer

Magnitude is \(P(8 + 3\sqrt{3})^{\frac{1}{2}}\). Tangent of the angle is \(\frac{7}{3}\).

Step 1 - Analyzing the Forces

Identify the forces acting along each side of the quadrilateral. The forces are: 2P along \(\overrightarrow{AB}\), P along \(\overrightarrow{BC}\), P along \(\overrightarrow{CD}\), and 2P along \(\overrightarrow{AD}\).

Step 2 - Represent Forces as Vectors

Define a coordinate system with origin at A. Represent the forces as vectors: \(\vec{F}_{AB} = (2P, 0)\), \(\rvec{F}_{BC} = P(-\cos(150^{\circ}), P\sin(150^{\circ})) = (-\frac{\sqrt{3}}{2}P, \frac{1}{2}P)\), \(\fvec{F}_{CD} = P(-1, 0)\), and \(\fvec{F}_{AD} = 2P(\cos(60^{\circ}), 2P\sin(60^{\circ})) = (P, \sqrt{3}P)\).

Step 3 - Sum of All Forces

Add the vectors \(\fvec{F}_{AB}, \vec{F}_{BC}, \rvec{F}_{CD}, \and \fvec{F}_{AD}\): \(\fvec{F}_{resultant}=(2P - \frac{\sqrt{3}P}{2} - P + P, 0 + \frac{1}{2}P + 0 + \sqrt{3}P) = (\frac{3P}{2}, \frac{7P}{2})\).

Step 4 - Magnitude of the Resultant Force

Calculate the magnitude of the resultant force using the Pythagorean theorem: \(\fvec{R}=\sqrt{(\frac{3P}{2})^2 + (\frac{7P}{2})^2} = P\sqrt{\frac{9}{4} + \frac{49}{4}} = P\sqrt{\frac{58}{4}} = P\sqrt{\frac{29}{2}} = P(8 + 3\sqrt{3})^{\frac{1}{2}}\).

Step 5 - Find Tangent of Angle with \overrightarrow{AB}\

The tangent angle the resultant force makes with \overrightarrow{AB}\ is found using the formula \(\tan(\theta)=\frac{\frac{7P}{2}}{\frac{3P}{2}} = \frac{7}{3}\).

Key Concepts

Vector Addition

Vector addition is an essential concept in physics, often used to determine the resultant force acting at a point. A vector represents a quantity with both magnitude and direction, such as force.
When multiple vectors act on a point, their combined effect can be found using vector addition. For example:

• Define each force as a vector.
• Break these vectors into their components, usually along the x and y axes.
• Add the corresponding components of each vector to find the resultant vector’s components.

In our quadrilateral problem, the forces acting along each side are:
\(\fvec{F}_{AB} = (2P, 0)\)
\(\rvec{F}_{BC} = P(-\frac{\text{\sqrt{3}}}{2}, \frac{1}{2})\)
\(\fvec{F}_{CD} = (-P, 0)\)
\(\fvec{F}_{AD} = (P, \text{\sqrt{3}})\)
Adding these together gives us the resultant vector.

Resultant Force

The resultant force is the single force that has the same effect on an object as all the individual forces acting together.

• To find it, we sum the vectors representing each of the acting forces.
• This involves adding the x-components and y-components separately to obtain the resultant vector's components.

In our example, the calculation is
\(\fvec{F}_{resultant}=(2P - \frac{\text{\sqrt{3}}P}{2} - P + P, 0 + \frac{1}{2}P + 0 + \text{\sqrt{3}}P)\)
This simplifies to:
\(\fvec{F}_{resultant}=(\frac{3P}{2}, \frac{7P}{2})\)

Force Vectors

Force vectors represent forces acting in specific directions and magnitudes. When we're dealing with multiple forces, we can think of each force as a vector.

• Each vector consists of an x-component and a y-component which describe its horizontal and vertical parts.
• We can use trigonometric functions to determine these components if the force directions are given in angles.

For instance, along the side \(BC\) of the quadrilateral, where the force is P:

• The x-component: \(P \times -\text{cos}(150^\text{\circ}) = -\frac{\text{\sqrt{3}}}{2}P\)
  
• The y-component: \(P \times \text{sin}(150^\text{\circ}) = \frac{1}{2}P\)
  

By representing all forces this way, we can easily add them to find the resultant force.

Trigonometry

Trigonometry bridges the gap between angles and lengths in geometry. In the context of forces, it helps break down force vectors into components.
When given an angle \(\theta\), components of a force \(F\) can be found using:

• \(F_x = F \times \text{cos}(\theta)\)
  
• \(F_y = F \times \text{sin}(\theta)\)
  

In our example, for a force P at an angle of 150°:

• The cosine component: \(\text{cos}(150^\text{\circ}) = -\frac{\text{\sqrt{3}}}{2}\)
  
• The sine component: \(\text{sin}(150^\text{\circ}) = \frac{1}{2}\)
  

Applying this to calculate the magnitude, we use the Pythagorean theorem:
\(\text{Magnitude} = \text{\sqrt}((\frac{3P}{2})^2 + (\frac{7P}{2})^2) = P\text{\sqrt}(\frac{58}{4}) = P(8 + 3\text{\sqrt{3}})^{\frac{1}{2}}\)
We then find the tangent of the angle the resultant force makes with a reference direction using:
\(\text{tan}(\theta) = \frac{\frac{7P}{2}}{\frac{3P}{2}} = \frac{7}{3}\)