Problem 25 - Fundamentals of Physics

Question

Force \(\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \mathrm{j}\) acts on a particle with position vector \(\vec{r}=(3.0 \mathrm{~m}) \mathrm{i}+(4.0 \mathrm{~m}) \hat{j}\) What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of \(\vec{r}\) and \(\vec{F} ?\)

Step-by-Step Solution

Answer

(a) \(50 \hat{k} \mathrm{N}\cdot\mathrm{m}\); (b) \(90^{\circ}\)."

Step 1: Understand Torque Formula

The torque \( \vec{\tau} \) on a particle is defined as the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \): \( \vec{\tau} = \vec{r} \times \vec{F} \). The cross product of two vectors \([a, b, c]\) and \([x, y, z]\) is given by \(\vec{a} \times \vec{b} = (bz - cy)\hat{i} + (cx - az)\hat{j} + (ay - bx)\hat{k} \).

Step 2: Compute Cross Product for Torque

Given \( \vec{r} = (3.0 \,\mathrm{m}) \hat{\mathrm{i}} + (4.0 \,\mathrm{m}) \hat{\mathrm{j}} \) and \( \vec{F} = (-8.0 \,\mathrm{N}) \hat{\mathrm{i}} + (6.0 \,\mathrm{N}) \hat{\mathrm{j}} \), compute the cross product \(\vec{\tau} = \vec{r} \times \vec{F} = (3 \times 6 - 4 \times (-8)) \hat{k} \). Thus, \(\vec{\tau} = (18 + 32) \hat{k} = 50 \hat{k} \).

Step 3: Compute Magnitudes of \(\vec{r}\) and \(\vec{F}\)

First, compute the magnitude of \(\vec{r}\): \( |\vec{r}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \,\mathrm{m} \). Next, compute the magnitude of \(\vec{F}\): \( |\vec{F}| = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = 10 \,\mathrm{N} \).

Step 4: Find the Dot Product for Angle Calculation

The dot product \( \vec{r} \cdot \vec{F} \) is computed as follows: \( \vec{r} \cdot \vec{F} = 3 \times (-8) + 4 \times 6 = -24 + 24 = 0 \).

Step 5: Calculate the Angle Between the Vectors

The cosine of the angle \(\theta\) between \(\vec{r}\) and \(\vec{F}\) is given by \(\cos\theta = \frac{\vec{r} \cdot \vec{F}}{|\vec{r}| \times |\vec{F}|} \). Since \(\vec{r} \cdot \vec{F} = 0 \), \(\cos\theta = 0\), which implies \(\theta = 90^{\circ} \).

Step 6: Conclusion

For part (a), the torque \(\vec{\tau}\) is \(50 \hat{k} \,\mathrm{N}\cdot\mathrm{m}\). For part (b), the angle between \(\vec{r}\) and \(\vec{F}\) is \(90^{\circ}\).

Key Concepts

Cross Product

To understand torque, one of the crucial aspects is the cross product. The cross product is a vector operation that results in another vector. This new vector is perpendicular to the plane formed by the original two vectors. For two 3D vectors \(\vec{a} = [a_1, a_2, a_3]\) and \(\vec{b} = [b_1, b_2, b_3]\), the cross product is given by:\[\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} + (a_3b_1 - a_1b_3)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}\]The result is a vector, often represented by \(\vec{k}\), which is orthogonal to both \(\vec{a}\) and \(\vec{b}\).In our exercise, we used the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\) to find torque \(\vec{\tau} = \vec{r} \times \vec{F}\). The calculation showed the z-component of the resulting vector since \(\vec{i}\) and \(\vec{j}\) components cancel out, leaving us with a torque \(50 \hat{k} \ \mathrm{N \cdot m}\). Understanding the cross product is essential for problems involving rotational motion.

Dot Product

To find the angle between two vectors, the dot product proves very useful. The dot product, unlike the cross product, results in a scalar. This operation is handy when calculating angles and projections.For two vectors \(\vec{a} = [a_1, a_2, a_3]\) and \(\vec{b} = [b_1, b_2, b_3]\), the dot product is:\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]When the result of the dot product is zero, \(\vec{a}\) and \(\vec{b}\) are perpendicular.In our problem, the position vector \(\vec{r}\) and the force \(\vec{F}\) resulted in a dot product of 0, indicating they are perfectly perpendicular. This directly helped in determining the angle as 90 degrees. Knowing how to calculate and interpret the dot product is fundamental in vector mathematics.

Angle Between Vectors

The angle between two vectors can tell you much about their spatial relationship. To derive the angle \(\theta\) between vectors, you use the formula:\[\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\]This formula demonstrates how the cosine of the angle relates to both the dot product and the magnitudes of the vectors.In our particular scenario, since the dot product \(\vec{r} \cdot \vec{F} = 0\), the angle is determined using the fact that \(\cos\theta = 0\). The only angle for which cosine is zero is \(90^{\circ}\), confirming the vectors are perpendicular.Understanding vector angles is crucial for problems involving directions and forces in physics. This principle applies broadly, from simple geometric interpretation to complex physical calculations.

Unit Vector Notation

Unit vector notation is a streamlined way to express vectors, making them easier to understand and calculate with. Each unit vector represents a direction in space: \(\hat{i}\) for the x-direction, \(\hat{j}\) for the y-direction, and \(\hat{k}\) for the z-direction. A vector expressed in unit vector notation might look like \(\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\). Here, each coefficient \(A_x, A_y, A_z\) corresponds to how much the vector extends in that particular direction.In our example, both the force \(\vec{F}\) and position vector \(\vec{r}\) were given in unit vector form:- \(\vec{F} = (-8.0 \mathrm{~N}) \hat{i} + (6.0 \mathrm{~N}) \hat{j}\)- \(\vec{r} = (3.0 \mathrm{~m}) \hat{i} + (4.0 \mathrm{~m}) \hat{j}\)This method simplifies the process of vector arithmetic, allowing easy calculations of operations like cross and dot products. Mastery of unit vector notation is key in both scientific and engineering contexts.