Question

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

Answer 1

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 $1{\rm{ kg}}$. Mathematically, we can write:
$1{\rm{ kgf}} = mg$……(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 $9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$.
Substitute $9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$ for g and $1{\rm{ kg}}$ for m in equation (1).

$$1{\rm{ kgf}} = \left( {1{\rm{ kg}}} \right)\left( {9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right)\\\ \Rightarrow 1{\rm{ kgf}} = 9.81{{{\rm{ kg}} \cdot {\rm{m}}} {\left/ {\vphantom {{{\rm{ kg}} \cdot {\rm{m}}} {{{\rm{s}}^2} \times \left( {\dfrac{{\rm{N}}}{{{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}}}} \right)}}} \right. } {{{\rm{s}}^2} \times \left( {\dfrac{{\rm{N}}}{{{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}}}} \right)}}\\\ \Rightarrow 1{\rm{ kgf}} = 9.81{\rm{ N}}$$

(ii) Gram force (gf) is the value of force due to gravity on a body of mass $1{\rm{ g}}$.
$1{\rm{ gf}} = {m_1}g$
Here ${m_1}$ is the mass of the body.
Substitute $9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}$ for g and $1{\rm{ g}}$ for ${m_1}$ in the above expression.

1{\rm{ gf}} = \left( {1{\rm{ g}}} \right)\left( {9.81{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right)\\\ \Rightarrow 1{\rm{ gf}} = 9.81{\rm{ g}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}} \times \left( {\dfrac{{{\rm{kg}}}}{{1000{\rm{ g}}}}} \right)\\\ \Rightarrow 1{\rm{ gf}} = 9.81 \times {10^{ - 3}}{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}} \times \left( {\dfrac{{\rm{N}}}{{{\rm{kg}} \cdot {{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}}}} \right)\\\ \Rightarrow 1{\rm{ gf}} = 981 \times {10^{ - 5}}{\rm{ N}}$$……(2) We know that the value of dyne in terms of newton is given as: $$1{\rm{ dyne}} = {10^{ - 5}}{\rm{ N}}$$ Substitute $$1{\rm{ dyn}}$$ for $${10^{ - 5}}{\rm{ N}}$$ in equation (2).

1{\rm{ gf}} = 981 \times \left( {1{\rm{ dyn}}} \right)\\
\therefore 1{\rm{ gf}} = 981{\rm{ dyn}}

**Therefore, the relation between one kgf is equal to $$9.81{\rm{ N}}$$ , and one gf is equal to $$981{\rm{ dyn}}$$.** **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.