**Newton’s 1 ^{st} law** of motion is often called the

**Law of Inertia**. Inertia is a property of matter that opposes any change in motion.

**Newton’s 1 ^{st} law** can be stated as such: an object at rest will remain at rest, an object in motion will continue moving at a constant speed in a straight line [unless acted on by some unbalanced (net) force]. Newton’s 1

^{st}law of motion describes what occurs when no unbalanced force acts on an object.

If the object is at rest to begin with, it will remain in static equilibrium. The study of bodies at rest is called **statics**. It is a subdivision of the larger category of **dynamics**, which is a subcategory of **mechanics**.

For an object to remain at rest or to continue its motion undisturbed, the sum of the forces acting on it must equal zero. Notice, it is not a requirement that NO forces be acting on it just that their sum = 0.

Σ F = 0

[NOTE: = Greek letter sigma represents the sum of]

Here is an example of a statics problem.

For the diagram below, construct a free body diagram that consists of W, T_{1} and T_{2}. Assign an angle for T_{1} with respect to the x-axis and an angle for T_{2} with respect to the x-axis

Image from http://mypages.iit.edu/~smart/acadyear/statics.htm

For example purposes, let’s say that W = 500. N and that angle 1 is 30^{o} and angle 2 is 50^{o}. Solve for the values of T_{1} and T_{2}.

Because the system is at rest (or moving with constant velocity), the system is in equilibrium.

Therefore F_{netx} = 0 and Right Forces = Left Forces

T_{1} cos 30 = T_{2} cos 50; 0.866 T_{1} = 0.643 T_{2}; T_{1} = 0.742 T_{2}

This is the best that it gets for the x-axis since there are two unknowns. Luckily, we have the y-axis to use for our second equation.

F_{nety} = 0 and Top Forces = Bottom Forces

T_{1} sin 30 + T_{2} sin 50 = W sin 90

0.500 T_{1} + 0.766 T_{2} = 500. N (Eq 1)

Substitution of T_{1} = 0.742 T_{2} into the equation yields…

0.500 (0.742 T_{2}) + 0.766 T_{2} = 500. N

0.371 T_{2} + 0.766 T_{2} = 500. N

1.137 T_{2} = 500. N

T_{2} = 440. N

Substitution into (Eq 1) yields ..

T_{1} = 440N (0.742) = 326 N

Notice that because T_{2} has a greater vertical component, it supports more of the weight. The x-component of a tension force is wasted when supporting a vertical weight!

(source)