December 10, Period 2 - Sign, 2d force, equilibrant problem. Period 2 - Inclined plane with and without friction

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1 Period 2 - Sign, 2d force, equilibrant problem Period 3 - Sign, 2d force, equilibrant problem 1) Newton s Second Law 2) Equilibrant is the opposite of the resultant of two forces 3) Components 4) Sin o= opposite/hypotenuse 5) Coso = adjacent/hypostenuse 6) When the ropes angles are equal, Ta=Tb 7) No acceleration, so when Newton s 2 nd law is applied the equations always equal zero 8) Resultant = hypotenuse of components of x and y added together to Oind a new triangle 9) Free body Diagram 10) Period 2 - Inclined plane with and without Lindsay Zipparo, Amanda Mesko, Jess Frank, Mike Flewelling Equilibrant. Sign Problem. 2d Force Period 6 - Sign, 2d force, equilibrant problem Equilibrant (force that causes the object to be in equilibrium) 1. Draw vectors 2. Find Components 3. Apply Newton s Second Law 4. Build resultant triangle 5. Find Magnitude and direction (this is the resultant) 6. The Equilibrant is the opposite force of the resultant. (For example: If after building a resultant triangle and getting a hypotenuse of 35N at an angle of 60 degrees up from the positive x axis in the +irst quadrant, the equilibrant is the exact same thing but it is in the third quadrant and down from the negative x axis. So in this speci0ic example the equilibrant would be 35N at 60 degrees down from the negative x axis). Sign Problem 1. Find components 2. Don t forget force of gravity 3. Apply Newton s second law (equations) 4. There is no acceleration 5. Different angles=bigger angle had larger tension 6. Different angles= smaller angle has smaller tension 7. Same angles=smaller the angle bigger the tension and bigger the angle smaller the tension 2d Force 1. To 0ind sum angle make two separate triangles with the two forces 2. Add components to make a resultant triangle 3. Use inverse of tan to 0ind angle

2 Period 3- Inclined plane with and without Period 4 - Inclined plane with and without Inclined Plane With and Without Friction Peter, Shemaine, Aaron Acceleration upward is 0 Zach O Neill, Ben Holzer, &/or Alex Jackson Inclined Plane w/ and w/out Friction. Normal Force Not Equal to Weight Concepts: 1. Acceleration Perpendicular is always 0 m/s^2 with inclined planes. 2. To Pind the force of you multiply mu and the normal force. 3. To Pind the force of gravity you multiply the mass (kg) by 9.8 m/s^2. 4. First you have to make a free body diagram. 5. Apply Newton s Second Law 6. Normal Force is equal to the perpendicular component of Force of Gravity. 7. You need to know theta and the mass. 8. To Pind the Fg you multiply the sine of theta by the Fg. 9. To Pind the Fg perpendicular you multiply the cosine of theta by Fg. Apply Newton s Second Law Fn equals Fg perpendicular Fg must be broken into components: Fg parallel and Fg perpendicular Fg equals mass times 9.81 Ff equals MEW times Fn Inclined angle equals the components triangle angle Draw F.B.D. Fg does NOT equal mass Period 6- Inclined plane with and without Period 2 - Pull an object along a flat plane with and without, with and Concepts for Inclined Plane with Friction Free body diagram FN is always up from the angle in which the mass is on Apply Newton s second law Build new triangle Solve cosine and sine (parallel and perpendicular) F=uk * FN After shifting the axis, theta is below parallel, and 90 degrees is perpendicular to parallel

3 Period 3 - Pull an object along a flat plane with and without, with and Period 4 - Pull an object along a flat plane with and without, with and Katrina Sterner, Shelby Naretto, Ben Talik Concepts:!lat plane pulling at an angle & without an angle with ; Hockey puck 1. Given angle. Break into components. 2. Newtons 3rd Law For every action, there is an equal and opposite reaction. 3. Acceleration due to gravity is 9.8 m/s^2. 4. Friction is opposite direction of motion. 5. FN=Fg 6. Vertical acceleration on a!lat plane is always Free Body Diagrams 8. Newtons 2 nd Law Brittany Hetzell Adele Ferris Erin O Shea Box sliding no : Free body diagram. Newton s Second Law. In the y direction the Newton s second law equals zero because there is no acceleration so the natural force and force of gravity are equal. Box sliding with : Free body diagram Newton s second law including in the x direction. In the y direction the Newton s second law equals zero because there is no acceleration so the natural force and force of gravity are equal. Solve for the force of or coefficient of, whichever necessary. Box being pulled at an angle without : Free body. Solve for components of the new triangle. Newton s second law including components in both x and y direction. In the y direction the Newton s second law is equal to zero because there is no acceleration so the natural force and force of gravity are equal. Box being pulled at an angle with : Same as without, just add to the free body diagram and Newton s second law in both the x and y direction. In the y direction the Newton s second law equals zero because there is no acceleration so the natural force and force of gravity are equal. Solve for the force of or coefficient of, whichever necessary. Hockey puck sliding on ice: Draw free body. There s no force pushing it forward. Little. In the y direction the Newton s second law equals zero because there is no acceleration so the natural force and force of gravity are equal. Solve for the force of or coefficient of, whichever necessary. Period 6 - Pull an object along a flat plane with and without, with and Period 2 - Atwood machine, modified atwood, Team 3 Flat Plane with/without Friction: No vertical acceleration Horizontal component ff + fx =m Pull at an angle: Horizontal force Force of gravity when there is an angle is not just m x 9.8 it is also vertical component of the triangle (Fy) To $ind the normal force, use the following equation: fg + fn + Fy=m(ay) then solve algebraically. Horizontal acceleration = horizontal force divided by mass Vertical acceleration = 0 Hockey puck problem: No Fa force, nothing is pushing it forward Ay in this case is 0 therefore Fn = Fg It will stop on the ice because there is kinetic

