## Statics. What You Will Learn. Motivation. Forces and Moments

Statics What You Will Learn    What forces and moments are, how they are different, and how they are related. How to draw a Free Body Diagram (FBD...
Author: Lester Rose
Statics What You Will Learn   

What forces and moments are, how they are different, and how they are related. How to draw a Free Body Diagram (FBD), and why you should for every problem. How to add forces and moments, and what you can learn by doing so.

Motivation The study of statics might seem more important for civil engineers than mechanical engineers; after all, mechanical engineers make things that move, so why bother studying how to analyze things that stand still? The fact is that not only do the tools of statics provide an excellent stepping stone to studying dynamics, fluid mechanics, and other mechanical discipline, they are useful in every day practice in their own right. In this section we will start by introducing the concepts of Forces and Moments. Nearly every facet of mechanical engineering depends on these concepts. We will then learn to draw Free Body Diagrams (FBD). These idealizations of the problem at hand are the first step in solving nearly any engineering problem. We will finish by showing how to add forces and moments together and what we can learn about real life problems by doing so.

Forces and Moments Two simple examples can be used to demonstrate the difference between these the concepts. All three represent the action of one body on another, but the type of action is different. 1) Force – The more straightforward of the two. Its meaning in engineering is closely aligned with the meaning in everyday use. The hammer in the picture below is applying a force to the nail. The force can be represented as a vector with the same direction as the motion of the head of the hammer at the time of impact.

Question: Any other examples of forces? (weight of an object, tapping a key) 1|Statics

2) Moment – Newton’s 2nd law is usually thought of as applying to straight line motion. It can be applied to rotational motion as well. Rather than linear force, we require the concept of a Moment. For now, think of it as rotational force. The Moment about a particular axis can be found using the following equation: 𝑀 = 𝑟×𝐹 Where r is a vector that connects the points where the force is applied to the axis of rotation. The nut in the figure at right has a moment applied to it by the spider wrench. It is important to always designate the point about which the moment is taken, that is what is the center of the rotation? Harder Question: Any other examples of moments? [Squeezing orange halves to make juice. Turning a doorknob. Turning the wheel of a car ] Example – Calculate a moment. The mechanic in the picture above is applying 50N with each hand, and his hands are 0.2m from the axis of the wrench. The nut is at point A F

A 0.2m

0.2m

F

The moment is: Σ𝑀 = 2 ⋅ 𝐹 × 𝑟 = 2 ⋅ 50 ⋅ 0.2 = 20 𝑁 ⋅ 𝑚 There is an important simplification in this problem: both forces were perpendicular to the r vectors. If they weren’t, we’d find the component of the vector that was perpendicular first, and then find the moment. (or we’d use the cross product, but we haven’t had time to show that to them)

Drawing Free Body Diagrams Drawing a FBD has many benefits:  It helps you to clearly identify what you know and don’t know about a problem  It helps you to clearly state your assumptions  A FBD makes it easier for others to follow your work  A FBD makes it easier for future-you to follow your work if you have to review it six months, or six years after you do it. In industries where safety and government regulations are extremely important, such as aviation or nuclear power, it is not uncommon to review 25 year old hand calculations in the event of a proposed design change or an accident.

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The procedure can be summarized as follows: 1) Select which parts can must have their geometry included in the sketch, and which parts can have their impact on the problem represented as forces or moments. This process is sometimes visualized as drawing a circle around the problem, so that things inside the circle a represented as geometry, and things outside the circle are represented as forces applied to the parts inside the circle 2) Sketch the geometry inside the circle. 3) Draw arrows to show where forces and moments are being applied, and label them with the magnitudes of the forces. Example – Box on a slope Step 1 – Draw the boundary. We won’t analyze the slope, we want know what is happening to the box. Perhaps we want to determine if it will slide, or how much force would be needed to make it start moving. Step 2 – Draw the geometry.

𝜽

Step 3 – Draw the vectors for any and all possible forces.

W Ff N The weight of the box can be modeled as the weight of the box applied to the middle of the box, assuming that the density of the contents of the box are uniform. Weight is equal to the product of mass and gravity: 𝑊 = 𝑚𝑔. There are two other forces. The first is the normal force. The normal force is a reaction force, and it always acts perpendicular to the surface. Finally, there will be a friction force, tangent to the surface. The direction will be opposite to motion or impending motion.

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Example – Bridge Assume  Support for the base at 2 points on each end  The bridge is so narrow that the cars drive down the middle, again leaving us with a symmetrical part. This is not such a big assumption. The method is easily extended to 3D, but is much harder to draw on a whiteboard, so just go with it.  A car is driving along the road, and it is accelerating, which puts a force on the road in the opposite direction. It is front wheel drive.

