June 17, Week 3 Today: Chapter 4, Forces Homework #3 is now - PowerPoint PPT Presentation

June 17, Week 3 Today: Chapter 4, Forces Homework #3 is now available. Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant I Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant I A x = positive A y = positive Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant II Quadrant I A x = positive A y = positive Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant II Quadrant I A x = negative A x = positive A y = positive A y = positive Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant II Quadrant I A x = negative A x = positive A y = positive A y = positive Quadrant III Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant II Quadrant I A x = negative A x = positive A y = positive A y = positive Quadrant III A x = negative A y = negative Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant II Quadrant I A x = negative A x = positive A y = positive A y = positive Quadrant III Quadrant IV A x = negative A y = negative Forces 17th June 2014

Quadrants Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components A x = − 1 m and A y = − 1 m . When your calculator is wrong, it’s always 180 ◦ off Quadrant II Quadrant I A x = negative A x = positive A y = positive A y = positive Quadrant III Quadrant IV A x = negative A x = positive A y = negative A y = negative Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ (b) θ = − 53 ◦ Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ (b) θ = − 53 ◦ (c) θ = 307 ◦ Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ (b) θ = − 53 ◦ (c) θ = 307 ◦ (d) θ = 233 ◦ Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ (b) θ = − 53 ◦ (c) θ = 307 ◦ (d) θ = 233 ◦ θ = 53 ◦ (e) Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ (b) θ = − 53 ◦ (c) θ = 307 ◦ (d) θ = 233 ◦ θ = 53 ◦ (e) Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? (a) θ = 127 ◦ 4 m − 3 m Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? − → r (a) θ = 127 ◦ 4 m − 3 m Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? − → r (a) θ = 127 ◦ 4 m − 3 m tan − 1 ( − 4 / 3) = − 53 ◦ ← wrong quadrant θ = − 53 ◦ + 180 ◦ = 127 ◦ Forces 17th June 2014

Component Exercise Which of the following is the correct standard angle for the vector with components x = − 3 m , y = 4 m ? − → r (a) θ = 127 ◦ 127 ◦ 4 m − 3 m tan − 1 ( − 4 / 3) = − 53 ◦ ← wrong quadrant θ = − 53 ◦ + 180 ◦ = 127 ◦ Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. y Assume: − → A and − → B x Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. y Assume: − → A and − → B Find the components of − → A A y x A x Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. y Assume: − → A and − → B Find the components of − → A B y Find the components of − → B A y x A x B x Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. y Assume: − → A and − → B Find the components of − → A B y Find the components of − → B A y Find the vector sum − → x R A x B x Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. y Assume: − → A and − → B Find the components of − → A B y R y Find the components of − → B A y Find the vector sum − → x R A x B x R x Forces 17th June 2014

Component Addition While we cannot add the magnitudes of vectors. We can add the components. y Assume: − → A and − → B Find the components of − → A B y R y Find the components of − → B A y Find the vector sum − → x R A x B x The components of − → R : R x R x = A x + B x R y = A y + B y Forces 17th June 2014

Projectile Motion Projectile Motion is one example of two-dimensional motion with a constant acceleration. Forces 17th June 2014

Projectile Motion Projectile Motion is one example of two-dimensional motion with a constant acceleration. Projectile - Any object that is launched into motion and then acted on by gravity only. Forces 17th June 2014

Projectile Motion Projectile Motion is one example of two-dimensional motion with a constant acceleration. Projectile - Any object that is launched into motion and then acted on by gravity only. Ignore air resistance again. Forces 17th June 2014

Projectile Motion Projectile Motion is one example of two-dimensional motion with a constant acceleration. Projectile - Any object that is launched into motion and then acted on by gravity only. Ignore air resistance again. Gravity pulls straight down, so it causes acceleration in the y -direction only. a x = 0 , a y = − g ( Down is negative ) Forces 17th June 2014

Launch Angle ( v x ) i and ( v y ) i are the components of the initial velocity vector. Usually, we are given the launch speed, v i and angle, θ . Forces 17th June 2014

Launch Angle ( v x ) i and ( v y ) i are the components of the initial velocity vector. Usually, we are given the launch speed, v i and angle, θ . − → v i = launch speed v i θ = launch angle θ Forces 17th June 2014

Launch Angle ( v x ) i and ( v y ) i are the components of the initial velocity vector. Usually, we are given the launch speed, v i and angle, θ . − → v i = launch speed v i θ = launch angle θ ( v x ) i = v i cos θ Forces 17th June 2014

Launch Angle ( v x ) i and ( v y ) i are the components of the initial velocity vector. Usually, we are given the launch speed, v i and angle, θ . − → v i = launch speed v i θ = launch angle θ ( v x ) i = v i cos θ ( v y ) i = v i sin θ Forces 17th June 2014