Example

The Scalar Product

For the vectors shown in (Figure), find the scalar product $$ \overset{\to }{A}·\overset{\to }{F}$$.

Strategy

From (Figure), the magnitudes of vectors $$ \overset{\to }{A} $$ and $$ \overset{\to }{F} $$ are A = 10.0 and F = 20.0. Angle $$ \theta $$, between them, is the difference: $$ \theta =\phi -\alpha =110\text{°}-35\text{°}=75\text{°}$$. Substituting these values into (Figure) gives the scalar product.

Solution

Example

Angle between Two Forces

Three dogs are pulling on a stick in different directions, as shown in (Figure). The first dog pulls with force $$ {\overset{\to }{F}}_{1}=(10.0\hat{i}-20.4\hat{j}+2.0\hat{k})\text{N}$$, the second dog pulls with force $$ {\overset{\to }{F}}_{2}=(-15.0\hat{i}-6.2\hat{k})\text{N}$$, and the third dog pulls with force $$ {\overset{\to }{F}}_{3}=(5.0\hat{i}+12.5\hat{j})\text{N}$$. What is the angle between forces $$ {\overset{\to }{F}}_{1} $$ and $$ {\overset{\to }{F}}_{2}$$?

Figure 2.28 Three dogs are playing with a stick.

Strategy

The components of force vector $$ {\overset{\to }{F}}_{1} $$ are $$ {F}_{1x}=10.0\,\text{N}$$, $$ {F}_{1y}=-20.4\,\text{N}$$, and $$ {F}_{1z}=2.0\,\text{N}$$, whereas those of force vector $$ {\overset{\to }{F}}_{2} $$ are $$ {F}_{2x}=-15.0\,\text{N}$$, $$ {F}_{2y}=0.0\,\text{N}$$, and $$ {F}_{2z}=-6.2\,\text{N}$$. Computing the scalar product of these vectors and their magnitudes, and substituting into (Figure) gives the angle of interest.

Solution

Significance

Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between them without knowing the specifics about the geographic directions the unit vectors represent. Here, for example, the +x-direction might be to the east and the +y-direction might be to the north. But, the angle between the forces in the problem is the same if the +x-direction is to the west and the +y-direction is to the south.

Example

The Work of a Force

When force $$ \overset{\to }{F} $$ pulls on an object and when it causes its displacement $$ \overset{\to }{D}$$, we say the force performs work. The amount of work the force does is the scalar product $$ \overset{\to }{F}·\overset{\to }{D}$$. If the stick in (Figure) moves momentarily and gets displaced by vector $$ \overset{\to }{D}=(-7.9\hat{j}-4.2\hat{k})\,\text{cm}$$, how much work is done by the third dog in (Figure)?

Strategy

We compute the scalar product of displacement vector $$ \overset{\to }{D} $$ with force vector $$ {\overset{\to }{F}}_{3}=(5.0\hat{i}+12.5\hat{j})\text{N}$$, which is the pull from the third dog. Let’s use $$ {W}_{3} $$ to denote the work done by force $$ {\overset{\to }{F}}_{3} $$ on displacement $$ \overset{\to }{D}$$.

Solution

Significance

The SI unit of work is called the joule $$ (\text{J})$$, where 1 J = 1 $$ \text{N}·\text{m}$$. The unit $$ \text{cm}·\text{N} $$ can be written as $$ {10}^{-2}\text{m}·\text{N}={10}^{-2}\text{J}$$, so the answer can be expressed as $$ {W}_{3}=-0.9875\,\text{J}\approx -1.0\,\text{J}$$.

Example

The Torque of a Force

The mechanical advantage that a familiar tool called a wrench provides ((Figure)) depends on magnitude F of the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force is applied. The distance R from the nut to the point where force vector $$ \overset{\to }{F} $$ is attached and is represented by the radial vector $$ \overset{\to }{R}$$. The physical vector quantity that makes the nut turn is called torque (denoted by $$ \overset{\to }{\tau })$$, and it is the vector product of the distance between the pivot to force with the force: $$ \overset{\to }{\tau }=\overset{\to }{R}\,×\,\overset{\to }{F}$$.

To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle $$ \phi =40\text{°} $$ and at a distance of 0.25 m from the nut, as shown in (Figure)(a). Find the magnitude and direction of the torque applied to the nut. What would the magnitude and direction of the torque be if the force were applied at angle $$ \phi =45\text{°}$$, as shown in (Figure)(b)? For what value of angle $$ \phi $$ does the torque have the largest magnitude?

Figure 2.31 A wrench provides grip and mechanical advantage in applying torque to turn a nut. (a) Turn counterclockwise to loosen the nut. (b) Turn clockwise to tighten the nut.

