EXSC 3500 (Biomechanics) CH 1 - Maintaining Equilibrium or Changing Motion PDF

Preview

Summary

The topics in this set of notes includes different types of forces, vectors, friction, graphical technique, trigonometric technique, free body diagrams, static analysis, review questions, and equations...

Description

CH 1 - Maintaining Equilibrium or Changing Motion Tuesday, December 5, 2017

10:46 PM

objectives When you finish this chapter, you should be able to do the following: • Define force • Classify forces • Define friction force • Define weight • Determine the resultant of two or more forces • Resolve a force into component forces acting at right angles to each other • Determine whether an object is in static equilibrium, if the forces acting on the object are known • Determine an unknown force acting on an object, if all the other forces acting on the object are known and the object is in static equilibrium

*******************************************************

What are Forces (PG 20) - Force - A push of a pull ○ Exerted by one objects on other objects ○ Comes in pairs (action and reaction) ○ Something that accelerates or deforms an object ○ Something that can cause an object to start, stop, speed up, slow down or change direction ○ Unit = Newton (N) ▪ One newton of force is defined as the force required to accelerate a 1 kg mass 1 m/s2 or 1.0 N = (1.0 kg)(1.0 m/s2) ▪ One newton of force is equal to 0.225 lb of force or 1 lb equals 4.448 N ○ A vector quantity ○ Characteristics: ○ Point of Application ○ Direction (line of action) ○ Sense (Whether it pushes or pulls along the line) - Vector - A mathematical representation of anything that is defined by its size or magnitude (a number) and its direction (its orientation) ○ The length of the arrow indicates the size of the force ○ The shaft of the arrow indicates its line of application ○ The arrowhead indicates its sense or direction along that line of application ○ One of the arrow's ends indicates the point of application of the force

Internal Forces (PG 21) - Internal Forces - Forces that act within the object or system whose motion is being investigated - Muscles pull on tendons which pull on bones - At joints, bones push on cartilage which pushes on other cartilage and bones - Tensile Forces - Pulling forces whose direction and point of application would tend to lengthen or stretch an object along the dimension coinciding with the line of action of the force

- Tension - State of an object as a result of forces pulling on it and producing tensile stress ○ Tensile Stress - Is axial stress that tends to pull molecules apart and stretch the object - Compressive Forces - Pushing force whose direction and point of application would tend to shorten or squeeze an object along the dimension coinciding with the line of action of the force - The structure fails and breaks when the tensile or compressive forces acting on a structure are greater than the internal forces the structure can withstand - Muscles - Structures that produce the forces that cause us to change our motion - The body is able to change its motion only if it can push or pull against some external object - External forces are solely responsible for changes in motion of the body's center of mass

External Forces (PG 21) - External Forces - Forces that act on an object as a result of its interaction with the environment surrounding it ○ Classified as: 1. Contact Forces (most forces) - Occur when objects are touching each other 2. Noncontact Forces - Occur even if the objects are not touching each other - Ex: The gravitational attraction of the Earth; magnetic forces and electrical forces - Weight of Gravity - The force of gravity acting on an object - Gravitational Acceleration (or the Acceleration due to gravity, abbreviated as g) = 9.81 m/s2 (or 32.2 ft/s2) ○ How much does 1 kg weigh? ○ Weigh of 1 kg * Force of gravity acting on 1 kg ○ (1 kg)(9.81 m/s2) = 1 kg weighs 9.81 N

- On Earth: ○ Mass is measured in kg ○ Weight is measured in newtons ○ The weight of an object (in newtons) is its mass (in kilograms) times the acceleration due to gravity (9.81 m/s2) ▪ W = mg - Contact Forces - Forces that occur between objects in contact with each other ○ The objects can be solid or fluid ○ Examples of Fluid Contact Forces: Air resistance and water resistance ○ Can be resolved into parts or components ▪ The component of that acts perpendicular to the surface of the objects in contact and the component of force that acts parallel to the surfaces in contact ○ 1st Component: Is called a normal contact force (or normal reaction force) ▪ Refers to the fact that the line of action of this force is perpendicular to the surfaces in contact ○ 2nd Component: Is called Friction ▪ Friction - The component of a contact force that acts parallel to the surfaces in contact; The magnitude of friction is the product of the coefficient of friction and the normal contact force (the component of the contact force acting perpendicula to the surfaces in contact) ▪ The line of action of friction is parallel to the two surfaces in contact and opposes motion or sliding between the surfaces - The reaction force from the ground pushes up on you and accelerates you up into the air

