Question

Write the relationship between-
(i) kgf and newton
(ii) gf and dyne

Answer 1

Hint:All the given units are units of force. Kgf is the gravitational metric unit, and Newton is S.I. unit of force, gf is a metric unit of force, and dyne is the unit of force in a centimeter-gram-second system. We will be establishing the relationship between all the given units by converting them into the S.I. unit system.

Complete step by step answer:
(i) One kilogram-force (kgf) is the value of force due to gravity on a body of mass $$1\mathrm{k}\mathrm{g}$$. Mathematically, we can write:
$$1\mathrm{k}\mathrm{g}\mathrm{f}=mg$$……(1)
Here m is the mass of the body, and g is the acceleration due to gravity.
We know that the value of acceleration due to gravity is $$9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2$$.
Substitute $$9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2$$ for g and $$1\mathrm{k}\mathrm{g}$$ for m in equation (1).
$$\begin{gathered} 1\mathrm{k}\mathrm{g}\mathrm{f}=\left(1\mathrm{k}\mathrm{g}\right)\left(9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2\right) \\ \Rightarrow 1\mathrm{k}\mathrm{g}\mathrm{f}=9.81\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{k}\mathrm{g}\cdot \mathrm{m}{\mathrm{s}}^2\times \left({\displaystyle \frac{\mathrm{N}}{\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2}}\right)}{\mathrm{s}}^2\times \left({\displaystyle \frac{\mathrm{N}}{\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2}}\right) \\ \Rightarrow 1\mathrm{k}\mathrm{g}\mathrm{f}=9.81\mathrm{N} \end{gathered}$$
(ii) Gram force (gf) is the value of force due to gravity on a body of mass $$1\mathrm{g}$$.
$$1\mathrm{g}\mathrm{f}=m_{1}g$$
Here $$m_1$$ is the mass of the body.
Substitute $$9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2$$ for g and $$1\mathrm{g}$$ for $$m_1$$ in the above expression.
$$\begin{gathered} 1\mathrm{g}\mathrm{f}=\left(1\mathrm{g}\right)\left(9.81\mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2\right) \\ \Rightarrow 1\mathrm{g}\mathrm{f}=9.81\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2\times \left({\displaystyle \frac{\mathrm{k}\mathrm{g}}{1000\mathrm{g}}}\right) \\ \Rightarrow 1\mathrm{g}\mathrm{f}=9.81\times {10}^{-3}\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2\times \left({\displaystyle \frac{\mathrm{N}}{\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2}}\right) \\ \Rightarrow 1\mathrm{g}\mathrm{f}=981\times {10}^{-5}\mathrm{N} \end{gathered}$$……(2)
We know that the value of dyne in terms of newton is given as:
$$1\mathrm{d}\mathrm{y}\mathrm{n}\mathrm{e}={10}^{-5}\mathrm{N}$$
Substitute $$1\mathrm{d}\mathrm{y}\mathrm{n}$$ for $${10}^{-5}\mathrm{N}$$ in equation (2).
$$\begin{gathered} 1\mathrm{g}\mathrm{f}=981\times \left(1\mathrm{d}\mathrm{y}\mathrm{n}\right) \\ \therefore 1\mathrm{g}\mathrm{f}=981\mathrm{d}\mathrm{y}\mathrm{n} \end{gathered}$$

Therefore, the relation between one kgf is equal to $$9.81\mathrm{N}$$ , and one gf is equal to $$981\mathrm{d}\mathrm{y}\mathrm{n}$$.

Note: It would be better if we remember converting units of force into different units of units.There are various systems of units such as the centimetre-gram-second (CGS) system, foot-pound-second (FPS) system, and system Internationale (S.I.) system. But the S.I. the unit system is mostly followed by mathematicians and scientists.