4 Brian Yankello, Brianne Durkosh, Chloe Delaney, Jeff Chmay Atwood problem, 1. Tension in string is same both sides 2. The heavier mass pulls on the lighter mass Period 4 - Atwood machine, modified atwood, Period 2 - projectile, trajectory and rocket problem 3. Acceleration on both sides are equal ModiFied Atwood problem 1. Force of gravity is equal to natural force 2. Tensions are equal 3. Acceleration in the X direction is the same as the acceleration in the Y direction Elevator 1. Picture 2. Force of the scale = mass x acceleration + Fg Helicopter 3. Force of the propeller must be greater than the force of the mass of the helicopter and the truck it s carrying. Team 4 (Hannah Levin, Justin Veshio, Aaron Singleton, Jessica Scioscia) Atwood Machine Modi?ied Atwood Machine Elevator Period 6 - Atwood machine, modified atwood, Atwood Machine: 1. Tension is the same 2. Acceleration is the same 3. Fg=mass times No horizontal force Modi?ied Atwood Machine: 1. Two free body diagrams 2. Have to use substitution 3. Only one free body diagram has a horizontal force Elevator: 1. Weightless feeling: best time to jump is at the top approaching the top 2. No horizontal force 3. Inertia is present (explains the weightless feeling)

5 Concepts for Projectile Problems Horizontal acceleration is 0 m/s^2 Velocity at the top in the y direction is 0 m/s Acceleration due to gravity is 9.8 m/s^2 The initial velocity in the y direction is 0 m/s Period 3 - projectile, trajectory and rocket problem Period 4 - projectile, trajectory and rocket problem The initial velocity in the x direction is equal to the 5inal velocity as it hits the ground assuming the surface is 5lat Concepts for Trajectory Problems Time to the top is half of the total time Velocity at the top in the y direction is 0 m/s Horizontal acceleration in the x direction is 0 m/s Acceleration due to gravity is 9.8 m/s^2 The initial velocity in the x direction is equal to the 5inal velocity as it hits the ground assuming the surface is 5lat The initial velocity in the y direction is equal but opposite to the 6inal velocity of the object Concepts for Rocket Problem No forces= Equilibrium Objects in motion remain in motion unless acted upon by an outside force Direction is not relevant Justin, CJ, Alexa 1. Gravity equals Velocity in the Y direction is zero at peak. 3. Velocity in the X direction is the same when it hits the ground, as it is when it starts. 4. The time it takes to get to the peak is half the total 5. To Kind the initial velocity in the x and y direction you have to break it down into components 6. Apply Newton s 2 nd law 7. If have the mass and you want to Kind the weight multiply the mass by When rocket is turned off there is no force on the rocket. It is in equilibrium. 9. To Kind time you use deltad=vi*t+ ½ a*t 2 but a=0 so vf=vi*t Projectile, Trajectory, and Rocket Problem: Projectile: Period 6 - projectile, trajectory and rocket problem 1. Horizontal vectors do not change what happens vertically 2. Vertical or y direction: Velocity = zero Acceleration = 9.81 m/s/s Velocity increases in the negative direction as it falls to the ground Use vf = vi + at to solve for Einal velocity 3. Horizontal or x direction: Acceleration = 0 m/s/s Horizontal velocity remains constant when in the air Use d = (vi)(t) + ½ (a)(t^2) to solve for horizontal displacement Trajectory: 1. Vertical or y direction: Find components to solve for initial velocity in y direction Acceleration = 9.81 m/s/s because free fall acceleration As it approaches top gets closer to zero As it travels from top to ground increases in the negative direction Vertical velocity of an object at peak = 0 m/s Time at highest point vf = vi + at Total time = (time at highest point)(2) 2. Horizontal or x direction: Acceleration = 0 m/s/s Horizontal velocity never changes in the air (remains constant) Horizontal displacement d = (vi)(t) + ½ (a)(t^2) Rocket Problem (Chapter 4): 1. No forces because it has constant velocity (equilibrium) 2. Objects in motion stay in motion when there are no outer forces (Newton s First Law)