Mathematics · Ch 6 — Applications of Vector Algebra

Application of Dot and Cross Product in Physics

6.3.4

Application of Dot and Cross Product in Physics

Definition 6.2 (Work). If a constant force $\vec F$ acts on a particle while it is displaced by $\vec d$ (from one point to another), the work done by the force is

w=\vec F\cdot\vec d.

Since $\vec F\cdot\vec d=|\vec F||\vec d|\cos\theta$ , the work is positive when the force has an acute angle with the displacement, zero when perpendicular, and negative when obtuse. When several forces act together, first find their resultant (vector sum) and dot that resultant with the displacement (Example 6.9: forces $2\hat i+5\hat j+6\hat k$ and $-\hat i-2\hat j-\hat k$ resolve to $\hat i+3\hat j+5\hat k$ ; with displacement $\vec{AB}$ from $(4,3,-2)$ to $(6,1,-3)$ , the work comes out to $9$ units). If the displacement's endpoint carries an unknown (Example 6.10), setting $\vec F\cdot\vec d$ equal to the given work done yields a linear equation for the unknown.

Definition 6.3 (Torque/Moment). If a force $\vec F$ acts on a particle at a point with position vector $\vec r$ relative to the point about which the turning effect is measured, the torque (or moment, also called the rotational force) is

\vec\tau=\vec r\times\vec F.

Being a cross product, torque is a vector, perpendicular to the plane of $\vec r$ and $\vec F$ ; its magnitude $|\vec\tau|=|\vec r||\vec F|\sin\theta$ measures the turning strength, and its direction cosines (found from $\vec\tau/|\vec\tau|$ ) describe the axis about which the force tends to rotate the body — this is exactly the everyday idea of a merry-go-round or a wrench turning a bolt (Example 6.11: force $\hat i+2\hat j-\hat k$ through the origin, torque taken about $(2,0,-1)$ , giving $\vec\tau=-\hat i-2\hat k$ , magnitude $\sqrt5$ ). …

Figure 6.13Fig. 6.13 - Triangle ABC with D, E, F the midpoints of the sides forming the medial triangle DEF, whose area is one-fourth the area of ABC.

Fig. 6.13 — Fig. 6.13 - Triangle ABC with D, E, F the midpoints of the sides forming the medial triangle DEF, whose area is one-fourth the area of ABC.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig. 6.13 - Triangle ABC with D, E, F the midpoints of the sides forming the medial triangle DEF, whose area is one-fourth the a …

Figure 6.14Fig. 6.14 - Work done by a force F-hat acting at an acute angle, a right angle and an obtuse angle to the displacement d, giving positive, zero and negative work respectively.

Fig. 6.14 — Fig. 6.14 - Work done by a force F-hat acting at an acute angle, a right angle and an obtuse angle to the displacement d, giving positive, zero and negative work respectively.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig. 6.14 - Work done by a force F-hat acting at an acute angle, a right angle and an obtuse angle to the displacement d, giving positive, zero and negative wo …

Figure 6.15Fig. 6.15 - Torque as a rotational force t = r x F illustrated on a merry-go-round: r is the position vector (radius), F the applied force in the plane of the disc, and the torque t points along the vertical axis.

Fig. 6.15 — Fig. 6.15 - Torque as a rotational force t = r x F illustrated on a merry-go-round: r is the position vector (radius), F the applied force in the plane of the disc, and the torque t points along the vertical axis.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig. 6.15 - Torque as a rotational force t = r x F illustrated on a merry-go-round: r is the position vector (radius), F the applied force in the plane of the disc, and the torque t points al …

Figure 6.16Fig. 6.16 - Torque about the point A(2,0,-1) of a force F whose line of action passes through the origin O; the position vector r = AO points from A to O.

Fig. 6.16 — Fig. 6.16 - Torque about the point A(2,0,-1) of a force F whose line of action passes through the origin O; the position vector r = AO points from A to O.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Fig. 6.16 - Torque about the point A(2,0,-1) of a force F whose line of action passes through the origin O; the position vector r = AO points …

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