- Understand what a force is and how force can affect the human body
- Be able to distinguish force and mass
- Understand different classifications, or types, of forces, and how each can differently affect the human body. These different force classifications include: (1) internal and external force, (2) contact and non-contact force, and (3) friction and normal force
- Understand the coefficients of static and dynamic friction and be able to use the following equation to quantify friction force, normal force, and/or the coefficient of friction: FF = FN × µ
- Be able to determine the resultant (net) effect of two or more forces using vector addition
- Be able to resolve a resultant (net) force into orthogonal (perpendicular) components using vector resolution
- Understand the concept of static equilibrium. Know how to determine whether an object is in static equilibrium by summing the forces that are acting on an object. Be able to determine an unknown force that is acting on a static object (if all other forces acting on the object are known).
The following math review may be helpful for you before learning some of the following material.Force and mass:Force is a push or a pull that can potentially accelerate an object, including the human body; forces can also deform anatomical structures. Force is a vector quantity that is measured in Newtons (N) or pounds (lb).
Mass is a scalar measure of inertia, or a measure of the quantity of matter for an object, including the human body. Mass is often measured in kilograms (kg).
Some helpful conversions include 1 N = 0.225 lb, 1 lb = 4.45 N, 1 kg = 2.214 lbs, and 1 N = 1 kg × 1 m/s 2 (F = ma).
Practice: Body weight is a force that is directed downward. If you know how much you weigh, in lbs, use the third aforementioned conversion (1 kg = 2.214 lbs) to calculate your body mass in kg. Then, use the final aforementioned conversion (1 N = 1 kg × 1 m/s 2), recalling that acceleration due to gravity on earth is -9.81 m/s 2, to calculate your body weight in N. It is important for you to remember that body weight is a force that results from gravity acting on your body mass, body mass is a measure of the quantity of matter within your body (body weight = body mass × acceleration due to the earth's gravity).
Force Classifications (different ways to think of force)
One way to consider forces is to classify them as either internal or external.
1. Internal forces:
act within the object, or the system, that is being investigated. Alone, these forces cannot alter the motion of the center of mass.
Practice: What are two internal forces that have acted on your body today?
2. External forces:
on an object, or system, as a result of its interaction with the external environment. These forces can alter the
motion of the center of mass.
Practice: What are two external forces that have acted on your body today?
A video that may or may not be relevant (probably not)...An additional way to consider force is to classify force as contact or non-contact.
1. Contact forces:
Most forces we will consider in this course will be contact forces (some internal and some external). These forces result as objects contact one another, and can include solids and fluids. We will often resolve resultant contact forces into two perpendicular force components that are classified as either (1) friction (F F) or (2) normal (F N). A ground reaction force is a good example of a contact force that can be resolved into friction and normal force.
Friction force (FF): the component of a contact force that acts parallel to the contacting surfaces. Although friction is often associated with opposition of motion, remember that friction often facilitates human motion. Friction can be static or dynamic, depending on whether the contacting objects are moving relative to one another. (Static friction > dynamic friction).
The coefficient of friction (µ = F F/F N) is a dimensionless value that depends on the material properties of the contacting surfaces. The coefficient differs depending on whether the circumstances are static or dynamic (static coefficient > dynamic coefficient).
Normal Contact Force (FN): the component of a contact force that acts perpendicular to the contacting surfaces.
Practice: The coefficient of static friction between a runner's shoe and the ground is 0.6. If the ground is applying a 2000-N vertical (normal) ground reaction force on the runner, what is the maximum horizontal force the runner can generate under his shoe before the shoe will start to slide and the runner may slip? Answer: 1200 N. What variables might this runner change in order to decrease her probability of slipping? Similarly, how might you decrease your chance of slipping on an icy sidewalk this semester?
2. Non-contact forces
These do not require contact between two objects. For this course, the non-contact force of primary concern is gravity, which, on the surface of the earth, causes a vertical acceleration of about -9.81 m/s/s.
Analysis of Multiple Simultaneous Forces: to consider the effect of multiple forces acting on an object, it is often useful to utilize a
free body diagram. Within a free body diagram, forces are represented as vectors that indicate point of force application, force magnitude, and force direction.