Strategy

We adopt the frame of reference shown in (Figure), where vectors $$ \overset{\to }{R} $$ and $$ \overset{\to }{F} $$ lie in the xy-plane and the origin is at the position of the nut. The radial direction along vector $$ \overset{\to }{R} $$ (pointing away from the origin) is the reference direction for measuring the angle $$ \phi $$ because $$ \overset{\to }{R} $$ is the first vector in the vector product $$ \overset{\to }{\tau }=\overset{\to }{R}\,×\,\overset{\to }{F}$$. Vector $$ \overset{\to }{\tau } $$ must lie along the z-axis because this is the axis that is perpendicular to the xy-plane, where both $$ \overset{\to }{R} $$ and $$ \overset{\to }{F} $$ lie. To compute the magnitude $$ \tau $$, we use (Figure). To find the direction of $$ \overset{\to }{\tau }$$, we use the corkscrew right-hand rule ((Figure)).

Solution

Significance

When solving mechanics problems, we often do not need to use the corkscrew rule at all, as we’ll see now in the following equivalent solution. Notice that once we have identified that vector $$ \overset{\to }{R}\,×\,\overset{\to }{F} $$ lies along the z-axis, we can write this vector in terms of the unit vector $$ \hat{k} $$ of the z-axis:

In this equation, the number that multiplies $$ \hat{k} $$ is the scalar z-component of the vector $$ \overset{\to }{R}\,×\,\overset{\to }{F}$$. In the computation of this component, care must be taken that the angle $$ \phi $$ is measured counterclockwise from $$ \overset{\to }{R} $$ (first vector) to $$ \overset{\to }{F} $$ (second vector). Following this principle for the angles, we obtain $$ RF\,\text{sin}\,(+40\text{°})=+3.2\,\text{N}·\text{m} $$ for the situation in (a), and we obtain $$ RF\,\text{sin}\,(-45\text{°})=-3.5\,\text{N}·\text{m} $$ for the situation in (b). In the latter case, the angle is negative because the graph in (Figure) indicates the angle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwise because $$ +(360\text{°}-45\text{°})=+315\text{°} $$ and $$ \text{sin}\,(+315\text{°})=\text{sin}\,(-45\text{°})$$. In this way, we obtain the solution without reference to the corkscrew rule. For the situation in (a), the solution is $$ \overset{\to }{R}\,×\,\overset{\to }{F}=+3.2\,\text{N}·\text{m}\hat{k}$$; for the situation in (b), the solution is $$ \overset{\to }{R}\,×\,\overset{\to }{F}=-3.5\,\text{N}·\text{m}\hat{k}$$.

Example

A Particle in a Magnetic Field

When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic field $$ \overset{\to }{B} $$ is a vector, the magnetic force $$ \overset{\to }{F} $$ is a vector, and the velocity $$ \overset{\to }{u} $$ of the particle is a vector. The magnetic force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we express as $$ \overset{\to }{F}=\zeta \overset{\to }{u}\,×\,\overset{\to }{B}$$. In this equation, a constant $$ \zeta $$ takes care of the consistency in physical units, so we can omit physical units on vectors $$ \overset{\to }{u} $$ and $$ \overset{\to }{B}$$. In this example, let’s assume the constant $$ \zeta $$ is positive.

A particle moving in space with velocity vector $$ \overset{\to }{u}=-5.0\hat{i}-2.0\hat{j}+3.5\hat{k} $$ enters a region with a magnetic field and experiences a magnetic force. Find the magnetic force $$ \overset{\to }{F} $$ on this particle at the entry point to the region where the magnetic field vector is (a) $$ \overset{\to }{B}=7.2\hat{i}-\hat{j}-2.4\hat{k} $$ and (b) $$ \overset{\to }{B}=4.5\hat{k}$$. In each case, find magnitude F of the magnetic force and angle $$ \theta $$ the force vector $$ \overset{\to }{F} $$ makes with the given magnetic field vector $$ \overset{\to }{B}$$.

Strategy

First, we want to find the vector product $$ \overset{\to }{u}\,×\,\overset{\to }{B}$$, because then we can determine the magnetic force using $$ \overset{\to }{F}=\zeta \overset{\to }{u}\,×\,\overset{\to }{B}$$. Magnitude F can be found either by using components, $$ F=\sqrt{{F}_{x}^{2}+{F}_{y}^{2}+{F}_{z}^{2}}$$, or by computing the magnitude $$ |\overset{\to }{u}\,×\,\overset{\to }{B}| $$ directly using (Figure). In the latter approach, we would have to find the angle between vectors $$ \overset{\to }{u} $$ and $$ \overset{\to }{B}$$. When we have $$ \overset{\to }{F}$$, the general method for finding the direction angle $$ \theta $$ involves the computation of the scalar product $$ \overset{\to }{F}·\overset{\to }{B} $$ and substitution into (Figure). To compute the vector product we can either use (Figure) or compute the product directly, whichever way is simpler.

Solution

Significance

Even without actually computing the scalar product, we can predict that the magnetic force vector must always be perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the magnetic force vector is the vector product $$ \overset{\to }{F}=\zeta \overset{\to }{u}\,×\,\overset{\to }{B} $$ and, by the definition of the vector product (see (Figure)), vector $$ \overset{\to }{F} $$ must be perpendicular to both vectors $$ \overset{\to }{u} $$ and $$ \overset{\to }{B}$$.