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

June 17, Week 3

Forces 17th June 2014

Today: Chapter 4, Forces Homework #3 is now available.

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles!

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m.

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive Quadrant II

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive Quadrant II Ax = negative Ay = positive

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive Quadrant II Ax = negative Ay = positive Quadrant III

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive Quadrant II Ax = negative Ay = positive Quadrant III Ax = negative Ay = negative

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive Quadrant II Ax = negative Ay = positive Quadrant III Ax = negative Ay = negative Quadrant IV

Quadrants

Forces 17th June 2014

Sometimes your calculator will be wrong in finding angles! Example: Find the magnitude and direction for the vector with components Ax = −1 m and Ay = −1 m. When your calculator is wrong, it’s always 180◦ off Quadrant I Ax = positive Ay = positive Quadrant II Ax = negative Ay = positive Quadrant III Ax = negative Ay = negative Quadrant IV Ax = positive Ay = negative

Component Exercise

Forces 17th June 2014

Which of the following is the correct standard angle for the vector with components x = −3 m, y = 4 m?

Component Exercise

Forces 17th June 2014

Which of the following is the correct standard angle for the vector with components x = −3 m, y = 4 m? (a) θ = 127◦

Component Exercise

Forces 17th June 2014

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◦

Component Exercise

Forces 17th June 2014

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◦

Component Exercise

Forces 17th June 2014

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◦

Component Exercise

Forces 17th June 2014

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◦ (e) θ = 53◦

Component Exercise

Forces 17th June 2014

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◦ (e) θ = 53◦

Component Exercise

Forces 17th June 2014

Which of the following is the correct standard angle for the vector with components x = −3 m, y = 4 m? (a) θ = 127◦ −3 m 4 m

Component Exercise

Forces 17th June 2014

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

Component Exercise

Forces 17th June 2014

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

Component Exercise

Forces 17th June 2014

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

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components.

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components. x y Assume: − → A and − → B

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components. x y Ax Ay Assume: − → A and − → B Find the components of − → A

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components. x y Ax Ay Bx By Assume: − → A and − → B Find the components of − → A Find the components of − → B

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components. x y Ax Ay Bx By Assume: − → A and − → B Find the components of − → A Find the components of − → B Find the vector sum − → R

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components. x y Ax Ay Bx By Rx Ry Assume: − → A and − → B Find the components of − → A Find the components of − → B Find the vector sum − → R

Component Addition

Forces 17th June 2014

While we cannot add the magnitudes of vectors. We can add the components. x y Ax Ay Bx By Rx Ry Assume: − → A and − → B Find the components of − → A Find the components of − → B Find the vector sum − → R The components of − → R: Rx = Ax + Bx Ry = Ay + By

Projectile Motion

Forces 17th June 2014

Projectile Motion is one example of two-dimensional motion with a constant acceleration.

Projectile Motion

Forces 17th June 2014

Projectile Motion is one example of two-dimensional motion with a constant acceleration. Projectile - Any object that is launched into motion and then acted

• n by gravity only.

Projectile Motion

Forces 17th June 2014

Projectile Motion is one example of two-dimensional motion with a constant acceleration. Projectile - Any object that is launched into motion and then acted

• n by gravity only.

Ignore air resistance again.

Projectile Motion

Forces 17th June 2014

Projectile Motion is one example of two-dimensional motion with a constant acceleration. Projectile - Any object that is launched into motion and then acted

• n by gravity only.

Ignore air resistance again. Gravity pulls straight down, so it causes acceleration in the y-direction only. ax = 0, ay = −g (Down is negative )

Launch Angle

Forces 17th June 2014

(vx)i and (vy)i are the components of the initial velocity vector. Usually, we are given the launch speed, vi and angle, θ.

Launch Angle

Forces 17th June 2014

(vx)i and (vy)i are the components of the initial velocity vector. Usually, we are given the launch speed, vi and angle, θ. − → vi θ vi = launch speed θ = launch angle

Launch Angle

Forces 17th June 2014

(vx)i and (vy)i are the components of the initial velocity vector. Usually, we are given the launch speed, vi and angle, θ. − → vi θ vi = launch speed θ = launch angle (vx)i = vi cos θ

Launch Angle

Forces 17th June 2014

(vx)i and (vy)i are the components of the initial velocity vector. Usually, we are given the launch speed, vi and angle, θ. − → vi θ vi = launch speed θ = launch angle (vx)i = vi cos θ (vy)i = vi sin θ

Summary

Forces 17th June 2014

Projectile Equations ax = 0 ay = −g (vx)f = (vx)i (vy)f = (vy)i − g∆t xf = xi + (vx)i∆t yf = yi + (vy)i∆t − 1

2g∆t2

(vx)i = vi cos θ (vy)i = vi sin θ

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first?

