Mechanics 101, part 2

In my last post, we covered some basic concepts and definitions in mechanics. Now, let's look at using some of these concepts in the context of statics. This is mainly to act as a primer for some of the more complex analyses which will come in later posts.

Simply Statics

 In statics, we're mainly concerned with the reactions (forces and moments) that result from applying a load (either a force or torque) to a rigid object. In this context, rigid means completely unable to deform regardless of the applied loads.

Let's start with possibly the simplest of static systems: a rigid ball sitting on a rigid surface. Because the ball and table are rigid, they only touch at a single infinitesimally small point.

Yep, it's a ball on a table.

 Let's say our ball has a weight, W. We can draw a schematic for the forces acting on the ball, called a free-body diagram (FBD). We can represent the weight of the ball as a downward force acting on the contact point, because the contact point is directly below the center of mass of the ball (for a sphere of uniform density, the center of the sphere is also the center of mass). The influence of the table can be represented simply as an upward reaction force, F\(_R\) acting on the contact point. We choose to use a set of reference axes such that the positive y-direction is upward and the positive x-direction is to the right.

Free-body diagram for the ball on the table.

Using the principles of statics, specifically static mechanical equilibrium, we can determine what the reaction force is in terms of the ball's weight. Static mechanical equilibrium means that the sum of all forces and moments acting on the body is zero. Because all of the forces act through the center of mass of the ball, we have no moments,  and since we have only vertical forces we only need to sum forces in that direction:
\[\sum{F_y} = -W  + F_R = 0\]
Therefore:
\[F_R = W\]
In other words, the magnitude of the reaction force is identical to the weight of the ball, but in the opposite direction. Had we gotten the assumed direction of the reaction force in our FBD wrong, we would have ended up with a sign difference.
This is a fairly trivial example, but still shows the basic process. Also, it is a demonstration of Newton's Third Law: for every force applied in a system in mechanical equilibrium, there is a reaction force of the same magnitude but opposite direction.

Let's move onto a more complicated problem: A simply supported rigid beam with an applied load. Our beam, of length L, and weight  W, has a force P applied at its midpoint (also the center of gravity):

Simply supported beam with applied load P

So what is the force at the two supports? The process is similar to the ball on the table:

• Draw free-body diagram
• Sum forces and moments, set the sums equal to 0.
• Solve for unknowns (profit!)

Our FBD is as follows:

FBD for simply supported beam

Which leads to the following equilibrium equation for the forces (again, only forces in the y-direction):

\[\sum{F_y} = F_1 + F_2 - W - P = 0\]

We have some freedom in choosing where we want to place the reference axis for the moments acting on the beam. One convenient place is at the center of the beam in the z-direction (out of the page), since the forces W &  P  would then not produce any moments (their moment arm or distance to the axis is zero), so we could then get an equation that only involved \(F_1\) and \(F_2\): 

\[\sum{M_z}\bracevert_{x=\frac{L}{2}} = -F_1 \frac{L}{2} + F_2 \frac{L}{2} = 0\]

Where we've used the standard right hand rule for assigning a direction to the moments. The rule works as follows: wrap the fingers of your right hand in the direction of the rotation the moment would cause, and then extend your thumb away from your hand. Your thumb indicates the 'direction' of the moment. In our example, since the positive z-axis is pointing out of the page, moments which act to cause a counter-clockwise rotation are positive. Clockwise rotation would be a negative moment.

We now have two independent equations for two unknowns, therefore we can solve these equations for the reaction forces:
\[\sum{M_z}\bracevert_{x=\frac{L}{2}} = F_2 \frac{L}{2} - F_1 \frac{L}{2} = 0\]
\[F_2 = F_1\]
\[\sum{F_y} = F_1 + F_2 - W - P = 0\]
\[ 2F_1 = W + P \rightarrow F_1 = F_2 = \frac{(W+P)}{2}\]
 So the reaction forces at the supports are both equal to half of the total load (P+W). Had the applied load P been offset closer to one support, that support's force would end up being higher. Try it out, and don't forget that the weight itself will cause a moment relative to P. I'll give you a hint: If you move the load P to be at position L/4 (closer to \(F_1\)), the relationship between the reaction forces is:
 \[F_2 = \frac{(F_1 + W)}{3}\]

One last static example: a tight-rope walker. What can we learn about the rope they are standing on from statics?

