TEXAS WOMAN'S UNIVERSITY
KINS 3591

Ground Reaction Force, Impulse, & Momentum

This lab was adapted from labs developed by Gerald Smith, Ph.D.,
of the Norwegian University of Sport and Physical Education

INTRODUCTION

Ground Reaction Force

In this lab the focus of study is vertical ground reaction force (GRF) and subsequent vertical motion of the body. Reaction Forces are equal and opposite to the forces applied by the body to the ground (Newton's 3rd law). In addition, Newton's 2nd law (F = ma) governs the resulting motion.

A force plate is a device which measures ground reaction force. It can measure forces in three-dimensions, but for this lab we will focus on the vertical component of the ground reaction force. A free body diagram of the plate and a subject is illustrated below:

For the case in which a person is stationary on the plate, the two forces (W and R) must be equal in magnitude and opposite in direction. If the person starts to move upward (positive acceleration of CM), then m*a is positive and the ground reaction force will increase in magnitude. With initiation of a downwards movement (negative acceleration of CM), m*a is negative and the ground reaction force will decrease below body weight.

In summary:

bullet         if a < 0, then R < W

bullet         if a = 0, then R = W

bullet         if a > 0, then R > W

For illustration of the relationship of GRF to the motion it causes, a short movie clip was created. Note that this is a very large file (about 1.5 Mb) and will require an extended download over a modem. It will be best viewed from a direct campus connection.

Play the movie through several times at normal speed to observe the motion and force relationships. Then advance it one frame at a time and observe where takeoff and landing occur on the GRF graph and how the body motion relates to the ground reaction force.

 

Impulse

Closely related to Ground Reaction Force is the concept of impulse. Newton's Second Law (F = ma) can be written in a form which includes the definition of acceleration:

wpeA.jpg (1584 bytes)

 

With algebraic rearrangement, this can be rewritten as:

wpeB.jpg (1346 bytes)

or

Impulse = Change of Momentum

The "Impulse" part of this expression is a new quantity in our study of mechanics. What it says, in essence, is that, in order to change the momentum of an object, a force must be applied for a period of time. The units of impulse are derived from the units of force and time: Newtons*seconds = Newton-Second = Ns.

The impulse-momentum relationship is an important means of determining what motion results from an applied force. In the case of constant force application, impulse is particularly easy to calculate. However, if force is not constant (which is typical of most real situations) a slightly different understanding must be used. Consider these force-time graphs.

In the first graph, force is constant. The impulse of the force is just the value of the force times the time of application. On the graph, this corresponds to the shaded rectangular area below the force line (2 Ns).

On the second graph, force is not constant. The impulse cannot be calculated based on a constant force times its period of application as was done in the first case. However, the idea of area under the force graph is still appropriate. If we could determine that area we would know the impulse (and consequently the change of momentum).

So, how can we determine the area under the force-time graph? Several ways of estimating the area under a graph are often used:

1.       Plot the graph on a grid and count the number of squares under the curve.

2.     Approximate the force-time curve using straight line segments. Then use formulas for the area of triangles and rectangles.

3.     Approximate the area as a set of vertical bars where the height of each bar is the average values of the data points at sides of the bar. The area of each bar (a rectangle) is then the product of its height and its width (time between measurements). The impulse is the sum of the areas of the bars. This animation illustrates this principle.

This is illustrated here for forces of 1, 1.2, 1.5, 1.9, 2.4 and 3.0 N at 0.0, 0.2, 0.4, 0.6, 0.8, & 1.0 s, respectively. The impulse of the first shaded bar can be found by first averaging the force of the two samples determining its height (1.2 and 1.5 N). The average force is 1.35 N over the interval of 0.2 seconds (0.2 to 0.4). Hence the impulse (shaded area) is 1.35 N times 0.2 seconds which is 0.27 Ns.

For the second shaded bar the average force is 2.15 N and the impulse would be 0.43 Ns of impulse. If you follow this procedure for the whole area under these six data points, the impulse is 1.8 Ns.

