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

Answer: The force vectors on the cables due to the load are: \(\vec{F_1} = 544.22\ \vec{i} - 1633.33\vec{k}\), \(\vec{F_2} = -408.33\ \vec{i} - 707.11\vec{j} -1414.22\vec{k}\), and \(\vec{F_3} = -408.33\ \vec{i} + 707.11\vec{j} -1414.22\vec{k}\).

Step-by-step solution

1. Position vectors of anchor points and the load point

To find position vectors of the anchor points, we will create vectors from the origin (0,0,0) to the anchor points. Similarly, we can find the position vector of the load point from its coordinates. For the first anchor point \((-2,0,0)\): \(\vec{A_1} = -2\vec{i}+0\vec{j}+0\vec{k}\) For the second anchor point \((1,\sqrt{3},0)\): \(\vec{A_2} = \vec{i}+\sqrt{3}\vec{j}+0\vec{k}\) For the third anchor point \((1,-\sqrt{3},0)\): \(\vec{A_3} = \vec{i}-\sqrt{3}\vec{j}+0\vec{k}\) For the load point \((0,0,-2\sqrt{3})\): \(\vec{L} = 0\vec{i}+0\vec{j}-2\sqrt{3}\vec{k}\)

2. Direction vector of each cable

To find the direction vectors of the cables, we will subtract the position vector of each anchor point from the position vector of the load point. Direction vector of cable 1: \(\vec{D_1} = \vec{L}-\vec{A_1}\) Direction vector of cable 2: \(\vec{D_2} = \vec{L}-\vec{A_2}\) Direction vector of cable 3: \(\vec{D_3} = \vec{L}-\vec{A_3}\) Now substitute the values of position vectors: \(\vec{D_1} = (0\vec{i}+0\vec{j}-2\sqrt{3}\vec{k}) - (-2\vec{i}+0\vec{j}+0\vec{k}) = 2\vec{i} - 2\sqrt{3}\vec{k}\) \(\vec{D_2} = (0\vec{i}+0\vec{j}-2\sqrt{3}\vec{k}) - (\vec{i}+\sqrt{3}\vec{j}+0\vec{k}) = -\vec{i} - \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}\) \(\vec{D_3} = (0\vec{i}+0\vec{j}-2\sqrt{3}\vec{k}) - (\vec{i}-\sqrt{3}\vec{j}+0\vec{k}) = -\vec{i} + \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}\)

3. Unit vectors of direction vectors

Now, we will find the unit vector for each direction vector. Unit vectors of direction vectors: \(\hat{d_1}=\frac{\vec{D_1}}{\left \| \vec{D_1} \right \|}\) \(\hat{d_2}=\frac{\vec{D_2}}{\left \| \vec{D_2} \right \|}\) \(\hat{d_3}=\frac{\vec{D_3}}{\left \| \vec{D_3} \right \|}\) Normalize the direction vectors: \(\hat{d_1}=\frac{2\vec{i} - 2\sqrt{3}\vec{k}}{\left \| 2\vec{i} - 2\sqrt{3}\vec{k} \right \|}\) \(\hat{d_1}=\frac{2\vec{i} - 2\sqrt{3}\vec{k}}{\sqrt{4+(-2\sqrt{3})^2}}=\frac{\vec{i} - \sqrt{3}\vec{k}}{2\sqrt{3}}\) \(\hat{d_2}=\frac{-\vec{i} - \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{\sqrt{1^2+\sqrt{3}^2+(-2\sqrt{3})^2}}=\frac{-\vec{i} - \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{4}\) \(\hat{d_3}=\frac{-\vec{i} + \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{\sqrt{1^2+\sqrt{3}^2+(-2\sqrt{3})^2}}=\frac{-\vec{i} + \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{4}\)

