Topic 02 – Forces

Topic 02 – Forces

Lesson Learning Outcomes

• Define forces
• Understanding the concept of vector quantities
• Explain the difference between internal and external forces
• Classify forces
• Define weight
• Define friction
• Be able to determine the resultant of two or more forces
• Resolve a force into component forces acting perpendicular to each other

Guiding Questions

• Why is it important to understand the different type of forces acting on a body?
• Am I able to give examples of internal and external forces in sporting situations?
• What is the difference between collinear and concurrent forces?
• Can I resolve forces and which methods should I use?

Definition

Force is defined as a push or a pull. Forces are exerted by objects on other objects. Forces come in pairs: The force exerted by one object on another is matched by an equal but oppositely directed force exerted by the second object on the first action and reaction. A force is something that accelerates or deforms an object. Mechanically speaking, something accelerates when it starts, stops, speeds up, slows down, or changes direction. So a force is something that can cause an object to start, stop, speed up, slow down, or change direction.

Unit of measurement

The SI unit of measurement for force is the newton, named in honor of the English scientist and mathematician Isaac Newton. The newton is abbreviated as N. One newton of force is defined as the force required to accelerate a 1 kg mass 1 m/s², or algebraically as follows:

• N = (1.0 kg)(1.0 m/s²)

Vectors

A force is what is known as a vector quantity. A vector is a mathematical representation of anything that is defined by its size or magnitude (a number) and its direction (its orientation). To fully describe a force, you must describe its size and direction.

Classification of forces

Forces can be classified as internal or external.

Internal forces are forces that act within the object or system whose motion is being investigated. Remember, forces come in pairs—action and reaction. With internal forces, the action and reaction forces act on different parts of the system (or body). Each of these forces may affect the part of the body it acts on, but the two forces do not affect the motion of the whole body because the forces act in opposition.

In sport biomechanics, the objects whose motion we are curious about are the athlete’s body and the implements manipulated by the athlete.

External forces are those forces that act on an object as a result of its interaction with the environment surrounding it. We can classify external forces as contact forces or noncontact forces. Most of the forces we think about are contact forces. These occur when objects are in contact with one another. Noncontact forces are forces that occur even if the objects are not touching each other. The gravitational attraction of the earth is a noncontact force. Other noncontact forces include magnetic forces and electrical forces.

Weight

In sports and exercise, the only noncontact force we will concern ourselves with is the force of gravity. The force of gravity acting on an object is defined as the weight of the object.

Earlier on in this guide, we defined 1 N of force as the amount of force that would accelerate a 1 kg mass 1 m/s². If the only force acting on an object is the force of the earth’s gravity, then the force of gravity will accelerate the object. This is the case when we drop something (if the force of air resistance can be ignored). Scientists have precisely measured this acceleration for various masses at various locations around the earth. It turns out to be about 9.81 m/s² (or 32.2 ft/s²) downward no matter how large or small the object is. This acceleration is called gravitational acceleration or the acceleration due to gravity and is abbreviated as g.

Now let’s see if we can figure out the weight of something if we know its mass. If a 1 N force accelerates a 1 kg mass 1 m/s2, then how large is the force that would accelerate a 1 kg mass 9.81 m/s2? Another way of asking this question is, How much does 1 kg weigh?

? N = (1 kg)(9.81 m/s²) = Weight of 1 kg = Force of gravity acting on 1 kg

If we solve this equation, we find that 1 kg weighs 9.81 N. On the earth, mass (measured in kilograms) and weight (measured in newtons) are proportional to each other by a factor of 9.81. The weight of an object (in newtons) is its mass (in kilograms) times the acceleration due to gravity (9.81 m/s²), or,

W = mg

Where

W = weight (measured in newtons),

m = mass (measured in kilograms), and

g = acceleration due to gravity = 9.81 m/s².

Contact forces

Contact forces are forces that occur between objects in contact with each other. The objects in contact can be solid or fluid. Air resistance and water resistance are examples of fluid contact forces. The most important contact forces in sport occur between solid objects, such as the athlete and some other object. For a shot-putter to put the shot, the athlete must apply a force to it, and the only way the athlete can apply a force to the shot is to touch it. To jump up in the air, you must be in contact with the ground and push down on it. The reaction force from the ground pushes up on you and accelerates you up into the air. To accelerate yourself forward and upward as you take a running step, you must be in contact with the ground and push backward and downward against it. The reaction force from the ground pushes forward and upward against you and accelerates you forward and upward.