A Brief Review of Vector Addition: this can be used to understand the net effect of multiple forces on a single object or system. It is easy to add two or more colinear forces: you can add colinear forces algebraically. If the forces are not colinear, however, you cannot algebraically sum each of the forces. Use the tip-to-tail method and trigonometry to add non-colinear forces.
Some Colinear Vector Addition Practice Problems:
Practice: A weight lifter (mass = 100 kg) holds a barbell that weighs (50 lbs). The vertical ground reaction force is 1000 N. What is the net force acting on the weight lifter? Answer: -203.5 N. Will the center of mass experience any vertical acceleration?
Practice: A 72-kg sprinter is acted upon by a 2200-N vertical ground reaction force. What is the net force acting on the sprinter? Answer: 1494 N. How will the center of mass be affected? (In other words, what will be the vertical acceleration of the runner's center of mass?)
Practice: Use vector addition to add the forces shown in the aforementioned free body diagram of the runner (normal and friction ground reaction forces, wind resistance, and body weight). Assume that air resistance is -15 N in a horizontal direction, body weight is -700 N, the anterior-posterior ground reaction force (friction) is 220 N, and the vertical ground reaction force (normal) is 820 N. What is the resultant force? Answer: 238 N. Also, at what angle, in relation to the horizontal axis, is this resultant force oriented? Answer: 30°.
Practice: draw a free-body diagram that shows three primary forces (quadriceps, patellar ligament, and femoral) that are applied to the patella. Use the radiograph, below, to help you. This free body diagram and the corresponding class discussion will help you better understand the data that were presented by Escamilla et al. (2009), MSSE, 41(4), 879-888.
Vector resolution is also important when considering forces acting on an object/system. Vector resolution allows you to resolve a resultant, or net, force into orthogonal (perpendicular) components.
Vector Resolution Practice: A 72-kg sprinter is acted upon by a resultant ground reaction force of 2124 N. The force acts at an angle that is 86 degrees above a fixed horizontal axis (the ground). Using vector resolution, what are the vertical and horizontal components of the this resultant ground reaction force? Answers: Vertical = 2119 N; Horizontal = 148 N What are the effects of this resultant ground reaction force? Upward vertical acceleration, and horizontal acceleration in the direction of the horizontal force.
Another Vector Resolution Practice Problem: This practice problem is made easier with the use of vector resolution. The answers to this problem are 68 N and 43°.
Static Equilibrium (ΣF = 0)
When an object is at rest and the forces acting on the object are in equilibrium, they result in a net force of zero. Although this concept won't be used often in this class (moving humans are not usually in static equilibrium), it is good to remember, as the concept sets a good foundation for the consideration of circumstances that do not involve static equilibrium. Use the concept of static equilibrium to determine the involved ground reaction force in the problem below.
Weightlifter example: The acceleration of this weightlifter is zero (static equilibrium). If the lifter's mass is 80kg and the mass of the barbell is 70kg, what must the vertical component of the ground reaction force be? Answer: 1471.5 N
Remember: Fgrf + Fbw + Fbb = 0
What would occur if the ground reaction force applied to the weightlifter was 2000 N?
How might this occur?
Additional Practice Problems:
- Using vector addition, calculate the resultant force acting on this gymnast
- Calculate the reaction forces applied to the child on a swing, assuming that the child is in static equilibrium (ΣF = 0). Calculate the resultant reaction force (FR), the two separate components of the resultant reaction force (FRX & FRY), and the orientation of the reaction force in relation to the horizontal (θ). Solutions for #1 and #2
- You are attempting to pull someone on a sled via a rope that is attached to the sled across a flat snowy field. Your pulling force is directed forward and upward, at an angle that is 30 degrees above the ground. The combined mass of the person on the sled and the sled is 58 kg, and the coefficient of static friction between the sled and the snow is 0.10. How much force must you pull with to start moving the sled? Answer: ~62N. A solution. Also, what variables might you change in order to increase your ability to pull the sled more easily?
- Complete practice problems 1-15, from Chapter 1 of the McGinnis textbook (2nd Edition).
- Complete the following practice problems.