- Frictional forces are primarily responsible for human locomotion

Friction (PG 23) - Frictional Force = Dry Friction (a.k.a. Coulomb Friction) - Static Friction (referred to as Limiting Friction) - When dry friction acts between two surfaces that are not moving relative to each other - Dynamic Friction (Other terms are Sliding Friction and Kinetic Friction) - When dry friction acts between two surfaces that are moving relative to each other

Friction and Surface Area (PG 24) - Dry friction arises due to the interaction of the molecules at the surface areas in contact - Pressure is force divided by area

Friction and Contacting Materials (PG 25)

- Friction is affected by the size of the normal contact force, but it is unaffected by the area in contact - Static friction is larger than dynamic friction ○ Static Friction: Fs = µsR ○ Dynamic Friction: Fd = µdR - Summary About Dry Friction: ○ Friction is a contact force that acts between and parallel to the two surfaces in contact ○ Friction is proportional to the normal contact force pushing the two surfaces together ○ Friction opposes relative motion or impending relative motion ○ Friction is proportional to the normal contact force pushing the two surfaces together ▪ As the normal contact force increases, the frictional force increases as well

- The Greek letter mu is the ratio of friction force to normal contact force

Addition of Forces: Force Composition (PG 26) The net force acting on an object is the sum of all the external forces acting on it

Friction in Sport and Human Movement (PG 26) - Leather and rubber have large coefficients of friction - Various external forces that can act on us in sport activities include: ○ Gravity ○ Friction ○ Contact Forces

Addition of Forces: Force Composition (PG 26) - The net force is the vector sum of all the external forces - The full description of a force includes its magnitude (how large is it?) and its direction (which way does it act?) - The Arrow: ○ Length of the shaft: Represents the magnitude of the force ○ Orientation of the arrow: Represents its line of application ○ Arrow Head: Represents its direction of action along that line - Forces are added using the process of vector addition, which results in a Resultant Force - Resultant Force - The vector sum of two or more forces; ○ The force that results from the vector addition of two or more forces - Net Force - The vector addition of all the external forces acting on an object

Colinear Forces (PG 26)

- Colinear Forces - Forces that have the same line of action ○ The forces may act in the same direction or in opposite directions along that line

- +100 means a positive direction

Concurrent Forces (PG 28) - Concurrent Forces - Forces that act through the same point and not along the same line ○ Two forces or more whose lines of action intersect at a single point - Ex: 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 th

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? ○ W= mg ▪ W = Represents weight in newtons ▪ m = Represents mass in kilograms ▪ g = Represents the acceleration due to gravity, or 9.81 m/s2 (can round to 10) ○ W = mg = (50 kg)(10 m/s2) = 500 kg m/s2 = 500 N which is the weight of the downward force that acts on the gymnast

- Finding the resultant force - First, draw the 20 N horizontal force acting to the right. - Now draw the 550 N upward force so that its tail begins at the head of the 20 N force. - Draw the 30 N horizontal force to the left so that the tail of this force begins at the head of the 550 N force. - Draw the 500 N downward force of gravity so that the tail of this force begins at the head of the 30 N force. Free-body diagram showing the external forces acting on a gymnast hanging from the horizontal bar.

- The resultant of the four forces can be represented by an arrow connecting the tail of the 20 N horizontal force (the first force in our drawing) with the head of the 500 N downward force (the last force in our drawing) - Forces On The Gymnast: ○ Horizontally: 20 N to the right and 30 N to the left ○ Vertically: 500 N downward and 550 N upward

- Resultant Force: ○ Consider the horizontal and vertical forces separately and determine what the horizontal resultant force is and what the vertica resultant force is. ▪ Horizontal Force (+ for towards the right and - for towards the left): □ 20 N + (−30 N) = 20 N − 30 N = −10 N (to the left) ▪ Vertically (+ for an upwards force and - for a downward force): □ (−500 N) + 550 N = +50 N (upward) ○ Expressed as a 10 N horizontal force acting to the left and a 50 N vertical force acting upward. ▪ Add the horizontal resultant force of 10 N and the vertical force of 50 N graphically to determine their resultant force. Draw the forces tip to tail, as shown in figure 1.10. ▪ Measure the angle ▪ 102 + 502 = 2600 ▪ = 50.9 = 51 N

- Sample Problem 1.1 ○ A spotter assists a weightlifter who is attempting to lift a 1000 N barbell. The spotter exerts an 80 N upward force on the barbel while the weightlifter exerts a 980 N upward force on the barbell. What is the net vertical force exerted on the barbell? ○ Assume that upward is the positive direction. The 80 N force and the 980 N force are positive, and the 1000N weight of the barbell is negative. Adding these up gives us the following - ∑F = (+80 N) + (+980 N) + (-1000 N) = 80 N + 980 N - 1000 N = +60 N (the vertical force acting on the barbell)