Friction

Frictional force opposes motion of sliding between two surfaces in contact. Frictional forces are primarily responsible for human locomotion, so an understanding of friction is important. Dry friction acts between the non-lubricated surfaces of solid objects or rigid bodies in contact and acts parallel to the contact surfaces. When dry friction acts between two surfaces that are not moving relative to each other, it is referred to as static friction. Static friction is also referred to as limiting friction when we describe the maximum amount of friction that develops just before two surfaces begin to slide. When dry friction acts between two surfaces that are moving relative to each other, it is referred to as dynamic friction. Other terms for dynamic friction are sliding friction and kinetic friction.

Contact Force

An increase in weight would increase the normal contact force acting between the two surfaces. This would increase the interactions of the molecules of the contacting surfaces, because they would be pushed together harder, increasing static friction force. Friction force is proportional to the normal contact force. As one increases, the other increases proportionally. This is true for both static and dynamic friction.

Surface Area

Dry friction, both static and dynamic, is not affected by the size of the surface area in contact. Dry friction arises due to the interaction of the molecules at the surface areas in contact. We have seen that if we press these surfaces together with greater force, the interactions of the molecules will be greater and friction will increase. It makes sense to say that if we increase the area of the surfaces in contact, we also increase the number of molecules that can interact with each other, and thus we create more friction. But if the force pushing the surfaces together remains the same, with the greater surface area in contact, this force is spread over a greater area, and the pressure between the surfaces will be less (pressure is force divided by area). Therefore, the net effect of increasing surface area is zero, and friction is unchanged.

Contacting Materials

Friction is affected by the characteristics of the surfaces in contact. Greater friction can be developed between softer and rougher surfaces than between harder and smoother surfaces.

Static and dynamic friction can be mathematically expressed as

Determining the resultant of two or more forces (Force Composition)

Forces are added using the process of vector addition. The result of vector addition of two or more forces is called a resultant force. The vector addition of all the external forces acting on an object is the net force. It is also referred to as the resultant force, because it results from the addition of all the external forces.

Colinear Forces

Colinear forces are forces that have the same line of action. The forces may act in the same direction or in opposite directions along that line.

Example: You are on a tug-of-war team with two others. You pull on the rope with a force of 100 N, and your teammates pull with forces of 200 N and 400 N. You are all pulling along the same line—the line of the rope. To find out the resultant of these three forces, we begin by graphically representing each force as an arrow, with the length of each arrow scaled to the size of the force.

If we measure the length of this arrow, it turns out to be seven times as long as the 100 N force. The resultant force must be a 700 N force acting to the right.

Now let’s consider the forces the opposing team exerts on the rope. That team also consists of three members. The forces they exert on the rope are to the left and are 200 N, 200 N, and 200 N, respectively. What is the resultant of these three forces?

Let’s arbitrarily say that forces acting to the right are positive. Then forces acting to the left must be considered negative. So the resultant force acting on the rope as a result of your team pulling on it to the right and the opposing team pulling on it to the left can now be determined algebraically by adding up the positive forces from your team and the negative forces from the opposing team.

100 N + 200 N + 400 N + (−200 N) + (−200 N) + (−200 N) = +100 N

Adding a negative number is just like subtracting it, so we could also write this as

100 N + 200 N + 400 N − 200 N − 200 N − 200 N = +100 N.

The positive sign associated with our answer of 100 N indicates that the resultant force acts in the positive direction. We set up our positive direction to the right, so the resultant force is a force of 100 N to the right.

Concurrent forces

If forces do not act along the same line but do act through the same point, the forces are concurrent forces. If we model objects as point masses, the forces acting on these objects will be considered colinear forces if they act along the same line, and concurrent forces if they do not act along the same line.

Now let’s consider a situation in which the external forces are not colinear but are concurrent. A gymnast is about to begin his routine on the high bar. He jumps up and grasps the bar, and his coach stops his swinging by exerting forces on the front and back of the gymnast’s torso. The external forces acting on the gymnast are the force of gravity acting on the mass of the gymnast, a horizontal force of 20 N exerted by the coach pushing on the front of the gymnast, a horizontal force of 30 N exerted by the coach pushing on the back of the gymnast, and an upward vertical reaction force of 550 N exerted by the bar on the gymnast’s hands. The gymnast’s mass is 50 kg. What is the net external force acting on the gymnast?

First, how large is the force of gravity that acts on the gymnast?

W = mg = (50 kg)(10 m/s2) = 500 kg m/s2 = 500 N.

This weight is a downward force of 500 N. We now have all the external forces that act on the gymnast.

Just as we did with the collinear forces, we can represent each force graphically with an arrow, scaling the length of the arrow to represent the magnitude of the force, orienting the arrow to show its line of application, and using an arrowhead to show its sense or direction.  As with the colinear forces, if we line up the forces tip to tail, we can find the resultant force

The head of the 500 N downward force and the tail of the 20 N horizontal force do not connect. The resultant of the four forces can be represented by an arrow connecting the tail of the 20 N horizontal force. The resultant is thus directed upward and slightly to the left. The size of the resultant force is indicated by the length of its arrow. Using the same scale that was used to construct the other forces in figure 1.9, we can estimate that the magnitude of the resultant force is about 51 N. We could describe the angle that the force makes with a vertical line or a horizontal line. Measuring clockwise from a vertical line, this force is about 11° from vertical. This angular description is much more precise than the description of the force as “upward and slightly to the left.”

