Often in physics and engineering, mechanics is divided into static systems ("statics") and moving or dynamic systems ("dynamics").

I'll add more examples over time. (12/8/17)

$$F_g = mg = 10\; Kg (9.8 \,m\cdot s^{-2}) = 98 \; N$$

The force of gravity pulling the box down the ramp is:

$$F_x = F_g \, sin(21˚) = 35.12 \; N$$

and the force vector pulling it into the ramp is:

$$F_y = F_g \, cos(21˚) = 91.49 \; N$$

For a ramp of angle less than 45˚, we expect more force into the ramp than down it.

Now we require that for the box not to move, the force of friction (which always opposes the motion or proposed motion) must be at least equal to F_x:

$$F_f \ge (F_x = 35.12 \; N)$$

Now the force of friction on the box is just the normal force, which mirrors F_y (Newton's third law), multiplied by the coefficient of friction, a unit-less quantity specific to the particular interface – in this case the interface between the surfaces of the block and ramp. We can rearrange to solve for μ.

$$F_f = \mu F_N \; \longrightarrow \; \mu = \frac{F_f}{F_N}$$

The coefficient is

$$\mu = \frac{35.12 \; N}{91.49 \; N} = 0.384$$

$$F_x \gt \mu \cdot F_N$$

Now if we use some trigonometry, we can express F_x and F_N in terms of F_g and θ:

$$F_x = F_g \, sin(\theta) \; \; \text{and} \; \; F_N = F_g \, cos(\theta)$$

Plugging those into the inequality gives:

$$F_g \, sin(\theta) \gt \mu \, F_g \, cos(\theta)$$

Now we see that F_g (and consequently the mass of the object) doesn't matter:

$$F_g \, sin(\theta) \gt \mu \, F_g \, cos(\theta)$$

Dividing both sides by cos(θ) gives

$$\frac{sin(\theta)}{cos(\theta)} \gt \mu$$

which gives us this very simple relation for the critical angle.

$$tan(\theta) \gt \mu$$

We could also write that as θ > tan^-1(μ). Here is a table of some critical angles as a function of mu.

Note: We should note here that we're referring to the coefficient of static friction. (See friction).

Plugging those into our vertical-components relation above gives us our second foundation equation:

$$T_1 \, sin(45˚) + T_2 \, sin(30˚) = 10 \; \; \; \text{(2)}$$

OK. Now we've got to find two tensions, and we've got two equations to do it. Let's take (1) and solve it for T₂:

$$T_2 = \frac{-T_1 \, cos(45˚)}{cos(30˚)}$$

Now we'll take that value of T₂ and plug it into (2):

$$T_1 \, sin(45˚) - \frac{T_1 \, cos(45)sin(30)}{cos(30)} = 10$$

We want to solve for T₁, which is in two terms, so we'll remove it as a factor:

$$T_1 \left( sin(45) - \frac{cos(45)sin(30)}{cos(30)} \right) = 10$$

All of the stuff in parentheses is a constant, so we'll just divide by it on both sides:

$$T_1 = \frac{10}{\left( sin(45) - \frac{cos(45)sin(30)}{cos(30)} \right)}$$

That gives

$$T_1 = 33.46 \; N$$

Now we can go back to our expression for T₂ in terms of T₁, plug in T₁ to get:

$$T_2 = \frac{-T_1 cos(45)}{cos(30)} = \frac{-33.46 cos(45)}{cos(30)}$$

Working out the arithmetic gives us both tensions:

$$T_2 = -27.32 \; N$$

Notice that (a) they're different, and (b) the wire with the larger angle (most straight) has the most tension. Also notice that the tension here is negative. That's a result of the fact that the component vectors are signed. We can regard the tension as positive in both wires because one isn't really pulling the the opposite direction as the other. We'll need the negative tension in what follows, though.

We should check these tensions and make sure that all of the forces are what we think they should be. The horizontal forces, F_1x and F_2x should sum to zero:

$$ \begin{align} F_{1x} &= T_1 \, cos(45˚) \\[5pt] &= 33.46 \, cos(45˚) = 23.66 \; N \\[5pt] F_{2x} &= T_2 \, cos(30˚) \\[5pt] &= -27.32 \, cos(30˚) = -23.66 \; N \\[5pt] &\; \; \rightarrow \bf F_{1x} + F_{2x} = 0 \end{align}$$

... and they do. The vertical forces, F_1y and F_2y should sum to the downward force of 10 N,

$$ \begin{align} F_{1y} &= T_1 \, sin(45˚) \\[5pt] &= 33.46 \, sin(45˚) = 23.66 \; N \\[5pt] F_{2y} &= T_2 \, sin(30˚) \\[5pt] &= -27.32 \, sin(30˚) = -33.66 \; N \\[5pt] &\; \; \rightarrow \bf F_{1y} + F_{2y} = 10 \; N \end{align}$$

$$2T sin(\theta) = F_g$$

Solving for the tension gives us:

$$T = \frac{F_g}{2 sin(\theta)}$$

Now let's analyze this solution. Notice that F_g and 2 are constants, so the variable expression is the sin(υ) in the denominator. Now as the angle theta approaches zero, so does the sine function, so the tension becomes infinite.

Here is a plot of the tension over angles from 90˚ (ropes hanging straight down and equally sharing half of F_g) and 0˚. Notice how rapidly the tension grows as the angle shrinks.

This is why we can never observe a wire stretched to a perfectly-horizontal line between two power poles.