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1 Vector Addition Using Vector Components = + R x = A x + B x B y R y = A y + B y R = (R x 2 + R y 2 ) 1/2 B x q = tan -1 (R y /R x )

2 Example 1.7: Vector has a magnitude of 50 cm and direction of 30º, and vector has a magnitude of 35 cm and direction 110º (both angles measured ccw from ). What is the resultant vector? Page 19 Adding Vectors Using Components R x = A x + B x = 50 cos(30 ) + 35 cos(110 ) = (-12.0) = 31.3 cm R y = A y + B y = 50 sin(30 ) + 35 sin (110 ) = = 57.9 cm R = (R x 2 + R y 2 ) 1/2 = ( ) 1/2 = 66 cm q = tan -1 (R y /R x ) = tan -1 (57.9/31.3) = 62.

3 Superposition of Forces: Resultant and Components of Force Vectors An example of superposition of forces: y In general, the resultant, or vector sum of forces, is: x θ = tan 1 R y R x direction

4 Motion in a Plane (2D) with Constant Acceleration A General Rule: Two sets of quantities are treated separately/independently. All the x-components of the quantities are related to each other in one set of kinematic equations: v x t = v 0x + a x t...(2.6) All the y-components of the quantities are related to each other in another set of kinematic equations: v y t = v 0y + a y t...(2.6) x t = x 0 + v 0x t a xt 2 (2.10) v 2 x = v 2 0x + 2a x x x 0 (2.11) y t = y 0 + v 0y t a yt 2 (2.10) v 2 y = v 2 0y + 2a y y y 0 (2.11) v av,x = 1 2 [v x t + v 0x ].. (2.7) v av,y = 1 2 [v y t + v 0y ].. (2.7) Important initial steps for solving 2D motion with constant acceleration: Set up a convenient x-y coordinate system. Identify the x and y components of initial position, initial velocity, and acceleration. Apply the rule set above and NEVER mix x-component and y-component quantities in the same kinematic equation.

5 Projectile Motion: The motion in a vertical plane of a point particle, given an initial velocity, under the influence of a constant gravitation acceleration, with other factors such as air friction and wind, etc., all neglected. 3.3 Projectile Motion v y = 0 at the maximum height A Summary of the Parameters (with the given coordinates) Acceleration: a x = 0 a y = - g = m s 2 Initial Conditions: x 0 = 0 y 0 = 0 v 0x = v 0 cos(q 0 ) v 0y = v 0 sin(q 0 ) q 0 constant a y constant v x since a x = 0 Other Examples airplane dropping a package motorcycle running off a cliff rock sliding off the edge of a roof etc. x-components v x t = v 0 cos(q 0 ) x t = v 0 cos(q 0 )t y-components v y t = v 0 sin(q 0 ) gt y t = v 0 sin(q 0 )t 1 2 gt2 v y 2 = (v 0 sin(q 0 )) 2 2gy

6 Projectile Motion x-components v x t = v 0 cos(q 0 ) x t = v 0 cos(q 0 )t y-components v y t = v 0 sin(q 0 ) gt y t = v 0 sin(q 0 )t 1 2 gt2 v y 2 = (v 0 sin(q 0 )) 2 2gy v y = 0 at the maximum height constant a y Examples 3.4&3.5: A home run hit Given: v 0 and q 0 Find: (a) x and y, and, v and q, at t = 2.00 s (b) t max and h max (c) Horizontal range R Solutions: (a) x t = v 0 cos(q 0 )t y t = v 0 sin(q 0 )t 1 2 gt2 v x t = v 0 cos(q 0 ) v y t = v 0 sin(q 0 ) gt v = (v x 2 + v y 2 ) 1 2 θ = tan 1 (v y /v x ) q 0 constant v x since a x = 0 (b) Set v y t = v 0 sin(q 0 ) gt = 0 to get t max = v 0 sin(q 0 )/g. Then, set v 2 y = (v 0 sin(q 0 )) 2 2gy = 0 to get h max = (v 0 sin(q 0 )) 2 /2g. (c) Set y t = v 0 sin(q 0 )t 1 2 gt2 = t(v 0 sin(q 0 ) 1 gt)= 0 2 to get the time of flight t f = 2v 0 sin(q 0 )/g = 2t max. Then, using x t = v 0 cos(q 0 )t to get the range R = 2v 2 0 sin(q 0 )cos(q 0 )/g = v 2 0 sin(2q 0 )/g

