#### xaktly | Physics | Mechanics

Static equilibrium

### Forces in static equilibrium

This section is a collection of worked examples in static equilibrium, situations in which a system is static (not moving), so that there is no acceleration and therefore no net forces in any direction.

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)

### Example 1

A 10 Kg object is at rest on a ramp inclined at 21˚ from the horizontal, as shown. Calculate the minimum coefficient of friction necessary for the block not to slide down the ramp.

The free-body diagram for this situation looks like this.

The gravitational force vector (Fg = mg) pulls the box toward the center of Earth. We break that force into components parallel to and perpendicular to the ramp – that coordinate system is much more convenient to the problem, and we call these Fx and Fy. Fg is

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

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 Fx:

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

Now the force of friction on the box is just the normal force, which mirrors Fy (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

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

### Example 2

A mass sits on a ramp inclined at an adjustable angle θ to the horizontal. If the coefficient of friction between the mass and the ramp is μ, calculate the angle at which the mass will begin to slide down the ramp. This is called the critical angle.

The picture is similar to the last example, but we don't need to know the mass of the object, and the angle will be a variable.

The free-body diagram looks like this. By the way, sometimes my students ask about why I sometimes draw two Fx vectors. Roll over/tap the image to see why.

In order for the mass to slide, we need the force of gravity down the ramp (the x-component of Fg) to be just greater than the force of friction, which always opposes the motion:

$$F_x \gt F_f$$

Now the force of friction is the normal force (the opposite of Fy) multiplied by the coefficient of friction, μ.

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

Now if we use some trigonometry, we can express Fx and FN in terms of Fg 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 Fg (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).

### Example 3

Calculate the tension in the two wires suspending a 10 N object from a horizontal ceiling. One wire makes a 45˚ angle with the ceiling and the other a 30˚ angle.

Here's a picture of the scenario. The vertical wire attached to the weight isn't necessary, but the tension in that segment must be 10 N.

We can break those tension vectors down into vertical and horizontal components. We'll label each component vector with its side (1 or 2) and with the appropriate coordinate (x for horizontal or y for vertical).

I've moved the angle labels according to the properties of parallel lines.

#### Horizontal vectors

Now we know that our object is not moving, so it's not moving horizontally. Thus the net force in the horizontal direction must be zero, so we have two ways of looking at that:

\begin{align} F_{1x} + F_{2x} &= 0 \\[5pt] F_{1x} &= -F_{2x} \end{align}

Here's how we find the horizontal components of the tension:

\begin{align} cos(45˚) &= \frac{F_{1x}}{T_1} \: \color{#E90F89}{\longrightarrow} \: F_{1x} = T_1 cos(45˚) \\[5pt] cos(30˚) &= \frac{F_{2x}}{T_2} \: \color{#E90F89}{\longrightarrow} \: F_{2x} = T_2 cos(30˚) \end{align}

Using our second relation above (F1x = - F2x), we have

$$T_1 \, cos(45˚) = T_2 \, cos(30˚) \; \; \; \text{(1)}$$

#### Vertical vectors

Now because the system isn't accelerating up or down, either, we know that the sum of all upward forces has to equal the downward force:

$$F_{1y} + F_{2y} = 10 \; N$$

Finding those vertical components of the tension looks like this:

\begin{align} sin(45˚) &= \frac{F_{1y}}{T_1} \rightarrow F_{1y} = T_1 \, sin(45˚)\\[5pt] sin(30˚) &= \frac{F_{2y}}{T_2} \rightarrow F_{2y} = T_2 \, sin(30˚) \end{align}

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, F1x and F2x 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, F1y and F2y 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}

### Example 4

Repeat the problem above with an unspecified mass, but this time with equal angles of pull from the wires or ropes. Find a formula for the tension as a function of the angle.

Here is the drawing for this problem. The tension in either rope/wire should be the same.

The vector diagram looks like this.

The mass is not accelerating to the left or right, so the horizontal components of each tension vector must add to zero. The sum of the two vertical components must equal the downward force, Fg.

$$2F_y = F_g$$

Some trigonometry gives us Fy in terms of the tension, T, and the angle, μ:

$$sin(\theta) = \frac{F_y}{T_1} \; \color{#E90F89}{\longrightarrow} \; F_y = T_1 sin(\theta)$$

Now we can plug that version of Fy into the previous equality to get

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

Solving for the tension gives us:

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

Now let's analyze this solution. Notice that Fg 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 Fg) 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.

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