1 / 156

1.61k likes | 1.91k Vues

Dynamics. Things to Remember from Last Year. 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. Newton's Laws of Motion.

Télécharger la présentation
## Dynamics

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Things to Remember from Last Year**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**Newton's Laws of Motion**1. An object maintains its velocity (both speed and direction) unless acted upon by a nonzero net force. 2. Newton’s second law is the relation between acceleration and force. 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. ΣF = ma**1**Which of Newton's laws best explains why motorists should buckle-up? A First Law B Second Law C Third Law D Law of Gravitation**2**You are standing in a moving bus, facing forward, and you suddenly fall forward. You can infer from this that the bus's A velocity decreased. B velocity increased. C speed remained the same, but it is turning to the right. D speed remained the same, but it is turning to the left.**3**You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes to an immediate stop. What force caused you to fall forward? A gravity B normal force due to your contact with the floor of the bus C force due to friction between you and the floor of the bus D There is not a force leading to your fall.**4**When the rocket engines on the spacecraft are suddenly turned off, while traveling in empty space, the starship will A stop immediately. B slowly slow down, then stop. C go faster. D move with constant speed**5**A net force F accelerates a mass m with an acceleration a. If the same net force is applied to mass 2m, then the acceleration will be A 4a B 2a C a/2 D a/4**6**A net force F acts on a mass m and produces an acceleration a. What acceleration results if a net force 2F acts on mass 4m? A a/2 B 8a C 4a D 2a**7**An object of mass m sits on a flat table. The Earth pulls on this object with force mg, which we will call the action force. What is the reaction force? A The table pushing up on the object with force mg B The object pushing down on the table with force mg C The table pushing down on the floor with force mg D The object pulling upward on the Earth with force mg**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 A the force on the truck is greater then the force on the car B the force on the truck is equal to the force on the car C the force on the truck is smaller than the force on the car D the truck did not slow down during the collision**Inertial Reference Frames**Newton's laws are only valid in inertial reference frames: An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.**Mass and Weight**MASS is the measure of the inertia of an object, the resistance of an object to accelerate. WEIGHT is the force exerted on that object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is: Mass is measured in kilograms, weight Newtons. FG = mg**Normal Force and Weight**FN The Normal Force, FN, is ALWAYS perpendicular to the surface. Weight, mg, is ALWAYS directed downward. mg**9**The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth? A Mass is less, weight is same B Mass is same, weight is less C Both mass and weight are less D Both mass and weight are the same**10**A 14 N brick is sitting on a table. What is the normal force supplied by the table? A 14 N upwards B 28 N upwards C 14 N downwards D 28 N downwards**Kinetic Friction**Friction forces are ALWAYS parallel to the surface exerting them. Kinetic friction is always directed opposite to direction the object is sliding and has magnitude: fK = μkFN v fK**11**A 4.0kg brick is sliding on a surface. The coefficient of kinetic friction between the surfaces is 0.25. What it the size of the force of friction? A 8.0 B 8.8 C 9.0 D 9.8**12**A brick is sliding to the right on a horizontal surface. What are the directions of the two surface forces: the friction force and the normal force? A right, down B right, up C left, down D left, up**Static Friction**Static friction is always equal and opposite the Net Applied Force acting on the object (not including friction). Its magnitude is: fS ≤ μSFN fS FAPP**13**A 4.0 kg brick is sitting on a table. The coefficient of static friction between the surfaces is 0.45. What is the largest force that can be applied horizontally to the brick before it begins to slide? A 16.33 B 17.64 C 17.98 D 18.12**14**A 4.0kg brick is sitting on a table. The coefficient of static friction between the surfaces is 0.45. If a 10 N horizontal force is applied to the brick, what will be the force of friction and will the brick move? A 16.12, no B 17.64, no C 16.12, yes D 17.64, yes**Tension Force**When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force, FT. FT a mg**15**A crane is lifting a 60 kg load at a constant velocity. Determine the tension force in the cable. A 568 N B 578 N C 504 N D 600 N**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: A 6F F 6 kg 4 kg B 4F C 3/5 F D 1/6 F E 1/4 F**Adding Orthogonal Forces**Since forces are vectors, they add as vectors. The simplest case is if the forces are perpendicular (orthogonal) with another. Let's add a 50 N force at 0o with a 40 N force at 90o.