FORCE AND FREE-BODY DIAGRAM

Forces and free-body diagrams - Giancoli, Ch4, section 7 visit the Physics Classroom for this topic
1. Describe force as the cause of deformation or velocity change.
· Force -- a push or a pull -- may do one of two things to an object. First, it can change the state of motion of the object à change its velocity. Second it can change the physical state of the object à sitting on a cushion does not put the cushion in motion, but it does change the shape of the cushion.
2. Identify the forces acting on an object and draw free-body diagrams representing the forces acting.
· Identify all forces acting on a single object and draw a vector diagram showing the forces. Separate diagrams are drawn for each object in the problem.
· Each force should be labeled by name or given a commonly accepted symbol. Vectors should have lengths approximately proportional to their magnitudes. The direction of the vector should also be labeled.
· Common symbols used for forces include the following. W (=mg) for the weight of an object, T (=tension) for strings or cords pulling an object, f (=friction) for the contact friction force exerted by one object on another, and N (=Normal) for the contact support force exerted by one object on another.
3. Resolve forces into components.
· Using the trig ratios and choosing your triangles carefully, find the perpendicular components that make up a force.
· Some hints about choosing triangles: if the direction of motion is known, choose this as one of the axes for finding force components. This also means that the direction perpendicular to the motion will be your other axis. OR, if the direction of the net force is known, choose this as one of the axes for finding force components.
4. Determine the resultant force in different situations.
· Once all forces acting have been broken down into perpendicular components (you get to decide what the two perpendicular directions are), add up the forces acting along each of the two axes that you chose to find the net force along each axis.
· It may be necessary to use the Pythagorean theorem to add the two perpendicular resultant forces together to find the net resultant force. You might also have to draw a new triangle and use tan-1 to find the angle for the direction of the net resultant force.
5. Describe the behavior of a linear spring and solve related problems.
· Springs exert what is called a restoring force on objects. The spring's force is directed opposite to the displacement of the spring. If I pull on a spring to the right, the spring pulls back to the left.
· The more a spring is stretched, the greater the force it exerts on objects.
· The two statements above can be combined in a force equation, known as Hooke's Law. F = -kx. The minus sign indicates the spring's force is opposite to the displacement, x. The fact that F is directly proportional to x shows that the force increases as x increases.

Newton's first law - Giancoli, Ch 4, section 2 visit the Physics Classroom for this topic
6. State Newton's first law of motion.
· Newton's first law of motion describes the effect of a force on an object.
· Every body continues in its state of rest or uniform motion unless it is compelled to change its state by a net external force acting on it.
· In physics the term uniform means "constant". The interpretation of uniform motion means constant motion -- both speed and direction. Uniform motion denotes moving in a straight line with constant speed.
7. Describe examples of Newton's first law.
· Reference demos done in class

Equilibrium - Giancoli, Ch 9, 1-2
8. State the condition for translational equilibrium.
· Translation refers to linear motion. There is also rotational equilibrium which we will encounter later.
· Translational equilibrium means to have a uniform state of motion. Objects may remain at rest or they may remain in uniform motion.
· Based on Newton's first law of motion, the condition for translational equilibrium to occur is that there may be no net external force acting on the object. Another way of saying it is Fnet = 0.
9. Solve problems involving translational equilibrium.
· Problems will state that certain forces (such as friction) are acting and that the object maintains a constant velocity. This is your clue that the object is in translational equilibrium and that Fnet = 0.



Newton's second law - Giancoli, Ch 4, Section 4, 6-7, 9-10 visit the Physics Classroom for this topic
10. State Newton's second law of motion.
· Newton's second law of motion describes how forces change motion.
· The change in motion (change in velocity) or acceleration is directly proportional to the net force applied.
· The change in motion (change in velocity) or acceleration is inversely proportional to the mass of the object.
· In equation form: a = Fnet / m
11. Solve problems involving Newton's second law.
· In practice, because of the vector nature of determining Fnet, the second law is usually written as Fnet = ma.
· This form of the equation gives us a method for applying the second law.
· First, draw a free body diagram for each object present in the problem.
· Second, find any vector components necessary to make all forces (or components) act in two perpendicular directions. Choose one of the directions to be the direction of motion.
· Write an expression for the net force in each of the two perpendicular directions for each object.
· Set each net force = ma for that object in that direction.
· Apply any conditions present -- e.g. objects linked by string both have the same acceleration.

Friction - Giancoli, Ch 4, section 8 visit the Physics Classroom for this topic
12. Describe the nature and properties of frictional forces.
· Friction is a force that occurs because of contact between two objects. Microscopically, even the smoothest surfaces appear quite rough and as two surfaces interact the ridges of one will "catch" on the valleys of the other, producing a loss of motion.
· The amount of friction depends on the size, shape, and mass of the objects involved.
· Materials are rated for the amount of friction they produce using a property called the coefficient of friction (m).
13. Distinguish between static and dynamic (sliding) friction.
· For objects in motion, the force of dynamic (sliding or kinetic) friction acts opposite to the direction of motion.
· The force of dynamic friction is directly related to the normal (support) force between the two surfaces.
· Static friction occurs when objects are in static equilibrium (Fnet = 0 and there is no motion), for example a book sitting on a table. I you push horizontally on the book, it will initially resist the force and remain at rest. This is the result of static friction, which opposes the force you are applying to the book. At some point you will push hard enough to get the book to move and dynamic friction will take over.
· The amount of static friction present will be less than or equal to some maximum value, which is directly related to the normal force between the two surfaces.
14. Define coefficient of friction.
· The coefficient of friction is a proportionality constant and is defined as the ratio of the friction force to the normal force between the two objects. m = f / N. Because there are two types of friction, there are two different coefficients, mk for kinetic friction and ms for static friction.
· For static friction and for kinetic friction
· Because of the nature of the two friction forces -- you almost always have to push harder to get a box to start to slide across the floor than you do to keep it moving -- ms is almost always greater than mk.
15. Solve static and dynamic problems involving friction.
· The key to problems that include friction is to remember that friction always acts opposite to the direction of motion.
· Since friction is parallel to the surface between two objects and its magnitude is related to the normal force (which acts perpendicular to the surface between the two objects), the equation for the magnitude of the friction (f = mN) will link the two equations that you wrote when applying Newton's second law.



Problems: GIANCOLI - Chapter 4 - #1, 3, 11, 15, 19, 21, 25, 27, 29, 37, 39, 43, 51a, 55