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VECTORS AND SCALARS A scalar quantity has only magnitude and is completely specified by a number and a unit. Examples are: mass (2 kg), volume (1.5 L), and frequency (60 Hz). Scalar quantities of the same kind are added by using ordinary arithmetic. A vector quantity has both magnitude and direction. Examples are displacement (an airplane has flown 200 km to the southwest), velocity (a car is moving at 60 km/h to the north), and force (a person applies an upward force of 25 N to a package). When vector quantities are added, their directions must be taken into account. A vector is represented by an arrowed line whose length is proportional to the vector quantity and whose direction indicates the direction of the vector quantity. The resultant, or sum, of a number of vectors of a particular type (force vectors, for example) is that single vector that would have the same effect as all the original vectors taken together. R VECTOR COMPONENTS A vector in two dimensions may be resolved into two component vectors acting along any two mutually perpendicular directions. 2.1 Draw and calculate the components of the vector F (250 N, 235o) Fx = F cos = 250 cos (235o) = - 143.4 N Fy = F sin = 250 sin (235o) = - 204.7 N Fx F Fy VECTOR ADDITION: COMPONENT METHOD To add two or more vectors A, B, C,… by the component method, follow this procedure: 1. Resolve the initial vectors into components x and y. 2. Add the components in the x direction to give Σx and add the components in the y direction to give Σy . That is, the magnitudes of Σx and Σy are given by, respectively: Σx = Ax + Bx + Cx… Σy = Ay + By + Cy… 3. Calculate the magnitude and direction of the resultant R from its components by using the Pythagorean theorem: R 2 x and tan 1 y x 2 y 2.2 Three ropes are tied to a stake and the following forces are exerted. Find the resultant force. A (20 N, 0º) B (30 N, 150º) C (40 N, 232º) A (20 N, 0) B (30 N, 150) C (40 N, 232) x-component 20 cos 0 30 cos 150 40 cos 232 y-component 20 sin 0 30 sin 150 40 sin 232 Σx = - 30.6 N Σy = -16.5 N R (30.6) 2 (16.5) 2 = 34.7 N tan 1 y x tan 1 16.5 = 28.3 30.6 Since Σx = (-) and Σy = (-) R is in the III Quadrant: therefore: 180 + 28.3 = 208.3 R (34.7 N, 208.3) 2.3 Four coplanar forces act on a body at point O as shown in the figure. Find their resultant with the component method. A (80 N, 0) B (100 N, 45) C (110 N, 150) D (160 N, 200) A (80 N, 0) C (110 N, 150) B (100 N, 45) D (160 N, 200) x-component 80 100 cos 45 110 cos 150 160 cos 200 y-component 0 100 sin 45 110 sin 150 160 sin 200 Σx = - 95 N Σy = 71 N R (95) 2 (71) 2 = 118.6 N tan 1 71 = 36.7 95 Since Σx = (-) and Σy = (+) R is in the II Quadrant, therefore: 180 - 36.7= 143.3 R (118.6 N, 143.3) R (5.7) 2 (3.2) 2 tan 1 = 6.5 N 3.2 = 29 5.7 Since Σx = (+) and Σy = (-) R is in the IV Quadrant, therefore: 360 - 29= 331 R (6.5 N, 331) AP PHYSICS LAB 2. VECTOR ADDITION Objective: The purpose of this experiment is to use the force table to experimentally determine the equilibrant force of two and three other forces. This result is checked by the component method. A system of forces is in equilibrium when a force called the equilibrant force is equal and opposite to their resultant force. Equipment Force Table Set DATA Table FORCE MASS (kg) FORCE (N) mg = m (9.8 m/s2) F1 F2 Equilibrant FE Resultant FR DIRECTION An object that experiences a push or a pull has a force exerted on it. Notice that it is the object that is considered. The object is called the system. The world around the object that exerts forces on it is called the environment. system FORCE Forces can act either through the physical contact of two objects (contact forces: push or pull) or at a distance (field forces: magnetic force, gravitational force). Type of Force and its Symbol Applied Force FA Description of Force An applied force is a force that is applied to an object by another object or by a person. If a person is pushing a desk across the room, then there is an applied force acting upon the desk. The applied force is the force exerted on the desk by the person. Direction of Force In the direction of the pull or push. Type of Force and its Symbol Normal Force FN Description of Force The normal force is the support force exerted upon an object that is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. The normal force is always perpendicular to the surface Direction of Force Perpendicular to the surface Type of Force and its Symbol Friction Force FF Description of Force The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force opposes the motion of the object. For example, if a book moves across the surface of a desk, the desk exerts a friction force in the direction opposite to the motion of the book. Direction of Force Opposite