Trigonometric Technique (PG 31)

- Pythagorean theorem ○ A2 + B2 = C2 ○ 102 + 502 = 2600 = 50.9 = 51 N ○ - Trigonometry tells us the ration among lengths of the sides of right triangles

- If the angle and the hypotenuse are known use the Sin equation, the adjacent side could be determined using the Tan equation - If the sides of the right triangle are known, then the inverse of the trigonometric function is used to compute the angle

- If the lengths of any two sides are known the arcsine, arccosine and arctangent functions should be used

- Arctan is abbreviated as tan-1 in the calculator

- Using the gymnast example from earlier:

Same as 90-78.7=11.3° for the other angle - If the forces are concurrent, you can add the forces to determine their resultant

Resolution of Forces (PG 33) - Consider the forces acting on a shot during the putting action. Imagine that at the instant shown, the athlete exerts a 100 N force on the shot at an angle of 60° above horizontal. The mass of the shot is 4 kg. What is the net force acting on the shot? ○ Determine the weight of the shot ▪ W = mg = (4kg)(10 m/s2) = 40 N

- Sample Problem 1.2

- The vertical ground reaction force (normal contact force) acting under a runner’s foot is 2000 N, while the frictional force is 600 acting forward. What is the resultant of these two forces?

- Draw the resultant force. Let the two known forces represent two sides of a box. - Draw the other two sides of the box. - The resultant force is the diagonal of this box, with one end at the point of application of the other two forces

Use the Pythagorean theorem to compute the size of the resultant force A2 + B2 = C2 (2000 N)2 + (600 N)2 = C2 4,000,000 N2 + 360,000 N2 = C2 4,360,000 N2 = C2j6 2088 N = C

Use the arctangent function to determine the angle of the resultant force with horizontal

arctan

Graphical Technique (PG 33) - Components - Parts that make up a system - Draw the 100 N force as a vector as shown below - Draw a box around the 100 N force so that the sides of the box align vertically or horizontally and so that the 100 N force runs diagonally through the box from corner to corner - Measure the lengths of these force vectors. The horizontal force component is about 50 N, and the vertical force component is abou 87 N. - A2 + B2 = C2 - 50 N2 + 87 N2 = 10069 ≈ 10,000 N2 - Include the 40 N weight of the shot as a downward force which would be subtracted from the 87 N upward component of the force exerted by the athlete ○ (-40 N) + 87 N = + 47 N (upward force on the shot)

- Add the 50 N horizontal component to the 47 N vertical force using the Pythagorean Theorem ○ A2 + B2 = C2 ○ (50 N)2 + (47 N)2 = C2 2500 N2 + 2209 N2 = C2 4709 N2 = C2

Trigonometric Technique (PG 36)

Horizontal Component: (100 N) cos(60) = (100 N)(0.500) = adjacent side = 50 N Vertical Component: (100)sin(60) = Opposite side = Vertical Component 100 sin(60) = 86.6 N

- Determine the net force by adding up all the horizontal forces and the vertical forces ○ 50 N is the horizontal component of the 100 N force ○ 40 N vertical force: weight of the shot ○ 86.6 N vertical force: upward vertical component of the 100 N force (50 N)2 + (46.6 N)2 = C2 4671.56 = C2 .5

- Force Resolution - The process of determining what two force components add together to make a resultant force

- Sample Problem 1.3 (PG 38) - The biceps muscle exerts a pulling force of 800 N on the radius bone of the forearm. The force acts at an angle of 30° to the radius in an anterior and superior direction. How large is the component of this force that pulls the radius toward the elbow join and how large is the component of this force that pulls perpendicular to the radius? ○ Step 1: Draw the force and show the angle it makes with the radius

○ Step 2: Draw a box around the force with sides parallel to and perpendicular to the radius. The 800 N force is the diagonal o the box. Two triangles are formed by the box and the 800 N diagonal. Choose the one with the side along the radius. This is the force triangle

○ Step 3: Use the cosine function to compute the force component pulling toward the elbow joint (the side of the triangle parallel to the radius). ▪

▪ ▪ ○ Step 4: Use the sine function to compute the force component pulling perpendicular to the radius (the side of the triangle perpendicular to the radius). ▪

▪ ▪

Static Equilibrium (PG 37) - Static Equilibrium - An object at rest ○ The state or condition of an object that results when the object is not moving and the net force and net torque acting on the object are zero

