Progressive Measurement and Vector Examples

Practice these examples in order: first build unit consistency and precision habits, then apply component-based vector methods to engineering-style force, displacement, work, and torque problems.

Field Tape Reading to SI Length

A steel tape gives a beam length of 15.5 ft15.5 \text{ ft}. Convert the measurement to meters using 1 ft=12 in1 \text{ ft} = 12 \text{ in}, 1 in=2.54 cm1 \text{ in} = 2.54 \text{ cm}, and 100 cm=1 m100 \text{ cm} = 1 \text{ m}.

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Construction Material Density Conversion

Normal-weight concrete is listed as 145. lb/ft3145. \text{ lb/ft}^3. Convert the density to kg/m3\text{kg/m}^3 using 1 lb=0.453592 kg1 \text{ lb} = 0.453592 \text{ kg} and 1 ft=0.3048 m1 \text{ ft} = 0.3048 \text{ m}.

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Force in Base SI Units

A cart with mass 12.0 kg12.0 \text{ kg} accelerates at 2.50 m/s22.50 \text{ m/s}^2. Find the force in newtons and show the equivalent base SI units.

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Dimensional Consistency of Work

Verify that the work relationship W=FdW = Fd is dimensionally consistent when FF is force and dd is displacement.

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Significant Figures in Measurements

Determine the number of significant figures in these measurements: (a) 0.00405 m0.00405 \text{ m}, (b) 120.0 kg120.0 \text{ kg}, and (c) 1.500×103 s1.500 \times 10^3 \text{ s}.

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Significant Figures in Area Density

A rectangular steel plate measures 4.52 cm4.52 \text{ cm} by 2.1 cm2.1 \text{ cm} and has a mass of 75.5 g75.5 \text{ g}. Calculate its area and area density.

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Percent Error in a Calibration Check

A survey rod segment with accepted length 2.000 m2.000 \text{ m} is measured as 1.998 m1.998 \text{ m}. Calculate the percent error of the measurement.

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Rectangular Components of a 250 N Force

A force of 250 N250 \text{ N} acts at 35.0∘35.0^\circ above the positive xx-axis. Find its xx- and yy-components.

Rectangular Components of a 250 N Force Diagram

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Resultant of Two Force Vectors

Two forces act at a joint: F⃗1=120 N\vec{F}_1 = 120 \text{ N} due east and F⃗2=80.0 N\vec{F}_2 = 80.0 \text{ N} at 30.0∘30.0^\circ north of east. Find the resultant magnitude and direction.

Resultant of Two Force Vectors Diagram

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Magnitude and Direction from Unit-Vector Form

A displacement vector is d⃗=(−6i^+8j^+3k^) m\vec{d} = (-6\hat{i} + 8\hat{j} + 3\hat{k}) \text{ m}. Find its magnitude and write the corresponding unit direction vector.

Magnitude and Direction from Unit-Vector Form Diagram

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Work from the Dot Product

A force F⃗=(5i^+2j^−3k^) N\vec{F} = (5\hat{i} + 2\hat{j} - 3\hat{k}) \text{ N} moves an object through d⃗=(3i^+4j^+1k^) m\vec{d} = (3\hat{i} + 4\hat{j} + 1\hat{k}) \text{ m}. Calculate the work done.

Work from the Dot Product Diagram

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Torque from the Cross Product

A force F⃗=(2i^−3j^+4k^) N\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N} is applied at position r⃗=(3i^+2j^+1k^) m\vec{r} = (3\hat{i} + 2\hat{j} + 1\hat{k}) \text{ m} from a pivot. Find the torque τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}.

Torque from the Cross Product Diagram

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