to the motion of the object Type of Force and its Symbol Air Resistance Force FD Description of Force Air resistance is a special type of frictional force that acts upon objects as they travel through the air. Like all frictional forces, the force of air resistance always opposes the motion of the object. This force will frequently be ignored due to its negligible magnitude. It is most noticeable for objects that travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas. Direction of Force Opposite to the motion of the object Type of Force and its Symbol Tensional Force FT Description of Force Tension is the force that is transmitted through a string, rope, or wire when it is pulled tight by forces acting at each end. The tensional force is directed along the wire and pulls equally on the objects on either end of the wire. Direction of Force In the direction of the pull Type of Force and its Symbol Gravitational Force (also known as Weight) Fg Description of Force The force of gravity is the force with which the earth, moon, or other massive body attracts an object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on an object on earth is always equal to the weight of the object. Direction of Force Straight downward FORCES HAVE AGENTS Each force has a specific identifiable, immediate cause called agent. You should be able to name the agent of each force, for example the force of the desk or your hand on your book. The agent can be animate such as a person, or inanimate such as a desk, floor or a magnet. The agent for the force of gravity is Earth's mass. If you can't name an agent, the force doesn't exist. agent A free-body-diagram (FBD) is a vector diagram that shows all the forces that act on an object whose motion is being studied. Directions: - Choose a coordinate system defining the positive direction of motion. - Replace the object by a dot and locate it in the center of the coordinate system. - Draw arrows to represent the forces acting on the system. FN FG FN FG FN FF FG FN FGY FGX FG FN FF FG FT FG FT1 FT2 FG FT1 FT2 FG FN FT FG FN1 FN2 FG FN FF FA FG FN FA θ FF FG FT FG FN FF θ FA FG FG FD FG 17.The ball has been punted by a football player. FG ARISTOTLE studied motion and divided it into two types: natural motion and violent motion. Natural motion: up or down. Objects would seek their natural resting places. Natural for heavy things to fall and for very light things to rise. Violent motion: imposed motion. A result of forces. GALILEO demolish the notion that a force is necessary to keep an object moving. He argued that ONLY when friction is present, is a force needed to keep an object moving. In the absence of air resistance (drag) both objects will fall at the same time. A ball rolling down the incline rolls up the opposite incline and reaches its initial height. As the angle of the upward incline is reduced, the ball rolls a greater distance before reaching its initial height. How far will the ball roll along the horizontal? FIRST LAW OF MOTION According to Newton's First Law of Motion: Isaac Newton (1642-1727) " If no net force acts on it, a body at rest remains at rest and a body in motion remains in motion at constant speed in a straight line." NEWTON'S FIRST LAW OF MOTION "An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force." There are two parts to this statement: - one which predicts the behavior of stationary objects and - the other which predicts the behavior of moving objects. The behavior of all objects can be described by saying that objects tend to "keep on doing what they're doing" (unless acted upon by an unbalanced force). If at rest, they will continue in this same state of rest. If in motion with an eastward velocity of 5 m/s, they will continue in this same state of motion (5 m/s, East). It is the natural tendency of objects to resist changes in their state of motion. This tendency to resist changes in their state of motion is described as inertia. Inertia is the resistance an object has to a change in its state of motion. The elephant at rest tends to remain at rest. Tablecloth trick: Too little force, too little time to overcome "inertia" of tableware. If the car were to abruptly stop and the seat belts were not being worn, then the passengers in motion would continue in motion. Now perhaps you will be convinced of the need to wear your seat belt. Remember it's the law - the law of inertia. If the truck were to abruptly stop and the straps were no longer functioning, then the ladder in motion would continue in motion. If the motorcycle were to abruptly stop, then the rider in motion would continue in motion. The rider would likely be propelled from the motorcycle and be hurled into the air. FIRST CONDITION FOR EQUILIBRIUM A body is in translational equilibrium if and only if the vector sum of the forces acting upon it is zero. Σ Fx = 0 Σ Fy = 0 Example of Free Body Diagram 300 600 A B Ay A 300 Ax B By 600 Bx 50 N Fg 1. Draw and label a sketch. 