Step 1 – Drawing our boundary. We won’t analyze the road. While it may move more than the patio from the previous example, we’re probably interested in how well we have to attach the bridge to the ground, so drawing our circle around that interface allows us to see what forces are being applied there. Step 2 – Drawing the geometry. Since there is a plane of symmetry, draw it from the side. Show the road support connections with little triangles.

𝐹𝐴𝑦

𝐹𝐵𝑦

𝐹𝐴𝑥 𝑊𝐶 𝐹𝐶𝑥

𝑊𝐷

W

5m 7m 10m 20m Step 3 – Draw the vectors for any and all possible forces. The weight of the car is spread over two wheels. The acceleration force is applied to the front wheel. The weight of the bridge is represented as a force in the middle. The connections to the road we’ll show as three different reaction forces. One each in the x and y directions on the left, but only in the y direction to the right. This is a VERY common assumption in academic problems. The problem is much easier to solve, and not much information is lost. You’ll see why in a little while. Also note that we assumed directions for the forces at A and B. If we guessed wrong, the forces we find will be negative.Image the bridge without cars, traffic or wind. It is just sitting there. Often we draw two diagrams: 1 for the geometry, and 1 for the forces. Also note that all dimensions start from a common origin.

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Example – Diving Board Case where diver stands motionless on end of board before the dive. Assume  Support for the base at 4 bolts  The diver stands in the middle of the board, at the end, so there is a plane of symmetry through the board – the force on each pair (front and back) of the bolts will be the same. Step 1 – Drawing our boundary. We can ignore the patio. It won’t move much compared to the diving board. Also, we’re probably interested in how strongly we have to attach the base to the ground, so drawing our circle around that interface allows us to see what forces are being applied there.

Step 2 – Drawing the geometry. Since there is a plane of symmetry, draw it from the side. Show the bolt connections with little triangles. 𝑊 𝑀𝐵 𝐹𝐴𝑥

0.2m

0.6m

𝐹𝐵𝑦

𝐹𝐵𝑥 𝐹𝐵𝑥

Step 3 – Draw the vectors for any and all possible forces. The weight of the diver is easy. The bolted connections we’ll show as a force AND a moment. Also note that we assumed the direction of the force and moment at B. If we guessed wrong, the force and moment we calculate will be negative.

Summation of Forces and Moments It should make intuitive sense that if an object at rest is subjected to two equal and opposite forces, then it will not move. Similarly, if a steering wheel of a car is pulled with equal force in two directions, it will not rotate. These are just special cases of Newton’s 2nd Law. The full version of the law is “force equals mass times acceleration” 𝐹 =𝑚⋅𝑎 As you might expect, in the case of statics, there is no movement, and thus no acceleration. So the law reduces to: 𝐹=0 5|Statics

Now we need to be clear that 𝐹 is the net, or total force, not just some of the forces applied to a part. It is more clear, and more traditional then to write it this way: Σ𝐹 = 0 𝐹 here are individual forces which are added together by the summation sign. We saw in the section on vectors that vectors can be added together easily, so this is no problem for us. The next step is to notice that the vectors are 3D. So we can also say that each component of the vectors must also be equal to zero. That is: Σ𝐹𝑥 = 0 Σ𝐹𝑦 = 0 Σ𝐹𝑧 = 0 That gives us three equations to use for any situation! What about rotation? Static means no rotation either. Following a similar line of thought (don’t want to bother explaining rotational inertia) we get: Σ𝑀𝑥 = 0 Σ𝑀𝑦 = 0 Σ𝑀𝑧 = 0 The sum of the moments about each of the axis are also zero. This gives us six equation in total. We will use these equations to solve many important problems. Example – Box on a slope Let’s return to the example of a box on a slope. Find the forces being applied to the box by the slope if the box is about to slip, but hasn’t yet started moving. For this example, let 𝑚 = 10𝑘𝑔 𝑔 = 9.81 𝑚 2 𝜃 = 30°. 𝑠 Step 1 – Draw a FBD. Check. Step 2 – Write the governing equations. Σ𝐹𝑥 = 0

Σ𝐹𝑦 = 0

Σ𝑀𝑧 = 0

Step 3 – Start applying the equations. We start by summing forces in the x direction, since there are only two forces with components in that direction. Σ𝐹𝑥 = 0 𝑁 cos(60°) − 𝐹𝑓𝑟 cos 30° = 0 cos(60°) 𝑁 𝐹𝑓𝑟 = 𝑁 = cos 30° 3 Now sum the forces in the y direction: Σ𝐹𝑦 = 0 𝑁

3

𝑚𝑔

𝑁 cos(30°) + 𝐹𝑓𝑟 cos 60° − 𝑊 = 0 𝑚𝑔 10 ⋅ 9.81 𝑁= = = 𝟖𝟒. 𝟗𝟓𝐍 1 3 1 1 cos(30°) + cos(60°) + ⋅ 3 2 3 2 And finally, we can use the value of N to solve for the force of friction: 𝑁 84.95 𝐹𝑓𝑟 = = = 𝟒𝟗. 𝟎𝟓𝐍 3 3

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Example – Bridge Let’s return to the example of a bridge. Let’s find the forces being applied to the road by the weight of the bridge, and the car that is crossing the bridge. For this example, let 𝑊𝐶 = 10𝑘𝑁 𝑊𝐷 = 5𝑘𝑁 𝑊 = 100𝑘𝑁 𝐹𝐶𝑥 = 1𝑘𝑁

Step 1 – Draw a FBD. Check.