Let's first assume they simply exert their weight at a point and for simplicity let's place them at the center of the rope. Their weight displaces the center of the rope by a distance d. The rope is held between two rigid supports separated by a length L. For now, we'll ignore the weight of the rope. Let's determine the force in the rope, then the forces at the supports.

Tight-rope walker. Note the high level of engineering artwork skill.

 Following the same steps, we first construct our FBD for the rope:

Tight-rope FBD

Where the vectors labeled T represent the tensile force (a force tending to cause elongation) in the rope. Because of the angle formed because of the displacement of the center of the rope, these tensile forces don't act in the same direction as the axes (as in our previous example). One way to deal with this is to redefine your reference axes to make the forces align more conveniently. However, this isn't always easy and so the general approach is to decompose the forces into components along each of the reference axes:

Decomposition of tensile force in left half of rope.

The angle \(\Theta\) is the angle of the rope relative to the horizontal (\(\Theta = \tan^{-1}{\left(\frac{d}{L/2}\right)}\)), and \(T_x\) and \(T_y\) represent the components of the tensile force along the x and y directions respectively. Now we can sum forces in each direction on the entire rope:
\[\sum{F_x} = -T_x + T_x = 0\]
\[\sum{F_y} = T_y + T_y - W = 0 \rightarrow T_y = \frac{W}{2}\]
Not surprisingly, the x-components of the tensile force cancels out (the rope isn't moving), though we don't yet know the magnitude of \(T_x\). The y-component of the tensile force is equal to half of the walker's weight, since both halves of the rope carry an equal portion of their weight. But what about the force along the rope, the tensile force T?

From the force decomposition, and some trigonometry we can write the tensile force in terms of the angle and the y-component:
\[T = \frac{T_y}{\sin{\Theta}} = \frac{W}{2 \sin{\Theta}}\]
Similarly, we can determine the magnitude of the x-component from the angle and the tensile force:
\[T_x = T\cos{\Theta}\]

There are some interesting implications here:

• For small angles (equivalent to small deflections of the rope), the force in the x-direction is significantly larger than the component in the y-direction. This means that the majority of the tensile force is actually acting to hold the rope between the supports, not to directly support the tight-rope walker's weight. 
• If the angle is zero, then the tight-rope walker's weight must also be zero. In other words, we've assumed the rope can't support a vertical load without a deflection. We've also assumed that the ends of the rope are free to rotate to accommodate the angle change. So statics isn't always about things that don't change shape.

So how about the supports? Since the system is symmetric, we'll only look at the left support. Using a method called the method of sections, where we separate our entire system into parts that are easier to analyze individually by drawing arbitrary dividing lines between the sections. For this example, we choose to section just off of the left support. From Newtons's third law, we know that at the section boundaries, the forces acting on each section must be equal in magnitude but opposite in direction. In other words, within each section we have equilibrium. The section for the left support looks like:

Section for left support.

Physically, this represents a beam that is fixed (can't move or rotate) at the bottom with the attached rope at the top. The part that is fixed can support forces and rotations in all directions. We'll run into this type of boundary condition again. The FBD for the left support is:

FBD for left support, with tensile force decomposed.

We can then write out the system of equilibrium equations:
\[\sum{F_x} = T_x - F_2 = 0 \rightarrow F_2 = T_x\]
\[\sum{F_y} = -T_y + F_1 = 0 \rightarrow F_1 = T_y\]
\[\sum{M_z\bracevert_{y=0}} = M - T_{x}h = 0 \rightarrow M = T_{x}h \]
We can substitute in for \(T_x\) and \(T_y\) in terms of \(T\) and \(\Theta\) to get the reaction forces at the base of the support:
\[F_1 = T \sin{\Theta}\]
\[F_2 = T \cos{\Theta}\]
\[M = (T \cos{\Theta})h\]
We'll explore what can happen at this support more in a later post when we focus on the basics of mechanics of materials and allow it to deform. It will actually be a rather useful system.

Now that we've covered some examples of statics, in the next post we'll look at examples of dynamics!