2. Draw and label vector force diagram. (FBD) 3. Label x and y components opposite and adjacent to angles. 2.4 A block of weight 50 N hangs from a cord that is knotted to two other cords, A and B fastened to the ceiling. If cord B makes an angle of 60˚ with the ceiling and cord A forms a 30° angle, draw the free body diagram of the knot and find the tensions A and B. A B By A 30º 60º y Ax Bx 50 N N1L ΣFx = B cos 60º - A cos 30º = 0 A cos 30 B cos 60 = 1.73 A ΣFy = B sin 60º + A sin 30º - 50 = 0 1.73 A sin 60º + A sin 30º = 50 1.5 A + 0.5 A = 50 A = 25 N B = 1.73 (25) = 43.3 N A = 25 N B = 43.3 N 2.5 A 200 N block rests on a frictionless inclined plane of slope angle 30º. A cord attached to the block passes over a frictionless pulley at the top of the plane and is attached to a second block. What is the weight of the second block if the system is in equilibrium? N1L FN FT θ y x 200 N ΣFx = FT - 200 sin 30º = 0 FT = 200 sin 30º = 100 N FT FG2 ΣFy = FT - FG = 0 FT = FG2 FG2 = 100 N To swing open a door, you exert a force. The doorknob is near the outer edge of the door. You exert the force on the doorknob at right angles to the door, away from the hinges. To get the most effect from the least force, you exert the force as far from the axis of rotation (imaginary line through the hinges) as possible. TORQUE Torque is a measure of a force's ability to rotate an object. Torque is Determined by Three Factors: • The magnitude of the applied force. • The direction of the applied force. • The location of the applied force. Each The 40-N of theforce 20-N The forces nearer the end forces produces has atwice different the of the wrench have greater torque torquetorques. as due does to the the direction 20-N force. of force. Magnitude Locationofof of force force Direction Force 20 N 2020 N 20NN 20 40NN 20 N 20 N The perpendicular distance from the axis of rotation to the line of force is called the lever arm of that force. It is the lever arm that determines the effectiveness of a given force in causing rotational motion. If the line of action of a force passes through the axis of rotation (A) the lever arm is zero. Torque Units for Torque Depends on the magnitude of the applied force and on the length of the lever arm, according to the following equation. r is measured perpendicular to the line of action of the force F t = Fr Units: Nm t = (40 N)(0.60 m) = 24.0 Nm 6 cm 40 N Applying a Torque Sign Convention: Torque will be positive if F tends to produce counterclockwise rotation. Torque will be negative if F tends to produce clockwise rotation. ROTATIONAL EQUILIBRIUM An object is in rotational equilibrium when the sum of the forces and torques acting on it is zero. First Condition of Equilibrium: Σ Fx = 0 and Σ Fy = 0 (translational equilibrium) Second Condition of Equilibrium: Στ = 0 (rotational equilibrium) By choosing the axis of rotation at the point of application of an unknown force, problems may be simplified. CENTER OF MASS The terms "center of mass" and "center of gravity" are used synonymously in a uniform gravity field to represent the unique point in an object or system that can be used to describe the system's response to external forces and torques. The concept of the center of mass is that of an average of the masses factored by their distances from a reference point. In one plane, that is like the balancing of a seesaw about a pivot point with respect to the torques produced. Center of Gravity The center of gravity of an object is the point at which all the weight of an object might be considered as acting for purposes of treating forces and torques that affect the object. The single support force has line of action that passes through the c. g. in any orientation. Examples of Center of Gravity Note: C. of G. is not always inside material. 2.6 A 300 N girl and a 400 N boy stand on a 16 m platform supported by posts A and B as shown. The platform itself weighs 200 N. What are the forces exerted by the supports on the platform? ΣF = 0 A + B - 300 - 200 - 400 = 0 A + B = 900 N Selecting B as the hinge ΣτB = 0 -A(12) +300(10) +200(4) - 400(4) = 0 - A12 + 3000 + 800 - 1600 = 0 2200 A = 183. 3 N 12 B = 900 - 183.3 = 716.7 N Selecting A as the hinge ΣτA = 0 - 300(2) - 200(8) + B(12) - 400(16) = 0 - 600 - 1600 + B12 - 6400 = 0 8600 = 716.7 N B 12 A = 900 - 716.7 = 183. 3 N 2.7 A uniform beam of negligible weight is held up by two supports A and B. Given the distances and forces listed, find the forces exerted by the supports. ΣF = 0 A - 60 - 40 + B = 0 A + B = 100 A B 3m 60 N 6m 2m 40 N ΣτA = 0 = - 60 (3) - 40(9) + B(11) = 0 540 B = 49.1 N 11 A = 100 - 49.1 = 50.9 N