𝐹𝐴𝑦

𝐹𝐵𝑦

𝐹𝐴𝑥 𝑊𝐶 𝐹𝐶𝑥

𝑊𝐷

W

5m 7m 10m 20m Step 2 – Write the governing equations. This is always good to do, so you don’t forget which tools you have available to you. It also makes it easier to review your work. Since this is a 2D problem, we can’t have out of plane displacement, or rotation that would have us leave the plane of the paper, so we are left with the following 3 equations. Σ𝐹𝑥 = 0 Σ𝐹𝑦 = 0 Σ𝑀𝑧 = 0 Now you can see why it was important to not have a reaction force in the x direction at B. We only have 3 equations, so we can only handle 3 unknowns (with this method). Step 3 – Start applying the equations. There is no “right” answer for which to use first, but sometimes picking one equation first will require less algebra. Experience will teach you how to pick the best method of attack. The x direction force balance looks easy, so let’s start with that. Σ𝐹𝑥 = 0 There are only two forces in the x direction! Σ𝐹𝑥 = 𝐹𝐶𝑥 − 𝐹𝐴𝑥 = 0 → 𝐹𝐴𝑥 = 𝐹𝐶𝑥 = 1𝑘𝑁 So the reaction force on the bridge is equal to the force of the car accelerating. Since the value is positive, we assumed the correct direction for the force. Now let’s find the moment about point A. We’re doing this because (based on experience) we can see that this equation will have only one variable, while the y direction force equation will have 2 variables. We should then start with the moment equation, before moving to the force. We could solve it in the other order, but there would be more algebra required. Σ𝑀𝐴 = 20 ⋅ 𝐹𝐵𝑦 − 5 ⋅ 𝑊𝐶 − 7 ⋅ 𝑊𝐷 − 10 ⋅ 𝑊 = 0 → 𝐹𝐵𝑦 =

5 ⋅ 10 + 7 ⋅ 5 + 10 ⋅ 100 = 54.25𝑘𝑁 20 7|Statics

Now we can use our final equation: Σ𝐹𝑦 = 0 Σ𝐹𝑦 = 𝐹𝐴𝑦 − 𝑊𝐶 − 𝑊𝐷 − 𝑊 + 𝐹𝐵𝑦 = 0 → 𝐹𝐴𝑦 = 10 + 5 + 100 − 54.25 = 60.75𝑘𝑁 This answer makes sense, because we would expect the force on the left to be larger than the force on the right. Example – Diving Board Revisit the diving board example. Keep the same assumptions. For this problem, let 𝑊 = 500𝑁

Step 1 – Step 1 – Draw a FBD. Check. 𝑊 𝑀𝐵

0.2m

𝐹𝐵𝑦

0.6m

𝐹𝐵𝑥

Step 2 – Write the governing equations. Σ𝐹𝑥 = 0 Step 3 – Start applying the equations.

Σ𝐹𝑦 = 0

Σ𝑀𝑧 = 0

Σ𝐹𝑥 = 𝐹𝐵𝑥 = 0 Σ𝐹𝑦 = 𝐹𝐵𝑦 − 𝑊 = 0 → 𝐹𝐵𝑦 = 𝑊 = 500𝑁 Take the moment around B. @ B Σ𝑀𝑧 = 0.6 ⋅ 𝑊 − 𝑀𝐵 = 0 → 𝑀𝐵 = 0.6 ⋅ 𝑊 = 0.6 ⋅ 500 = 300𝑁 ⋅ 𝑚 We have now found the reaction forces at the base of the diving board, and are now ready to look at how to attach the board.

Conclusion This was a whirlwind tour of statics. While there are many applications and special cases, the important fact is that with a basic understanding of forces and moments, and the ability to draw a good free body diagram, many important questions can be answered. Using the concept of summation of forces and moments is one of the most basic and useful ideas in engineering. Thorough understanding of the examples presented here will allow you to learn a lot about many practical problems. Suggested References Vector Mechanics for Engineers – Statics – Any Edition – F.P. Beer and E.R. Johnston (library has 3 different editions) 8|Statics