3Force and Acceleration Acceleration causesVWhat causes acceleration? ForceForce: an interaction that causes accelerationKinematics: Study of motion without consideration of force (Describe motion)Mechanics: Study of force and motion (Explain motion)
4Newton’s First Law of Motion Law of Inertia An object with no net force acting on it remains at rest or moves with constant velocity in a straight line.No net force is needed to keep a constant velocity.A net force is needed to change a velocity.Net force = 0 Object at Equilibrium.A special case of Second Law of Motion.
5Interpretation of Newton’s First Law If no net force acts on a body, we can always find a reference frame in which that body has no acceleration.Inertial Frame of Reference:Fnet = 0 on a body, then a = 0 Inertial FrameFnet = 0 on the frameThe frame moves with constant velocity.Newton’s First Law: Law of InertiaInertia: Resistance to forceInertia Mass for now
6Newton’s Second Law of Motion The acceleration of a body is directly proportional to the net force on it and inversely proportional to its mass.F or Fnet: net force, total force, resultant forceForce is the cause, and acceleration is the effect.Unit of Force:F = ma [F] =[m] [a]
7Practice: An experimental rocket sled can be accelerated at a constant rate from rest to 1600 km/h in 1.8 s. What is the magnitude of the required average force if the sled has a mass of 500 kg.
8ForceForce is a vector.The acceleration component along a given axis is caused only by the sum of the force components along that same axis and not affected by force components along any other axis.
9Example: Three forces act on a particle that moves with unchanging velocity of v = (2 m/s)i – (7 m/s)j. Two of the forces are F1 = (2N)i + (3N)j + (-2N)k and F2 = (-5N)i + (8N)j + (-2N)k. What is the third force?
10Two kinds of mass Two methods to measure or use mass Inertial mass: measure mass by comparing acceleration of this object to acceleration of an object of known mass when the same force applied to both of them.Gravitational mass: measure mass by comparing gravitational force on this object to gravitational force on an object of known mass at the same location
11But what exactly is mass? mass weight or sizeintrinsic characteristichow much a body resist to forcea characteristic of a body that relates a force on the body to the resulting acceleration of the body
12Free Body Diagram Force Diagram Draw simple diagram Draw all forces acting on the object being consideredIgnore all forces this object acting on other objectsDraw forces starting from center of object or at points of actionMake sure each force giver can be identified
13Free-Body Diagram Example vN: Normal force, force of incline supporting boxDo not confuse velocity with forces.T: tension, force of person pulling on boxf: friction, force of incline surface opposing motionf: friction, force of incline surface opposing motionW: Weight of object, gravity of Earth pulling on box
14Weight or gravityWeight or gravity: gravitational force the Earth pulling on object around itAlways straight downwardm: mass of objectg = 9.8 m/s2 near surface of Earth: acceleration due to gravityWhen it is the only force acting on an object,(acceleration due to gravity) free fallHigher elevation, smaller gLower latitude (closer to equator), smaller g
15Normal Force (N, but not Newton) Given by the surface in contact to support the object.No contact No normal forceNo tendency to move into surface No normal forceAlways perpendicular to the surface in contact. (Normal = perpendicular)Points from surface to the objectNFappNN
16Two kinds of frictionsFriction: force opposing the motion or the tendency of motion between two rough surfaces that are in contactStatic friction:Kinetic (or sliding) friction:Or simply:More next chapter.
17TensionTension is the force the person pulling on the box through the string. (Or string pulling on box.)Tension is along the string and points away from the object of consideration.Pulley can be used to change direction of tension.Reading of spring scale gives magnitude of tension in string.T
19What is the tension? Same force diagram as before. Answer: b) 200 N Consider:reading =tension =Wall?Wall200N200N200NSame force diagram as before.
20Weight and Apparent Weight Weight is force of gravity of Earth pulling on the object.NApparent weight is the reading on the apparatus.TApparent weight is the normal force the scale exerting on the objectApparent weight is the tension the spring exerting on the objectApparent weight does not have to be the same as the weight.
21Example: A weight-conscious penguin with a mass of 15 Example: A weight-conscious penguin with a mass of 15.0 kg rests on a bathroom scale. What are a) the penguin’s weight W and b) the normal force N on the penguin? c) What is the reading on the scale, assuming it is calibrated in weight units?Let upward = +NW
22Practice: An object is hung from a spring balance attached to the ceiling of an elevator. The balance reads 64 N when the elevator is standing still. What is the reading when the elevator is moving upward a) with a constant speed of 7.6 m/s and b) with a speed of 7.6 m/s while decelerating at a rate of 2.4 m/s2?+W=64NTa) v = 7.6 m/s =constant T = ?Define upward as the positive direction.W=mgb) a =-2.4 m/s2, T = ?This apparent weight is different from the weight.
23Newton’s Third Law of Motion When one object exerts a force on a second object, the second object also exerts a force on the first object that is equal in magnitude but opposite in direction.Action and reaction forces always exist together.Action and reaction forces are of the same kind of force. (Both gravitational forces or both electromagnetic forces.)Action and reaction forces act on two different objects and therefore do not cancel out each other when only one object is considered.
