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Chapter 12: Problem 45

Find the components of the vertical force \(\mathbf{F}=\langle 0,-10\rangle\) in the directions parallel to and normal to the following planes. Show that the total force is the sum of the two component forces. A plane that makes an angle of \(\pi / 3\) with the positive \(x\) -axis

Short Answer

Expert verified
Answer: The components of the force vector parallel to the plane are given as $\mathbf{F}_1 = \langle -5\sqrt{3}/2, -15/2 \rangle$ and $\mathbf{F}_2 = \langle -5\sqrt{3}/2, 5/2 \rangle$. When summed together, they result in the total force vector $\mathbf{F}_{total} = \langle 0, -10 \rangle$, which is equal to the given force vector $\mathbf{F}$.

Step by step solution

01

Find the direction vectors for the plane

Since the plane makes an angle of \(\pi/3\) with the positive \(x\)-axis, one direction vector parallel to the plane is the unit vector in the direction of \(\pi/3\) angle, which can be defined as: \(\mathbf{v}_1 = \langle \cos(\pi/3), \sin(\pi/3) \rangle = \langle 1/2, \sqrt{3}/2 \rangle\). Another vector parallel to the plane can be perpendicular to \(\mathbf{v}_1\) and have the same \(z\) component as \(\mathbf{v}_1\). Therefore, considering the dot product of vectors is zero, we can choose \(\mathbf{v}_2 = \langle -\sqrt{3}/2, 1/2 \rangle\).
02

Find the projections of force vector on the direction vectors

The projection of the force vector onto a direction vector can be calculated using dot product: Projection of \(\mathbf{F}\) on \(\mathbf{v}_1 = \frac{\mathbf{F} \cdot \mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{\langle 0,-10\rangle \cdot \langle 1/2, \sqrt{3}/2\rangle }{\| \langle 1/2, \sqrt{3}/2\rangle \|} = -5\sqrt{3}\) Projection of \(\mathbf{F}\) on \(\mathbf{v}_2 = \frac{\mathbf{F} \cdot \mathbf{v}_2}{\|\mathbf{v}_2\|} = \frac{\langle 0,-10\rangle \cdot \langle -\sqrt{3}/2, 1/2\rangle }{\| \langle -\sqrt{3}/2, 1/2\rangle \|} = 5\)
03

Find the component force vectors of vertical force vector

Now, the component force vectors can be found by multiplying each projection by its corresponding unit direction vector: Component of force parallel to \(\mathbf{v}_1: \mathbf{F}_1 = -5\sqrt{3} \cdot \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = -5\sqrt{3} \langle 1/2, \sqrt{3}/2 \rangle = \langle -5\sqrt{3}/2, -15/2 \rangle\) Component of force parallel to \(\mathbf{v}_2: \mathbf{F}_2 = 5 \cdot \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|} = 5 \langle -\sqrt{3}/2, 1/2 \rangle = \langle -5\sqrt{3}/2, 5/2 \rangle\)
04

Show that the total force is the sum of the component forces

To find the total force, we can add the component force vectors: \(\mathbf{F}_{total} = \mathbf{F}_1 + \mathbf{F}_2 = \langle -5\sqrt{3}/2, -15/2 \rangle + \langle -5\sqrt{3}/2, 5/2 \rangle = \langle -10\sqrt{3}/2, -10/2 \rangle = \langle 0, -10 \rangle\) We can clearly see that the total force \(\mathbf{F}_{total}\) is equal to the given force vector \(\mathbf{F}\). Thus, the total force is indeed the sum of the two component forces.

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Most popular questions from this chapter

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3,0\rangle, \mathbf{v}=\langle 1,1,1\rangle\)

Consider the parallelogram with adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\). a. Show that the diagonals of the parallelogram are \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). b. Prove that the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v}=0\). c. Show that the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the sides.

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Consider the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) (in any number of dimensions). Use the following steps to prove that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\). a. Show that \(|\mathbf{u}+\mathbf{v}|^{2}=(\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+\) \(2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Use the Cauchy-Schwarz Inequality to show that \(|\mathbf{u}+\mathbf{v}|^{2} \leq(|\mathbf{u}|+|\mathbf{v}|)^{2}\). c. Conclude that \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}|\). d. Interpret the Triangle Inequality geometrically in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\).

Prove that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$$
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