Slide 1 / 156 Slide 2 / 156 Things to Remember from Last Year Dynamics · Newton's Three Laws of Motion · Inertial Reference Frames · Mass vs. Weight · Forces we studied: weight / gravity normal force tension friction (kinetic and static) · Drawing Free Body Diagrams · Problem Solving Slide 3 / 156 Slide 4 / 156 Newton's Laws of Motion 1 Which of Newton's laws best explains why motorists should buckle-up? 1. An object maintains its velocity (both speed and First Law A direction) unless acted upon by a nonzero net force. Second Law B 2. Newton’s second law is the relation between Third Law C acceleration and force. ΣF = ma Law of Gravitation D 3. Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first object. Slide 5 / 156 Slide 6 / 156 2 You are standing in a moving bus, facing forward, 3 You are standing in a moving bus, facing forward, and you suddenly fall forward. You can infer from and you suddenly fall forward as the bus comes to this that the bus's an immediate stop. What force caused you to fall forward? A velocity decreased. gravity A velocity increased. B normal force due to your contact with the floor B speed remained the same, but it is turning of the bus C to the right. force due to friction between you and the floor C speed remained the same, but it is turning of the bus D to the left. There is not a force leading to your fall. D
Slide 7 / 156 Slide 8 / 156 4 When the rocket engines on the spacecraft are 5 A net force F accelerates a mass m with an suddenly turned off, while traveling in empty acceleration a. If the same net force is applied to space, the starship will mass 2m, then the acceleration will be stop immediately. 4a A A B slowly slow down, then stop. B 2a C go faster. a/2 C a/4 move with constant speed D D Slide 9 / 156 Slide 10 / 156 6 A net force F acts on a mass m and produces an 7 An object of mass m sits on a flat table. The Earth acceleration a. What acceleration results if a net pulls on this object with force mg, which we will force 2F acts on mass 4m? call the action force. What is the reaction force? a/2 A The table pushing up on the object with force A mg 8a B The object pushing down on the table with B 4a C force mg D 2a The table pushing down on the floor with force C mg The object pulling upward on the Earth with D force mg Slide 11 / 156 Slide 12 / 156 Inertial Reference Frames 8 A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the car Newton's laws are only valid in inertial reference frames: the force on the truck is greater then the A force on the car An inertial reference frame is one in which Newton’s the force on the truck is equal to the force first law is valid. This excludes rotating and B accelerating frames. on the car the force on the truck is smaller than the C force on the car the truck did not slow down during the D collision
Slide 13 / 156 Slide 14 / 156 Normal Force and Weight Mass and Weight F N MASS is the measure of the inertia of an object, the resistance of an object to accelerate. The Normal Force, F N , is ALWAYS WEIGHT is the force exerted on that object by perpendicular to gravity. Close to the surface of the Earth, where the the surface. gravitational force is nearly constant, the weight is: mg F G = mg Weight, mg, is ALWAYS directed downward. Mass is measured in kilograms, weight Newtons. Slide 15 / 156 Slide 16 / 156 9 The acceleration due to gravity is lower on the 10 A 14 N brick is sitting on a table. What is the Moon than on Earth. Which of the following is normal force supplied by the table? true about the mass and weight of an astronaut on the Moon's surface, compared to Earth? 14 N upwards A 28 N upwards B Mass is less, weight is same A 14 N downwards C Mass is same, weight is less B 28 N downwards D Both mass and weight are less C Both mass and weight are the same D Slide 17 / 156 Slide 18 / 156 Kinetic Friction 11 A 4.0kg brick is sliding on a surface. The coefficient of kinetic friction between the surfaces Friction forces are is 0.25. What it the size of the force of friction? v ALWAYS parallel to the surface exerting f K A 8.0 them. 8.8 B Kinetic friction is 9.0 C always directed opposite to 9.8 D direction the object is sliding and has magnitude: f K = μ k F N
