Module 8 - 1 Newton’s First Law of Motion Aristotle said that force was necessary to make an object move with constant velocity. Example: Pull a box across the table. Must pull to keep it going. (Aristotle was fixated on friction)
Module 8 - 2 Newton’s First Law of Motion Newton: asked what would happen if friction could be eliminated. “Every body continues in its state of rest or uniform speed in a straight line unless acted on by a nonzero net force.”
Module 8 - 3 Newton’s Second Law of Motion The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object Notice that mass is a measure of an object’s resistance to acceleration. We usually write the equation as
Module 8 - 4 Notes on Forces Units: FORCE: newton (N) 1 N = kg · m /s 2 MASS: kilogram (kg) Since this is a vector equation, it can be written in component form:
Module 8 - 5 Example 4-1 (9) A 0.140-kg baseball traveling 35.0 m/s strikes the catchers mitt, which, in bringing the ball to rest recoils backward 11.0 cm. What was the average force applied by the glove on the ball? Force is in opposite direction to velocity. First find a: (to the left)
Module 8 - 6 Newton’s Third Law of Motion Law of Action - Reaction Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. Examples: skater leans on wall wall exerts an equal but opposite force on skater Earth exerts a force on moon moon exerts equal but opposite force on earth
Module 8 - 7 Newton’s Third Law of Motion Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. Is important to realize that these forces act on different things and thus they don’t cancel.
Module 9 - 1 Weight Galileo told us that all objects experience an acceleration due to gravity of g and Newton gave us F = ma. We can combine these to ideas to realize that the weight of an object is the force of attraction that the earth exerts on objects and it can be written It should be noted that weight is a force and thus the proper units are newtons (N) or pounds. It is technically incorrect to say that something weights 2.0 kg because that is the mass of the object.
Module 9 - 2 Application: Ropes Tension: when the man pulls on the rope, the tension in the rope transmits the force to the box. The tension is 100 N which exerts a 100 N force upward on the box and a 100 N downward force on the hand.
Module 9 - 3 Steps in Solving Problems 1.Draw free-body diagram for every object that is ”free” 2.Select coordinate system such that one of the axis is along the direction of acceleration 3.Write out the equations of motion for the x and y coordinate: 4.Step 2 should guarantee that the sum of the forces in all but one direction equals zero. 5.Solve the equations simultaneously
Module 9 - 4 Application: Normal Force When a box rests on a table, the table must exert enough upward force to support the box, otherwise, the table will collapse. This upward force is called the normal force F N because it is normal to the surface. When we push down with a force of 40 N the normal force will increase by 40 N.
Module 9 - 5 Free-Body Diagrams Essential part of solution Vital tool to understand problem Forces are the only vectors on free-body diagrams If there are two objects, each of them will have a free-body diagram If there are two objects, label each mass properly: m 1 and m 2 Select a coordinate system such that the acceleration direction is along one axis Then apply Newton’s Second Law (two equations for each object):
Module 9 - 6 Example 4-2 A 65-kg woman ascends in an elevator that briefly accelerates at 1.0 m/s 2 upward when leaving a floor. She stands on a scale that reads in N. When acceleration is zero, the scale reads her weight:
Module 9 - 7 Application: Ropes and Pulleys A pulley changes the direction of the tension in the rope. If the pulley is frictionless and massless then the tension in the left rope is the same as the right
Module 9 - 8 Example 4-3 Two masses hang from a massless, frictionless pulley as shown. Draw free-body diagram for each of the masses. Derive a formula for the acceleration of the masses. Assume m 1 = 0.250 kg and m 2 = 0.200 kg.
Module 8 - 1 Example 4-3 Two masses hang from a massless, frictionless pulley as shown. Draw free-body diagram for each of the masses. Calculate the acceleration of the masses and the tension. Assume m 1 =0.250 kg and m 2 = 0.200 kg.
Module 9 - 9 Comments on Example 4-12 in Book Each box has the same acceleration a A = a B = 1.82 m/ s 2 F T is not equal to F P F P = 40 N F T = (12 kg) ( 1.82 m/s 2 ) = 22 N Treat as a single mass:
Module 10 - 1 Kinetic Friction Friction results when two surfaces slide across each other because even the smoothest surfaces have some roughness. Kinetic Friction results when a body slides across a surface. It is proportional to the normal force between the surfaces: where k is a unit-less number called the coefficient of kinetic friction.
Module 10 - 2 Static Friction If we gradually increase the applied force by adding water, the static friction force matches it until the object starts to move. Once it is sliding, the friction is kinetic and is constant. Static Friction: arises as a result of an external force even when the body is not yet moving:
Module 10 - 3 Example 4-4 From the data in the graph, determine µ k and µ k. Just before the box starts to move
Module 10 - 5 Inclined Planes An inclined plane exerts a normal force F N which is perpendicular to the surface. There may also be a frictional force which opposes the motion. It should also be noted that the angle between the weight and the normal is the same as the angle of the incline .
Module 10 - 6 Example 4-5 A block of wood rests on a wooden board. Derive the equations of motion
Module 10 - 7 Example 4-5 A block of wood rests on a wooden board. One end of the board is raised until the block starts to slip. Determine the coefficient of static friction if θ = 25 0 when it starts to slip. At the point where it starts to slip a ≈ 0 and = s.
Module 10 - 8 One 2.80 kg paint bucket (m 1 ) is hanging by a massless cord from a 3.50 kg paint bucket (m 2 ), also hanging by a massless cord. If the two buckets are pulled upward with an acceleration of 0.700 m/s 2 by the upper cord, calculate the tension in each cord. Example 4- 6