**Adding Orthogonal Forces**1. Draw the first force vector, 50 N at 0o, beginning at the origin. 90o 180o 0o 270o**Adding Orthogonal Forces**1. Draw the first force vector, 50 N at 0o, beginning at the origin. 2. Draw the second force vector, 40 N at 90o, with its tail at the tip of the first vector. 90o 180o 0o 270o**Adding Orthogonal Forces**1. Draw the first force vector, 50 N at 0o, beginning at the origin. 2. Draw the second force vector, 40 N at 90o, with its tail at the tip of the first vector. 3. Draw Fnet from the tail of the first force to the tip of the last force. 90o 180o 0o 270o**Adding Orthogonal Forces**We get the magnitude of the resultant from the Pythagorean Theorem. c2 = a2 + b2 Fnet2 = (50N)2 + (40N)2 = 2500N2 + 1600N2 = 4100N2 Fnet = (4100N2)1/2 = 64 N 90o 180o 0o 270o**Adding Orthogonal Forces**We get the direction of the net Force from the inverse tangent. tan(θ) = opp / adj tan(θ) = (40N) / (50N) tan(θ) = 4/5 tan(θ) = 0.8 θ = tan-1(0.80) = 39o Fnet = 64 N at 39o 90o 180o 0o 270o**Decomposing Forces into Orthogonal Components**90o Let's decompose a 120 N force at 34o into its x and y components. F θ 180o 0o 270o**Decomposing Forces into Orthogonal Components**90o We have Fnet, which is the hypotenuse, so let's first find the adjacent side by using cosθ = adj / hyp cosθ = Fx / Fnet Fx = Fnet cosθ = 120N cos34o = 100 N F θ 180o 0o Fx 270o**Decomposing Forces into Orthogonal Components**90o Now let's find the opposite side by using sinθ = adj / hyp sinθ = Fy / Fnet Fy = Fnet sinθ = 120N sin34o = 67 N Fy F θ 180o 0o Fx 270o**Adding non-Orthogonal Forces**Now that we know how to add orthogonal forces. And we know now to break forces into orthogonal components, so we can make any force into two orthogonal forces. We combine those two steps to add any number of forces together at any different angles.**Adding non-Orthogonal Forces**Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o First, we will do it graphically, to show the principle for how we will do it analytically. Then, we will do it analytically, which is easy once you see why it works that way.**Adding non-Orthogonal Forces**90o First, here are the vectors all drawn from the origin. F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 180o 0o 270o**Adding non-Orthogonal Forces**90o Now we arrange them tail to tip. F3 F2 F1 180o 0o F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 270o**Adding non-Orthogonal Forces**90o Then we draw in the Resultant, the sum of the vectors. F3 F2 F F F1 F2 F3 F1 180o 0o F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 270o**Adding non-Orthogonal Forces**90o F3 Now, we graphically break each vector into its orthogonal components. F3y F3x F F2 F2y F2x F1 F1y F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 180o 0o F1x 270o**Adding non-Orthogonal Forces**90o Then, we remove the original vectors and note that the sum of the components is the same as the sum of the vectors. F3y F3x F F2y F2x F1y 180o 0o F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o F1x 270o**Adding non-Orthogonal Forces**90o This is the key result: The sum of the x- components of the vectors equals the x- component of the Resultant. As it is for the y- components. F3y F F2y Fy F1y F3x F1x F2x F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o 180o 0o Fx 270o**Adding non-Orthogonal Forces**This explains how we will add vectors analytically. 1. Write the magnitude and direction (from 0 to 360 degrees) of each vector. 2. Compute the values of the x and y components. 3. Add the x components to find the x component of the Resultant. 4. Repeat for the y-components. 5. Use Pythagorean Theorem to find the magnitude of the Resultant. 6. Use Inverse Tangent to find the Resultant's direction.**Adding non-Orthogonal Forces**First make a table and enter what you know. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o**Adding non-Orthogonal Forces**Then calculate the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o**Adding non-Orthogonal Forces**Then add up the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o**Adding non-Orthogonal Forces**Then use Pythagorean Theorem and tan-1 to determine Fnet. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o**Adding non-Orthogonal Forces**Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o**Adding non-Orthogonal Forces**Now use the same procedure to add these forces. DO NOT IGNORE NEGATIVE SIGNS F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o

More Related