Slide 19 / 156 Slide 20 / 156 Static Friction 12 A brick is sliding to the right on a horizontal surface. Static friction is What are the directions of the two surface forces: always equal and the friction force and the normal force? f S F APP opposite the Net Applied Force right, down A acting on the object (not including right, up B friction). left, down C Its magnitude is: left, up D f S ≤ μ S F N Slide 21 / 156 Slide 22 / 156 13 A 4.0 kg brick is sitting on a table. The coefficient 14 A 4.0kg brick is sitting on a table. The coefficient of static friction between the surfaces is 0.45. of static friction between the surfaces is 0.45. If a What is the largest force that can be applied 10 N horizontal force is applied to the brick, what horizontally to the brick before it begins to slide? will be the force of friction and will the brick move? 16.33 A 16.12, no A 17.64 B B 17.64, no 17.98 C 16.12, yes C 18.12 D 17.64, yes D Slide 23 / 156 Slide 24 / 156 Tension Force 15 A crane is lifting a 60 kg load at a constant velocity. Determine the tension force in the cable. When a cord or rope pulls on an A 568 N object, it is said to be under 578 N tension, and the force it exerts is B called a tension force, F T . F T 504 N C a 600 N D mg
Slide 25 / 156 Slide 26 / 156 16 A system of two blocks is accelerated by an applied force of magnitude F on the frictionless horizontal surface. The tension in the string between the blocks is: Two Dimensions A 6F 6 kg 4 kg F B 4F C 3/5 F D 1/6 F E 1/4 F Slide 27 / 156 Slide 28 / 156 Adding Orthogonal Forces Adding Orthogonal Forces 1. Draw the first force vector, 50 N at 0 o , Since forces are vectors, they add as vectors. beginning at the origin. The simplest case is if the forces are perpendicular (orthogonal) with another. Let's add a 50 N force at 0 o with a 40 N force at 90 o . Slide 29 / 156 Slide 30 / 156 Adding Orthogonal Forces Adding Orthogonal Forces 1. Draw the first force 1. Draw the first force vector, 50 N at 0 o , vector, 50 N at 0 o , beginning at the origin. beginning at the origin. 2. Draw the second force 2. Draw the second force vector, 40 N at 90 o , with its vector, 40 N at 90 o , with its tail at the tip of the first tail at the tip of the first vector. vector. 3. Draw F net from the tail of the first force to the tip of the last force.
Slide 31 / 156 Slide 32 / 156 Adding Orthogonal Forces Adding Orthogonal Forces We get the direction of the net We get the magnitude of the Force from the inverse tangent. resultant from the Pythagorean Theorem. tan( θ) = opp / adj c 2 = a 2 + b 2 tan( θ) = (40N) / (50N) 2 = (50N) 2 + (40N) 2 F net tan( θ) = 4/5 = 2500N 2 + 1600N 2 = 4100N 2 tan( θ) = 0.8 F net = (4100N 2 ) 1/2 θ = tan -1 (0.80) = 39 o = 64 N F net = 64 N at 39 o Slide 33 / 156 Slide 34 / 156 Decomposing Forces into Orthogonal Decomposing Forces into Orthogonal Components Components We have F net , which is the Let's decompose a 120 N hypotenuse, so let's first force at 34 o into its x and y find the adjacent side by components. F using F cosθ = adj / hyp θ θ cosθ = F x / F net F x F x = F net cosθ = 120N cos34 o = 100 N Slide 35 / 156 Slide 36 / 156 Decomposing Forces into Orthogonal Adding non-Orthogonal Forces Components Now that we know how to add orthogonal forces. Now let's find the opposite side by using And we know now to break forces into orthogonal F y components, so we can make any force into two sinθ = adj / hyp F orthogonal forces. sinθ = F y / F net θ We combine those two steps to add any number of Fy = F net sinθ forces together at any different angles. = 120N sin34 o F x = 67 N
Slide 37 / 156 Slide 38 / 156 Adding non-Orthogonal Forces Adding non-Orthogonal Forces First, here are the vectors all drawn Let's add together these three forces: from the origin. F 1 = 8.0 N at 50 o F 2 = 6.5 N at 75 o F 3 = 8.4 N at 30 o First, we will do it graphically, to show the principle for how we will do it analytically. F 1 = 4.0 N at 50 o F 2 = 6.5 N at 75 o Then, we will do it analytically, which is easy once F 3 = 8.4 N at 30 o you see why it works that way. Slide 39 / 156 Slide 40 / 156 Adding non-Orthogonal Forces Adding non-Orthogonal Forces Now we arrange Then we draw in the F 3 F 3 them tail to tip. Resultant, the sum of the vectors. F 2 F 2 F F F 1 F 2 F 3 F 1 F 1 F 1 = 4.0 N at 50 o F 1 = 4.0 N at 50 o F 2 = 6.5 N at 75 o F 2 = 6.5 N at 75 o F 3 = 8.4 N at 30 o F 3 = 8.4 N at 30 o Slide 41 / 156 Slide 42 / 156 Adding non-Orthogonal Forces Adding non-Orthogonal Forces Then, we remove F 3 Now, we graphically the original vectors F 3y break each vector F 3y and note that the into its orthogonal sum of the F 3x F 3x components. components is the F F 2 same as the sum of F F 2y F 2y the vectors. F 2x F 2x F 1 F 1y F 1y F 1 = 4.0 N at 50 o F 1 = 4.0 N at 50 o F 1x F 1x F 2 = 6.5 N at 75 o F 2 = 6.5 N at 75 o F 3 = 8.4 N at 30 o F 3 = 8.4 N at 30 o
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