4. Forces on cables

Now, we will calculate the force due to the load on each cable. Since the load is 500kg, the weight due to the load is: Weight = mass x gravity = 500kg x 9.8 m/s² = 4900 N As the weight is evenly distributed among the three cables, the force on each cable is: Force on each cable = \(\frac{4900 N}{3}\) = 1633.33 N Now, multiply the unit vector with the force on each cable to find the force vector: Force vector of cable 1: \(\vec{F_1} = 1633.33\ \hat{d_1} = 1633.33\left(\frac{\vec{i} - \sqrt{3}\vec{k}}{2\sqrt{3}}\right)\) Force vector of cable 2: \(\vec{F_2} = 1633.33\ \hat{d_2} = 1633.33\left(\frac{-\vec{i} - \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{4}\right)\) Force vector of cable 3: \(\vec{F_3} = 1633.33\ \hat{d_3} = 1633.33\left(\frac{-\vec{i} + \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{4}\right)\) Now, the force vectors on the cables due to the load can be simplified: \(\vec{F_1} = 1633.33 \frac{\vec{i} - \sqrt{3}\vec{k}}{2\sqrt{3}} = 544.22\ \vec{i} - 1633.33\vec{k}\) \(\vec{F_2} = 1633.33\frac{-\vec{i} - \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{4} = -408.33\ \vec{i} - 707.11\vec{j} -1414.22\vec{k}\) \(\vec{F_3} = 1633.33\frac{-\vec{i} + \sqrt{3}\vec{j} - 2\sqrt{3}\vec{k}}{4} = -408.33\ \vec{i} + 707.11\vec{j} -1414.22\vec{k}\) So, the vectors describing the forces on the cables due to the load are \(\vec{F_1} = 544.22\ \vec{i} - 1633.33\vec{k}\), \(\vec{F_2} = -408.33\ \vec{i} - 707.11\vec{j} -1414.22\vec{k}\), and \(\vec{F_3} = -408.33\ \vec{i} + 707.11\vec{j} -1414.22\vec{k}\).

Key concepts

Forces

Forces are interactions that cause objects to change their state of motion or shape. In the context of this exercise, we are focusing on the forces exerted by cables, which keep a load suspended. A force can be understood as a push or pull acting upon an object resulting from its interaction with another object. Here, these forces are represented by vectors that not only depict the direction but also the magnitude of force.

• The load has a mass of 500 kg, and the gravitational force acting on it is calculated using the weight formula: Weight = mass x gravity. This gives us a total force of 4900 N (newtons) due to gravity.
• This force is evenly distributed across the three cables, which means each cable experiences a force of approximately 1633.33 N.

To find the force on each cable, you multiply the force by their respective direction unit vectors. This tells us precisely how each cable contributes to holding the load in place.

Equilibrium

Equilibrium occurs when all forces acting on a body cancel each other out, resulting in no net force, and thus no motion. In our scenario, the load hanging from the cables is in equilibrium because the forces from the three cables balance the force of gravity acting on the load.

• The concept of equilibrium is essential in engineering and physics as it ensures stability, especially when dealing with suspended structures like bridges and buildings.
• Equilibrium requires that the vector sum of forces is zero. Here, this means the vector forces exerted by the cables, when combined, exactly oppose the weight of the load.

Understanding equilibrium helps us design systems that can support specified loads without collapsing or moving.

Vector Magnitude

Vector magnitude, in simple terms, is the 'size' or length of a vector, which quantifies how much force is being applied in the vector's direction. To find the magnitude of a vector, we use the formula:\[ \text{Magnitude} = \sqrt{i^2 + j^2 + k^2} \]Where \(i\), \(j\), and \(k\) are the vector components in the three-dimensional plane.

• In the exercise, we initially calculated the direction vectors of the cables, then found their magnitudes to determine the unit vectors necessary to quantify each cable's force direction and force magnitude.
• Accurate calculation of magnitude is vital for ensuring the individual forces are correctly partitioned and applied, ensuring equilibrium.

Understanding vector magnitudes leads to precise calculations in physical systems, ensuring safety and reliability.

Direction Vectors

Direction vectors are vectors that point from one point to another and represent the direction and path along which a force is applied. In this solution, direction vectors are used to represent the connection between the anchor points and the load.

• To initially determine the direction vectors of the cables, we subtract the position vector of each anchor point from the load's position vector. This shows us how the load is "tugging" on each cable.
• These direction vectors are then normalized to unit vectors, which inform us of the exact direction of the forces by stripping away any magnitude factor and concentrating on the 'directionality' alone.

Direction vectors are useful in various applications, not just in physics but also in fields like computer graphics, where they dictate movement directions or the orientation of objects.