24Action and Reaction Forces In general, the force acting on the object under consideration is the action force, and the other is the reaction.ActionReactionForce between box and ground:Action force: ground supporting the box, FgbReaction force: box pushing on the ground, Fbg
25How To Find Reaction Force Identify the force giver and receiver of the force.Reversing the order of force giver and receiver, you will get the reaction force.Example:Action force: a bReaction force: b a
27Example (2) N: normal force, force table supporting the box forces on boxWb: weight of box, force Earth pulling on boxAre N and W a pair of action and reaction force?No2 different objectssame fundamental force
28Example (3) forces on table Ngt: normal force of ground supporting tableforces on tableN: Force of box pushing on tableNWt: weight of table, gravitational force Earth pulling on table
29Example (4) forces on earth Wt: reaction force to weight of table, gravitational force table pulling on EarthWtWb: reaction force to weight of box, force box pulling on EarthWbforces on earthNtg: force of table pushing on ground (Earth), reaction force of Ngt.
30Example (5)NNgtWbWtNWtWbNtgAction and reaction forces always exist in pairs and act on different objects.
31Example: 109-27 A 40 kg girl and an 8 Example: A 40 kg girl and an 8.4 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? c) How far from the girl’s initial position do they meet?N1N2m2m1TTW1W2
33Practice: An 80-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2. The mass of the parachute is 5.0 kg. a) What upward force is exerted on the open parachute by the air? b) What downward force is exerted by the person on the parachute?What if we did force diagrams for individual masses?fChoose downward = + direction. m1 = 80 kg, a = 2.5 m/s2, m2 = 5.0 kg, M = m1 + m2 = 85 kg+a) f = ?W = MgChute & personfb) T = ?Chute onlyTW2 = m2g
34Another Approach: An 80-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2. The mass of the parachute is 5.0 kg. a) What upward force is exerted on the open parachute by the air? b) What downward force is exerted by the person on the parachute?+fChoose downward = + direction. m1 = 80 kg, a = 2.5 m/s2, m2 = 5.0 kg, M = m1 + m2 = 85 kga) f = ?Chute only?TW2 = m2gTPerson onlyW1 = m1g
35Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 N Example: Two blocks, 3.0kg and 5.0kg, are connected by a light string and pulled with a force of 16.0N along a frictionless surface. Find (a) the acceleration of the blocks, and (b) the tension between the blocks.+N1N2TTF16.0N3.0kg5.0kgW2W1Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 NConsider only forces in the horizontal direction.m1:m2:Also applies
36Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 N Another Approach: Two blocks, 3.0kg and 5.0kg, are connected by a light string and pulled with a force of 16.0N along a frictionless surface. Find (a) the acceleration of the blocks, and (b) the tension between the blocks.NN2T3.0kg5.0kg16.0NFW2WLet right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 NTake the two masses as a system. Thenm1 and m2:Then we can apply 2nd on either m1 or m2 to find T.m2:
37Example: Two blocks, one of mass 5. 0 kg and the other of mass 3 Example: Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as to the right. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find a) the tension in the rope, and b) the acceleration of the blocks.+TT3+5W2Let downward = + for m1 = 5 kg, and upward = + for m2 = 3 kg.Then two masses will have the same acceleration, a = ?. And the tensions will be the same, T = ?W1m1:m2:
38Another Approach: Two blocks, one of mass 5 Another Approach: Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as to the right. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find a) the tension in the rope, and b) the acceleration of the blocks.+3TW25+Take the two masses as a single system. Let clockwise as the positive direction of the motions.W1Then the net force accelerating the system is:And this net force is acceleration a total mass ofSo the acceleration of the system isThen we apply Newton’s second law on one of the mass to find the tension.Let downward = + for 5kg. Then
39Example: In Fig. 5-66, a force F of magnitude 12 N is applied to a FedEx box of mass m2 = 1.0 kg. The force is directed up a plane tilted by = 37o. The box is connected by a cord to a UPS box of mass m1 = 3.0 kg on the floor. The floor, plane, and pulley are frictionless, and the masses of the pulley and cord are negligible. What is the tension in the cord?vF1.0 kg3.0 kg37oy+N1xN2FTW1W2xTW2W2ySet up the coordinates as in the diagram. Also decompose W2 into the x and y directions as W2x and W2y.
40Continues … N1 W1 T x y + W2 F N2 Sub into the third eqn, we have W2x W2yW2xContinues …Sub into the third eqn, we have
41Practice: A 5.00 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force T = 12.0N at an angle = 25.0o above the horizontal. a) What is the acceleration of the block? b) The force F is slowly increased. What is its value just before the block is lifted (completely) off the floor? c) What is the acceleration of the block just before it is lifted (completely) off the floor?yNTTyxW=mgTxb) T = ?Right before the force is large enough to lift off the blockN =0,ay =