In general, this third method is most commonly used in determining the impulse of biomechanical forces.

METHODS

Each student will perform two jumps on the force plate:

With hands on hips from a squat position, perform a maximal vertical jump without counter movement.

With hands on hips from an upright position, perform a maximal vertical jump with counter movement.

During each jump, keep your hands on your hips throughout the jump. When you land, stand up but do not step off the force plate until directed to do so. You will receive a force-time file on disk for each jump. Import your data into Excel. You should have a spreadsheet with [at least] these columns.

Time (s)

GRF (N)

0.00

 

0.01

 

0.02

 

0.03

 

etc.

 

Next, create two graphs of Ground Reaction Force vs Time (one for each jump). Use Scatterplot graphing mode, straight lines without dots. Draw and label lines on the graphs to represent the following:

bullet         Body Weight line

bullet         Takeoff (TO)

bullet         Landing (L)

bullet         Flight Time (FT)

 

ANALYSIS

A vertical jump force-time curve has been included for illustration. A body weight (BW) line has been added to the graph. The total force acting on the jumper is the ground reaction force minus gravitational force (BW). Thus to calculate the impulse, one must first subtract off body weight. In effect, this is as if the zero point for the force axis was shifted upward. The shaded regions of the graph sum to produce the impulse before takeoff and the impulse after landing. The regions below the BW line are "negative areas."

In equation form this can be written out as follows: 

Let Fi be the force at some instant in time. The total force acting on the person at that instant will be Fi - Body Weight (BW). The impulse of the force for samples Fi and Fi+1 would be:

indexa1.jpg (4706 bytes)

Note that this is the area or Impulse under just one strip or one interval from point i to point i+1. The Total Impulse involves summing each of these strips up to the time of takeoff.

Add three more columns to your spreadsheet, so that it looks like this:

Time (s)

GRF (N)

GRF-BW (N)

Average Force (N)

Impulse (N*s)

0.00

 

 

 

 

0.01

 

 

 

 

0.02

 

 

 

 

0.03

 

 

 

 

etc.

 

 

 

 

To determine body weight (BW), average the GRF's (Column 2) from the first 20-30 frames of data. This average is your Body Weight (don't worry if it doesn't match your body weight precisely). After computing the Body Weight Corrected Ground Reaction Forces (GRF - BW), graph them vs. Time (two graphs, one for each jump). Note the remarkable similarity between these graphs and the ones created above.

Determine the following information for each jump:

bullet         Impulse up to the point of takeoff

bullet         Impulse from the landing point to the end of the data

bullet         Takeoff velocity

bullet         Calculated flight time (from projectile kinematics)

bullet         Measured flight time (count the number of frames where GRF = 0)

bullet         Calculated flight height (from projectile kinematics)

GENERAL QUESTIONS:

Answer the following questions in a few paragraphs. Attach your force-time graphs (four of them) to these written responses and turn in to your lab instructor at the beginning of lab next week. Do not attach the force-time data to your lab after graphing the data (it will kill too many trees).

1.       Compare and contrast the force-time curves for the two vertical jumps.

2.     What does a longer flight time imply about jump height?

3.     What does a higher jump height imply about flight time?

4.     Explain possible reasons why calculated and measured flight times may not be in close agreement (w/in 0.1 s).

5.     Report your values for squat jump takeoff & landing impulse and countermovement jump takeoff & landing impulse.

6.     Compare the takeoff and landing impulses (i.e., compare SJ takeoff impulse to SJ landing impulse, CMJ takeoff impulse to CMJ landing impulse). Account for differences.

What is due at beginning of next class period:

bullet         Printout of Impulse calculations for both jumps

bullet         Printout of force-time graphs

bullet         Answers to General Questions

Updated 08/23/2005 by Gary Christopher, MS, ATC
Texas Woman's University, Department of Kinesiology, College of Health Sciences