Vectors and Scalars — Question 6

Question 6

Printed arrow length must be interpreted through its scale

Two calibrated diagrams use parallel north references. In diagram I, an eastward displacement arrow measures 3.0 cm on the original paper and the scale is 2.0 m per centimetre. In diagram II, an eastward displacement arrow measures 6.0 cm and the scale is 1.0 m per centimetre. The measurements and scales are exact.

A third diagram shows an eastward velocity arrow 3.0 cm long at a scale of 2.0 m/s per centimetre. The sketches below are schematic; use the stated paper measurements rather than measuring this page.

1. Recover all three represented magnitudes and identify which physical vectors can be equal.

2. Diagram I is photocopied at 150% linear size, including its labels. Find the new printed arrow length and the correct new scale in metres per centimetre. What error results if the old written scale is used unchanged?

3. Why would a drawn scale bar copied together with the arrow remain useful, while a literal “2.0 m per cm” label becomes wrong after resizing?

4. Convert the first displacement magnitude to physical centimetres. Explain why changing physical units does not change the vector, while confusing paper centimetres with physical centimetres does.

Question 6 - Solution

Printed arrow length must be interpreted through its scale

(a) Multiply paper length by the relevant calibration.

$$D_I=(2)(3)=6\,\mathrm m=D_{II}=(1)(6),\quad\boxed{\vec D_I=\vec D_{II}},\quad v=(2)(3)=6\,\mathrm{m/s}$$ The first two arrows are equal physical displacements despite their different paper lengths. The third is a velocity, so it cannot equal either displacement even with the same numerical magnitude.

(b) Resizing the symbol requires inverse resizing of its scale.

$$\ell'=1.5(3)=4.5\,\mathrm{cm},\quad k'=\frac6{4.5}=\frac43\,\mathrm{m/cm},\quad\frac{9-6}{6}=50\%$$ The enlarged arrow still represents 6 m. Keeping the old calibration would report 9 m, an excess of 3 m or 50 percent.

(c) Relative drawn lengths survive a common enlargement.

$$\frac{\ell'_{\mathrm{arrow}}}{\ell'_{\mathrm{bar}}}=\frac{1.5\ell_{\mathrm{arrow}}}{1.5\ell_{\mathrm{bar}}}=\frac{\ell_{\mathrm{arrow}}}{\ell_{\mathrm{bar}}}$$ A copied scale bar and copied arrow enlarge together, preserving their length ratio. A numerical label per actual ruler centimetre must instead be updated.

(d) Separate unit conversion from drawing calibration.

$$6\,\mathrm m\left(\frac{100\,\mathrm{cm}}{1\,\mathrm m}\right)=600\,\mathrm{cm}\ne3\,\mathrm{cm}\text{ on paper}$$ Physical unit conversion preserves the displacement. Paper centimetres describe the symbol and must be multiplied by its calibration before comparing with physical lengths.