# Mass vs. Weight

** Mass ** and ** Weight ** are two often misused and misunderstood terms in mechanics and fluid mechanics.

The fundamental relation between mass and weight is defined by Newton's Second Law . Newton's Second Law can be expressed as

F = m a(1)

where

F= force (N, lb_{f})

m= mass (kg, slugs )

a= acceleration (m/s^{2}, ft/s^{2})

### Mass

** Mass ** is a measure of the amount of material in an object, being directly related to the number and type of atoms present in the object. Mass does not change with a body's position, movement or alteration of its shape, unless material is added or removed.

- an object with mass
*1 kg*on earth would have the same mass of*1 kg*on the moon

Mass is a fundamental property of an object, a numerical measure of its inertia and a fundamental measure of the amount of matter in the object.

- mass electron
*9.1095 10*^{ -31 }kg - mass proton
*1.67265 10*^{ -27 }kg - mass neutron
*1.67495 10*^{ -27 }kg

### Weight

** **

** Weight ** is the ** gravitational force ** acting on a body mass. The generic expression of Newton's Second Law * (1) * can be transformed to express weight as a force by replacing the acceleration - * a * - with the acceleration of gravity - * g * - as

F_{ g }= m a_{ g }(2)

where

F= gravitational force - or weight (N, lb_{ g }_{f})

m= mass (kg, slugs (lb_{m}))

a_{ g }= acceleration of gravity on earth (9.81 m/s^{2}, 32.17405 ft/s^{2})

#### Example - The Weight of a Body on Earth vs. Moon

The acceleration of gravity on the moon is approximately * 1/6 * of the acceleration of gravity on the earth. The weight of a body with mass * 1 kg * on the earth can be calculated as

* F _{ g_ } _{ earth } = (1 kg) *

*(9.81 m/s*

^{2})* = 9.81 N *

The weight of the same body on the moon can be calculated as

* F _{ g_ } < _{ moon } = (1 kg) ( *

*(9.81 m/s*

^{2}) / 6)* = 1.64 N *

The handling of mass and weight depends on the systems of units used. The most common unit systems are

- the International System -
*SI* - the British Gravitational System -
*BG* - the English Engineering System -
*EE*

One newton is

- ≈ the weight of one hundred grams -
*101.972 gf (g*_{ F }) or 0.101972 kgf (kg_{ F }or kilopond - kp (pondus is latin for weight)) - ≈ halfway between one-fifth and one-fourth of a pound -
*0.224809 lb or 3.59694 oz*

### The International System - * SI *

In the SI system the mass unit is the * kg * and since the weight is a force - the weight unit is the * Newton * ( * N * ). Equation * (2) * for a body with * 1 kg * mass can be expressed as:

F_{ g }= (1 kg) (9.807 m/s^{2})

= 9.807 (N)

where

9.807 m/s^{2}= standard gravity close to earth in the SI system

As a result:

- a
*9.807 N*force acting on a body with*1 kg*mass will give the body an acceleration of*9.807 m/s*^{2} - a body with mass of
*1 kg*weights*9.807 N*

- More about the SI System - A tutorial introduction to the SI-system.

### The Imperial British Gravitational System - * BG *

The British Gravitational System (Imperial System) of units is used by engineers in the English-speaking world with the same relation to the * foot - pound - second * system as the * meter - kilogram - force second * system (SI) has to the * meter - kilogram - second * system. For engineers who deals with forces, instead of masses, it's convenient to use a system that has as its base units * length, time, and force *, instead of * length, time and mass *.

The three base units in the Imperial system are * foot, second * and * pound-force *.

In the BG system the mass unit is the * slug * and is defined from the Newton's Second Law * (1) * . The unit of mass, the * slug *, is derived from the pound-force by defining it as the mass that will accelerate with * 1 foot per second per second * when a * 1 pound-force * acts upon it:

* 1 lb _{f} = (1 slug) (1 ft/s^{2}) *

In other words, * 1 lb _{f} (pound-force) * acting on

*1 slug*of mass will give the mass an acceleration of

*1 ft/s*

^{2}.

The weight (force) of the mass can be calculated from equation * (2) * in BG units as

F_{ g }(lb_{f}) = m (slugs) a_{ g }(ft/s^{2})

With standard gravity - * a _{ g } = 32.17405 ft/s^{2}* - the weight (force) of

*1 slug*mass can be calculated as

* F _{ g } = (1 slug) ( 32.17405 ft/s^{2}) *

* 32.17405 lb _{f} *

### The English Engineering System - * EE *

In the English Engineering system of units the primary dimensions are are * force, mass, length, time and temperature. * The units for force and mass are defined independently

*the basic unit of mass is**pound-mass**(lb*_{m})*the unit of force is the**pound**(lb) alternatively**pound-force**(lb*_{f}).

In the EE system * 1 lb _{f} of force * will give a mass of

*1 lb*a standard acceleration of

_{m}*32.17405 ft/s*.

^{2}Since the EE system operates with these units of force and mass, the Newton's Second Law can be modified to

F = m a / g_{ c }(3)

where

g_{ c }= a proportionality constant

or transformed to weight (force)

F_{ g }= m a_{ g }/ g_{ c }(4)

The proportionality constant g _{ c } makes it possible to define suitable units for force and mass. We can transform (4) to

1 lb_{f}= (1 lb_{m}) (32.174 ft/s^{2}) / g_{ c }

or

g_{ c }= (1 lb_{m}) (32.174 ft/s^{2}) / (1 lb_{f})

Since * 1 lb _{f} * gives a mass of

*1 lb*an acceleration of

_{m}*32.17405 ft/s*and a mass of

^{2}*1 slug*an acceleration of

*1 ft/s*, then

^{2}

1 slug = 32.17405 lb_{m}

### Example - Weight versus Mass

The mass of a car is * 1644 kg * . The weight can be calculated:

F_{ g }= (1644 kg) (9.807 m/s^{2})

= 16122.7 N

= 16.1 kN

- there is a force (weight) of * 16.1 kN * between the car and the earth.

*1 kg gravitation force = 9.81 N = 2.20462 lb*_{f}

### Weight Converter

### Kg to lb Converter

## Related Topics

### • Basics

Basic engineering data. SI-system, unit converters, physical constants, drawing scales and more.

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### • Mechanics

The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.

### • Statics

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