Trigonometric Technique

Using this method, the resultant force can be expressed as a 10 N horizontal force acting to the left and a 50 N vertical force acting upward. Draw the forces tip to tail, as shown in figure Now draw the resultant by connecting the tail of the 10 N horizontal force with the tip of the 50 N vertical force. By Pythagorean Theorem, we know that the resultant force is about 51 N. We can determine the length of the unknown side of the triangular figure and the angles other than the right-angle using trigonometry.

In these equations, θ, which is pronounced “theta,” represents the angle; opposite refers to the length of the side of the triangle opposite the angle theta; adjacent refers to the length of the side of the triangle adjacent to the angle theta; and hypotenuse refers to the length of the side of the triangle opposite the right angle. The term sin refers to the word sine; cos refers to the word cosine; and tan refers to the word tangent. Any modern scientific calculator includes the functions for sine, cosine and tangent.

If the sides of the right triangle are known, then the inverse of the trigonometric function is used to compute the angle:

Now let’s go back to the resultant forces acting on the gymnast. Let’s determine the angle

between the 51 N resultant force (the hypotenuse of thetriangle) and the 10 N horizontal force (the adjacent side).The 50 N vertical force is the side opposite the angle.

Using the equation gives the following:

To determine the angle θ, we use inverse of the tangent function or the arctangent. On most scientific calculators, the arctangent function is the second function for the tangent key and is usually abbreviated as tan−1 or atan. Using a calculator (make sure its angle measure is programmed for degrees rather than radians), we find that θ = arctan (5) = 78.7°. The angles in a triangle add up to 180°. In a right triangle, one angle is 90°, so the sum of the other two angles is 90°. The other angle in this case is thus 11.3° (i.e., 90° − 78.7°). This is pretty close to the value we arrived at earlier using the graphical method.

Resolution of Force

The process of determining what two force components add together to make a resultant force is called force resolution.

What if the external forces acting on the object are not colinear and do not act in a vertical or horizontal direction? Imagine that an athlete exerts a 100 N force on the shot at an angle of 60° above horizontal. The mass of the shot is 4 kg.

Using the rough approximation for g, the shot weighs

W = mg = (4 kg)(10 m/s²) = 40 N.

Using a graphical technique, the resultant force is approximately 68N as shown in the figure below

In the shot-putting problem, we have one vertical force, the shot’s weight, but the force from the athlete is acting both horizontally and vertically. It is pushing upward and forward on the shot. Because this 100 N force acts to push the shot both horizontally and vertically, perhaps it can be represented by two different forces: a horizontal force and a vertical force.

In the above figure, measure the lengths of these force vectors. The horizontal force component is about 50 N, and the vertical force component is about 87 N.

To complete the original problem, we would include the 40 N weight of the shot as a downward force. This would be subtracted algebraically from the 87 N upward component of the force exerted by the athlete. The resulting vertical force acting on the shot would be

(−40 N) + 87 N = +47 N.

A 47 N force acts upward on the shot. We still have the 50 N horizontal component of the force exerted by the athlete. If we add this to the 47 N vertical force, using the Pythagorean theorem, we get

A² + B² = C²

(50 N)² + (47 N)² = C²

2500 N² + 2209 N² = C²

4709 N² = C²

C = 68.6 N

Besides the Pythagorean theorem, there are other relationships between the sides and the angles of a right triangle. Some of these relationships can be described by the sine, cosine, and tangent functions.

To complete the problem we need to know the direction of this net external force. We can use a trigonometric relationship to determine the angle this net force (68.4 N) makes with horizontal. The force triangle is made up of the 46.6 N upward force (the opposite side); the 50 N horizontal force (the adjacent side); and the resultant of these two forces, the net force of 68.4 N (the hypotenuse). The angle we are trying to determine is between the net force and the horizontal force. Let’s use the arctangent function to determine the angle:

The net external force acting on the shot is a force of 68.4 N acting forward and upward at an angle of 43° above horizontal. This is almost identical to the answer we got using the graphical techniques.

Summary

• Force is a push or a pull and one type of force, friction is especially important in understanding movement in sport.
• Friction is dependent on the size of the normal contact force.
• Collinear forces act along the same line while concurrent forces act through the same point.
• A vector quantity is one which possess both size and direction.
• There are a number of ways to resolve vectors depending on which information is available: graphical, using Pythagoras theorem and trigonometric methods for right-angled vector problems.