7 3.4 Uniform Circular Motion An object moving along a circular path with a constant speed v (magnitude of velocity) Velocity and Acceleration Vectors in Uniform Circular Motion Velocity: Accelerarion: tangent to the circle with constant magnitude v 1 = v 2 = v a rad = v2, always pointing toward the center of the circle. R v = റs v R റa av = v = v റs t R t റv = v R റs v റa = lim = v lim റs = v റv t 0 t R t 0 t R a rad = റa = v R റv = v2 R

8 The Component Form: σ റF = m റa σ F x = ma x and σ F y = ma y Example: An object of mass 5.0 kg is acted upon by two forces, F A and F B. F A is directed toward east and has a magnitude 3.0 N. F B is directed toward north and has a magnitude 4.0 N. (a) Draw diagram that describes this situation. (b) Set up a coordinate system. (c) Calculate the components, the magnitude, and the direction of the net force. (d) Calculate the components and the magnitude of the acceleration. y റF B O റF θ റF A x (a) Sketch a diagram that describes this situation. (b) Set up a coordinate system. (c) F x = F Ax = 3.0 N F y = F By = 4.0 N F = ( ) 1/2 = 5.0 N θ = tan = (d) a x = F x = 3.0 = 0.60 m 5.0 m/s2 a y = F y = 4.0 = 0.80 m/s2 m 5.0 a = ( ) 1/2 = 1.0 m/s 2 or a = F/m = 1.0 m/s 2

9 Newton s Second Law: in vector form σ റF = m റa in component form σ F x = ma x σ F y = ma y Strategy for Solving Newton s Law Problems Isolate the bodies in a system. Analyze all the forces acting on each body and draw one free-body diagram for each body. Based on the free-body diagram, set up a most convenient x-y coordinate system. Break each force into components using this coordinates. For each body, sum up all the x-components of the forces to an equation: σ F x = ma x. For each body, sum up all the y-components of the forces to an equation: σ F y = ma y. Use these equations to solve for unknown quantities. Example 1: A box of known mass m is resting on a leveled table surface. Find all the forces acting on this box and their actionreaction counterparts. y n W m Two forces act on the box: weight (known) normal force W = mg n Sum up the y component forces: n + ( mg) = 0. Therefore, n = mg.

10 Example: Box of mass m on a frictionless incline Given: m and q Find: n and a y n Solutions: Weight: W = mg W x = mgsinq W y = mgcosq x-axis: mgsinq = ma a = gsinq y-axis: n mgcosq = 0 n = mgcosq x q q mg Another Question: Near the bottom of the incline, the box is given an initial velocity of a known magnitude v 0 pointing up the incline. What distance will it slide before it turns around and slides downward? Solution: The acceleration is given above a = gsinq Use the kinematic equation v 2 2 v 2 1 = 2a(x 2 x 1 ) (x 2 x 1 ) = v 2 1 /2a = v 2 0 /2gsinq The distance that it will slide up the incline is v 2 0 /2gsinq.

11 Example: Box of mass m is pushed up an incline by a force F parallel to the incline. The coefficient of kinetic friction between the box and the incline is μ k. y f k n F x Given: F, m, μ k, and q Find: n and a Solutions: Weight and its components W = mg W x = mgsinq W y = mgcosq x-axis: F - mgsinq - f k = ma f k = μ k n = μ k mgcosq q q mg a = (F - mgsinq - μ k mg)/m y-axis: n mgcosq = 0 n = mgcosq What if some other quantities are given and you are asked to calculate some different quantities? Consider Problem 9 on Term Exam 1 from 2015.

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