Problem 75 - Contemporary Precalculus

Question

Do Exercise 74 when the weight is 600 pounds and the angles are \(28^{\circ}\) and \(38^{\circ}\).

Step-by-Step Solution

Answer

Question: In the given exercise, a weight of 600 pounds is being supported by two tension forces acting at angles of 28 degrees and 38 degrees, respectively. Determine the magnitude of the tension forces, T1 and T2, required to maintain equilibrium. Answer: The tension forces required to maintain equilibrium are T1 ≈ 397.7 pounds and T2 ≈ 441.8 pounds.

Step 1: Identify the Forces

In this exercise, we have three forces acting on a point: the weight W acting vertically downward, and the two tension forces T1 and T2 acting at angles \(28^{\circ}\) and \(38^{\circ}\), respectively. The weight is given as 600 pounds.

Step 2: Break Forces into Components

Break the two tension forces into their horizontal (x) and vertical (y) components. We have: T1_x = T1 * cos(28^{\circ}) T1_y = T1 * sin(28^{\circ}) T2_x = T2 * cos(38^{\circ}) T2_y = T2 * sin(38^{\circ})

Step 3: Set up Equations for Equilibrium

For the system to be in equilibrium, the sum of the horizontal components and the sum of the vertical components must both be equal to zero. Hence, we have the following equations: 1) \(T1_x + T2_x = 0\) 2) \(T1_y + T2_y - W = 0\) Substitute the expressions from Step 2 to get: 1) \(T1 * cos(28^{\circ}) + T2 * cos(38^{\circ}) = 0\) 2) \(T1 * sin(28^{\circ}) + T2 * sin(38^{\circ}) = 600\)

Step 4: Solve the Equations

We now have a system of two equations with two unknowns (T1 and T2). We can solve this system using any method (substitution, elimination, or matrices). Using the elimination method, we can multiply Equation (1) by sin(28) and Equation (2) by cos(28). After that, we subtract Equation (1) from Equation (2) to eliminate T1 and solve for T2: (sin(28) * Equation 1) = (\(T1 * sin(28) * cos(28^{\circ}) + T2 * sin(28) * cos(38^{\circ}) = 0\)) (cos(28) * Equation 2) = (\(T1 * sin(28) * cos(28^{\circ}) + T2 * cos(28) * sin(38^{\circ}) = 600 * cos(28)\)) Subtracting Equation (1) from Equation (2), we get: \(T2 * (sin(28) * cos(38) - cos(28) * sin(38)) = 600 * cos(28)\) Then, divide by \((sin(28) * cos(38) - cos(28) * sin(38))\) to get the value of T2: \(T2 = \frac{600 * cos(28)}{sin(28) * cos(38) - cos(28) * sin(38)}\) Now that we have T2, we can substitute its value into the first equation to get T1: \(T1 * cos(28^{\circ}) + \frac{600 * cos(28) * cos(38^{\circ})}{sin(28) * cos(38) - cos(28) * sin(38)} = 0\) Solving for T1: \(T1 = -\frac{600 * cos(38)}{cos(28) (sin(28) * cos(38) - cos(28) * sin(38))}\)

Step 5: Calculate the Tension Forces

Now that we have expressions for T1 and T2, we can plug in the given angles to their expressions to find their magnitudes: \(T1 = -\frac{600 * cos(38)}{cos(28) (sin(28) * cos(38) - cos(28) * sin(38))} \approx 397.7 \, \text{pounds}\) \(T2 = \frac{600 * cos(28)}{sin(28) * cos(38) - cos(28) * sin(38)} \approx 441.8 \, \text{pounds}\) Thus, the tension forces are T1 \(\approx 397.7\) pounds and T2 \(\approx 441.8\) pounds.

Key Concepts

Force Components

In physics, understanding the concept of force components is crucial when analyzing the effects of forces acting at angles. This is particularly true in problems involving equilibrium, where we often deal with forces that are not aligned with the standard coordinate axes.
To break down a force into its components, we project it onto the horizontal (x-axis) and vertical (y-axis). This projection makes use of trigonometric functions from the branch of mathematics called trigonometry. For a force \(F\) making an angle \(\theta\) with the horizontal, the horizontal and vertical components are calculated as:

• Horizontal Component: \(F_x = F \cdot \cos(\theta)\), which represents the force parallel to the x-axis.
• Vertical Component: \(F_y = F \cdot \sin(\theta)\), which indicates the force parallel to the y-axis.

In the exercise provided, the tension forces \(T1\) and \(T2\) are resolved into their horizontal and vertical components using the cosines and sines of their respective angles with the horizontal. This technique facilitates the application of equilibrium conditions to find the unknown forces. Being adept at discerning and calculating force components is a vital skill for solving physics problems involving multiple forces.

Trigonometry in Physics

Trigonometry is not just a subject that is confined to mathematics; it is an indispensable tool in physics as well. It provides a method to relate angles with the sides of triangles. Since many physical scenarios can be modeled using right-angled triangles, trigonometry becomes essential for solving these problems. In the exercise, the angles given are key to finding the force components.
For instance, when tension forces act at angles to a point, their horizontal and vertical components form right angles with each other, leading to a right-angled triangle. The use of sine and cosine:

• \textbf{Cosine (cos)}: In a right-angled triangle, it is the ratio of the adjacent side to the hypotenuse, so it is used to find horizontal force components.
• \textbf{Sine (sin)}: It is the ratio of the opposite side to the hypotenuse, used for vertical force components.

In our problem, these trigonometric functions help in breaking down the tensions \(T1\) and \(T2\) into components along the x and y-axis. Without a good grasp of trigonometry, solving such problems would be significantly more difficult.

Equilibrium Equations

Equilibrium equations are at the heart of statics, a branch of mechanics that deals with forces in a system that is not moving. When an object is in equilibrium, the sum of all forces acting on it must be zero, meaning there is no unbalanced force to cause acceleration. This principle can be applied in two dimensions by setting up equilibrium equations for the horizontal and vertical components of the forces.
In the context of our exercise, the equilibrium conditions can be expressed as:

• For horizontal forces: \(\sum F_x = 0\), indicating that the sum of all horizontal components equals zero.
• For vertical forces: \(\sum F_y = W\), where W is the downward weight, and the sum of all upward force components must equal this weight.

Using these conditions, we can write down two equations that allow us to solve for the unknowns in our problem – in this case, the tensions \(T1\) and \(T2\). By plugging the force components into these equilibrium equations, we can manipulate and solve the system of equations to find the magnitude of each tension force acting on the system. Understanding how to establish and solve equilibrium equations is vital for anyone studying physics or engineering as it is fundamental to analyzing forces in any static structure or system at rest.