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (a) The first tennis ball.

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (a) The first tennis ball. (b) The second tennis ball.

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (a) The first tennis ball. (b) The second tennis ball. (c) It depends on how fast the first is launched.

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (a) The first tennis ball. (b) The second tennis ball. (c) It depends on how fast the first is launched. (d) They hit at the same time.

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (a) The first tennis ball. (b) The second tennis ball. (c) It depends on how fast the first is launched. (d) They hit at the same time. (e) None of these are correct.

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (a) The first tennis ball. (b) The second tennis ball. (c) It depends on how fast the first is launched. (d) They hit at the same time. (e) None of these are correct.

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (d) They hit at the same time. For each ball: yf = yi + (vy)i∆t − 1

2g∆t2

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (d) They hit at the same time. For each ball: yf = yi + (vy)i∆t − 1

2g∆t2

For each ball: yf = 0 and yi = h = the height at which they are launched

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (d) They hit at the same time. For each ball: yf = yi + (vy)i∆t − 1

2g∆t2

For each ball: yf = 0 and yi = h = the height at which they are launched For each ball: (vy)i = 0 since a horizontal vector (one to the right) has no y-component. − → vi

Projectile Motion Exercise

Forces 17th June 2014

Two tennis balls are started from the same height above the ground at the exact same time. The first tennis ball is launched horizontally with some unknown speed. The second tennis ball is simply

• dropped. Ignoring air resistance, which of the two hits the ground

first? (d) They hit at the same time. For each ball: yf = yi + (vy)i∆t − 1

2g∆t2

For each ball: yf = 0 and yi = h = the height at which they are launched For each ball: (vy)i = 0 since a horizontal vector (one to the right) has no y-component. − → vi For each ball: Their vertical motion is the same ⇒ the same ∆t!

Dynamics

Forces 17th June 2014

Dynamics - Why objects move.

Dynamics

Forces 17th June 2014

Dynamics - Why objects move. Sir Isaac Newton (1642-1727) British Physicist, In 1687 he published the Philosophiæ Naturalis Principia Mathematica. The Principia details how all motion can be explained by one

• f three simple statements = Newton’s

Three Laws of Motion.

Exceptions

Forces 17th June 2014

These three laws are still in use; however, they need modification in two situations:

Exceptions

Forces 17th June 2014

These three laws are still in use; however, they need modification in two situations: When an object’s velocity approaches the speed of light

Exceptions

Forces 17th June 2014

These three laws are still in use; however, they need modification in two situations: When an object’s velocity approaches the speed of light - Einstein’s theory of relativity.

Exceptions

Forces 17th June 2014

These three laws are still in use; however, they need modification in two situations: When an object’s velocity approaches the speed of light - Einstein’s theory of relativity. Motion of atomic-sized objects

Exceptions

Forces 17th June 2014

These three laws are still in use; however, they need modification in two situations: When an object’s velocity approaches the speed of light - Einstein’s theory of relativity. Motion of atomic-sized objects - Quantum Mechanics (Also started by Einstein).

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force.

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces:

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact.

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact. Long-range Forces - Act at a distance. Examples = gravity, magnetism.

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact. Long-range Forces - Act at a distance. Examples = gravity, magnetism. Weight - − → w, Force on an object due to gravity.

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact. Long-range Forces - Act at a distance. Examples = gravity, magnetism. Weight - − → w, Force on an object due to gravity. Unit of Force:

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact. Long-range Forces - Act at a distance. Examples = gravity, magnetism. Weight - − → w, Force on an object due to gravity. Unit of Force: U. S. Customary: Pound (lb)

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact. Long-range Forces - Act at a distance. Examples = gravity, magnetism. Weight - − → w, Force on an object due to gravity. Unit of Force: U. S. Customary: Pound (lb)

• S. I. : Newton (N),

Force

Forces 17th June 2014

Underlying all three of Newton’s Laws is the concept of force. Force, − → F - Push or Pull exerted by some agent Two types of forces: Contact Forces - Two objects in contact. Long-range Forces - Act at a distance. Examples = gravity, magnetism. Weight - − → w, