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Introduction to Marine Engineering

Subsection 2.4.5 Mass vs. Weight

One important force which we experience every day is weight. A common misconception is that mass and weight are the same. They are not: weight is a force, not a mass.
Figure 2.4.3. Weight vs. Mass
Weight is the force of gravity - the attraction between an object and the earth. The force exerted on an object is proportional to the mass of the object according to Newton’s second law, which states:
\begin{equation} W=mg\text{,}\tag{2.4.1} \end{equation}
where \(W\) is weight, \(m\) is mass, and \(g\) is the local acceleration due to gravity. Note that the weight of an object depends on both its mass and the acceleration of gravity acting on it.
The value of g depends on where you are: on earth the constant is greater than it is on the moon, so you weigh less on the moon than you do on earth. In outer space, the constant is zero, making you weightless there. The mass of an object will not change when it moves to another location, but its weight may. For example, on the earth’s surface, a one kilogram mass weighs about 2.2 pounds, but on Mars, the same mass would weigh only about 0.8 pounds, and on Jupiter it would weigh roughly 5.5 pounds.
One reason why people often confuse weight with mass is because in the English system and on the surface of the earth, the gravitational constant g is equal to 1, and so one pound of mass has a weight of one pound of force. Move to the moon, and the pound of mass now weighs about six tenths of a pound. On earth in the SI system,
\begin{equation*} g = \aSI{9.8}\text{.} \end{equation*}
Note the units of \(g\text{.}\) This is a metric quantity; appropriate unit conversions must be made when working with US units. See the following examples.

Example 2.4.4. Weight 1.

What is the mass in kg of an object that weighs 150 N on Earth?
Answer.
The object has a calculated mass of 15.3 kg.
Solution.
The problem supplies the weight
\begin{equation*} w = \N{150} \end{equation*}
and its location, Earth, which has a gravitational acceleration of
\begin{equation*} g = \aSI{9.8}\text{.} \end{equation*}
Solving (2.4.1) for mass and substituting gives:
\begin{align*} m \amp= W/g\\ \amp= \frac{\N{150}}{\aSI{9.8}}\\ \amp= 15.3 \frac{\mathrm{N}}{\mathrm{m}/\mathrm{s}^2}\\ \amp= 15.3 \frac{\left[\mathrm{kg\ m}/\mathrm{s}^2\right]}{\mathrm{m}/\mathrm{s}^2}\\ \amp= \kg{15.3} \end{align*}
Note that we used the conversion
\begin{equation*} \N{1} = 1 \mathrm{kg\ m}/\mathrm{s}^2\text{.} \end{equation*}

Example 2.4.5. Weight 2.

What is the mass of in kg an object that weighs \(\lb{175}\) on Earth?
Answer.
The object has a calculated mass of 79.4 kg.
Solution.
Since the weight is given in pounds-force, first convert it to newtons and then proceed as in the previous example.
\begin{equation*} w= \lbf{175} \times \left[\frac{\N{4.4482}}{\lbf{1}}\right] = \N{778.4} \end{equation*}
\begin{equation*} m = W/g = \frac{\N{7778.4}}{\aSI{9.8}}= \kg{79.43} \end{equation*}
A person who weighs 175 pounds on a scale, experiences a force of approximately 778.4 Newtons, has a calculated mass of 79.4 kilograms.