Question

\(A\) 500-kg load hangs from three cables of equal length that are anchored at the points \((-2,0,0),(1, \sqrt{3}, 0)\) and \((1,-\sqrt{3}, 0) .\) The load is located at \((0,0,-2 \sqrt{3})\). Find the vectors describing the forces on the cables due to the load.

Short answer

Question: Determine the forces acting on the three cables holding a 500 kg load in equilibrium. Answer: The forces on each cable are: - For Cable A: \(\vec{F_A} = f \cdot \frac{1}{4}(2,0,-2\sqrt{3})\) - For Cable B: \(\vec{F_B} = f \cdot \frac{1}{4}(-1,-\sqrt{3},-2\sqrt{3})\) - For Cable C: \(\vec{F_C} = f \cdot \frac{1}{4}(-1,\sqrt{3}, -2\sqrt{3})\) where \(f = \frac{1}{3}(500 \cdot 9.8)\).

Step-by-step solution

Find the position vectors of each cable

To find the position vectors, we need to subtract the anchoring point coordinates from the load coordinates. Let's call these vectors \(\vec{PA}, \vec{PB}\) and \(\vec{PC}\) for the points A, B, and C respectively. $$ \vec{PA} = (0, 0, -2\sqrt{3}) - (-2, 0, 0) = (2, 0, -2\sqrt{3}) \\ \vec{PB} = (0, 0, -2\sqrt{3}) - (1, \sqrt{3}, 0) = (-1, -\sqrt{3}, -2\sqrt{3}) \\ \vec{PC} = (0, 0, -2\sqrt{3}) - (1, -\sqrt{3}, 0) = (-1, \sqrt{3}, -2\sqrt{3}) $$

Normalize the position vectors

To normalize a vector, we divide each component of the vector by its magnitude. Let's call the normalized vectors \(\vec{uA}, \vec{uB},\) and \(\vec{uC}\) for the points A, B, and C, respectively. $$ \vec{uA} = \frac{\vec{PA}}{\Vert \vec{PA} \Vert} = \frac{(2,0,-2\sqrt{3})}{\sqrt{2^2+0^2+(-2\sqrt{3})^2}} = \frac{1}{4}(2,0,-2\sqrt{3}) \\ \vec{uB} = \frac{\vec{PB}}{\Vert \vec{PB} \Vert} = \frac{(-1,-\sqrt{3},-2\sqrt{3})}{\sqrt{(-1)^2+(-\sqrt{3})^2+(-2\sqrt{3})^2}} = \frac{1}{4}(-1,-\sqrt{3},-2\sqrt{3}) \\ \vec{uC} = \frac{\vec{PC}}{\Vert \vec{PC} \Vert} = \frac{(-1,\sqrt{3},-2\sqrt{3})}{\sqrt{(-1)^2+(\sqrt{3})^2+(-2\sqrt{3})^2}} = \frac{1}{4}(-1,\sqrt{3}, -2\sqrt{3}) $$

Use equilibrium conditions to find the forces on each cable

In equilibrium, the sum of all forces acting on the load must be equal to zero. Since the load has a mass of 500 kg and the acceleration due to gravity is 9.8 m/s², the downward force is 500 kg x 9.8 m/s². We can divide this force into 3 equal forces acting along the direction of the cables. Let F_A, F_B, and F_C be the forces on each cable. $$ \vec{F_A} + \vec{F_B} + \vec{F_C} = (0, 0, 500 \cdot 9.8) $$ Since the force is divided equally among the cables, let \(f = \frac{1}{3}(500 \cdot 9.8)\). The forces acting along the direction of cables are \(f \cdot \vec{uA}\), \(f \cdot \vec{uB}\), and \(f \cdot \vec{uC}\). Therefore, we have: $$ f\cdot\vec{uA} + f\cdot\vec{uB} + f\cdot\vec{uC} = f\left(\vec{uA} + \vec{uB} + \vec{uC}\right) = (0, 0, 500 \cdot 9.8) $$ Now, we can find the forces on each cable: $$ \vec{F_A} = f\cdot\vec{uA} = f \cdot \frac{1}{4}(2,0,-2\sqrt{3}) \\ \vec{F_B} = f\cdot\vec{uB} = f \cdot \frac{1}{4}(-1,-\sqrt{3},-2\sqrt{3}) \\ \vec{F_C} = f\cdot\vec{uC} = f \cdot \frac{1}{4}(-1,\sqrt{3}, -2\sqrt{3}) $$

Key concepts

Position Vectors

In the context of our exercise, position vectors are crucial for understanding the spatial relationships between the load and the anchor points. Imagine number lines stretching through space in three dimensions—these are your axes. A position vector tells us where along these axes a certain point is located, relative to a specified origin.

In practical terms, the position vector is determined by subtracting the coordinates of the origin from the point in question. For the load suspended by the cables, we are interested in the position vectors that extend from the load itself to the anchor points. These vectors are essential because they provide the directional components of the forces we're trying to find, determined by the paths the cables take from their anchors to the load.

Normalize Vectors

When dealing with vectors, particularly within the realms of physics and engineering, we often need to focus solely on direction, distancing ourselves from magnitude. That is where normalizing a vector comes in—it gives us a direction vector with a magnitude of one.

Normalizing is done by dividing each component of the vector by the vector's magnitude. In our exercise, the magnitude is the square root of the sum of the squares of each component. This gives us the unit vectors that retain the direction of our position vectors, but with a standardized length. These normalized vectors then allow us to distribute forces equally and determine the tension in each cable.

Equilibrium Conditions

Equilibrium is a state where all the forces acting on an object are balanced, leading to no movement or a constant velocity—a concept vital in statics and dynamics. In our scenario, the equilibrium conditions dictate that the vector sum of all forces on the load must equal zero for it to remain at rest.

This means that the downward gravitational force, resulting from the mass of the load and the acceleration due to gravity, must be counterbalanced exactly by the combined tension forces in the cables. We can split this gravitational force into components that align with the normalized direction of each cable, setting the stage to calculate how much each cable contributes to upholding the structure's equilibrium.

Cable Tension

In any scenario involving suspended loads, cable tension accounts for the force exerted along a cable to support the load. Understanding the forces within the cables is key to ensuring the structure's integrity.

In our exercise, once we've normalized the position vectors and acknowledged the equilibrium conditions, we can compute the tension in each cable. The tension indirectly reveals the cable's capacity and whether it can withstand the strain without failing. As we found the tension in our three cables, we conferred that each must handle one-third of the total load's weight, showcasing a fairly common distribution approach in systems designed to balance loads symmetrically.

Vector Addition

Finally, vector addition is the process that allows us to combine several vectors into one resultant vector. Vectors are added together by summing their corresponding components, which, in the context of our exercise, is used to determine whether the forces on the load are in equilibrium.

With the forces along each cable represented by vectors, their addition—in our case, the combination of the three tension forces—should result in a vector that exactly opposes the weight of the load. If the sum is equivalent to the gravitational force (in the opposite direction), then our load remains suspended in equilibrium, a fact we've used to guide our calculations and ensure we've distributed the weight correctly among the cables.