Free-Body Diagrams (PG 39)

- A woman in skates is standing still on the ice. The woman’s mass is 50 kg. What external forces act on her? - Center of Gravity - The imaginary point in space through which the force of gravity acts on that object ○ The point at which the entire weight of the body may be assumed to be concentrated ○ The point above which the torques created by the weights of the various body parts balance ○ The point of balance of the body

Static Analysis (PG 39) - If an object is not moving it is in static equilibrium - In Static Equilibrium: ○ Acceleration is zero ○ The sum of all external forces acting on the object is zero ○ Equation: ∑F = 0

- The Skater: ○ Mass = 50 kg ○ Weight = 50 * 9.81 = 490.5 N or 50 * 10 = 500 N ○ Equilibrium Equation for the skater: ▪ ∑F = R + (-500 N) = R - 500 N = 0 □ R represents the reaction force from the ice ▪ R - 500 N = 0 ▪ R - 500 N + 500 N = 0 + 500 N ▪ R = +500 N (the reaction force from the ice in the upward direction) - External forces must be equal in size but act in opposite directions Example: An 80 kg weightlifter has lifted 100 kg over his head and is holding it still. He and the barbell are in static equilibrium. What is the reaction force from the floor that must act on the weightlifter’s feet? Draw the free-body diagram first. Should the barbell be included in the free-

body diagram of the weightlifter? ○ Three external forces act on the weightlifter: 1. The reaction force from the floor acting upward on the athlete’s feet 2. The weight of the athlete acting downward through the athlete’s center of gravity 3. The reaction force from the barbell acting downward on the athlete’s hands.

- The weight of the barbell is ○ W' = mg = (100 kg)(10 m/s2) = -1000 N - The reaction force exerted by the hands on the barbell: ○ ∑F = R' + (-1000 N) = R' - 1000 N = 0 ○ R' - 1000 N + 1000 N = 0 + 1000 N ○ R' = +1000 N ○ This force is equal in magnitude but opposite in direction to the force exerted by the barbell on the hands ▪ Hands on the barbell: 1000 N upward ▪ Reaction force exerted by the barbell, hands must be a force of 1000 N acting downward - Reaction force from the floor: ○ Weight of the athlete ▪ W = mg = (80 kg)(- 10 m/s2) = -800 N - Reaction force from the floor: ○ ∑F = R + (-1000 N) + (-800 N) = R - 1800 N = 0 ○ R - 1800 N + 1800 N = 0 + 1800 N ○ R = +1800 N (acting upward on the weightlifter's feet) - External Forces: ○ Vertical: ▪ Reaction from the floor acting upward on the athlete's feet - Noncontact Forces: ○ Gravity of the barbell ○ Gravitational forces as the weight of the athlete: 800 N ○ The weight of the barbell: 1000 N ○ Reaction Force From the Floor: ▪ ∑F = R + (-800 N) + (-1000 N) = R - 1800 N = 0 R - 1800 N + 1800 N = 0 + 1800 N R = 1800 N (upward on the weightlifter)

- Unknowns: The reaction force from the floor and the reaction force from the barbell - The external forces acting on the barbell are the weight of the barbell and the reaction force from the hands acting upward on the barbell.

- Example: Brian, a 200 kg strongman competitor, is competing in the truck pull event in a strongman competition. Brian wears a harness attached to a cable, and the cable is attached to a truck and trailer. Brian is also using his hands to pull on a rope that is attached to an immovable object in front of him. Brian is attempting to move the truck, but it hasn’t yet started moving and neither has Brian. Brian is in a state of static equilibrium. Brian pulls forward and slightly upward on the cable with a force of 2200 N acting a an angle of 14° above horizontal. Brian also pulls on the rope with a horizontal force of 650 N. How much force does the ground exe against Brian’s feet?

- The force of gravity acting on Brian ○ W = mg = (200 kg)(-9.81 m/s2) = -1962 N - Force of the cable from the truck exerted against Brian ○ The force Brian exerts on the cable is 2200 N ○ Forces come in pairs and are in equal and opposite directions ○ The force the cable exerts against Brian is 2200 N but in a downward angle of 14° below the horizontal - The force the rope exerts against Brian ○ Brian pulls back on the rope with a force of 650 N ○ The force the rope exerts against Brian is also 650 N, but it pulls forward on Brian in the horizontal direction

- Equilibrium Equations of External Forces: ○ Horizontal: ∑Fx = 0 ○ Vertical: ∑Fy = 0 - Resolve this 2200 N force into its vertical and horizontal components. Do thi...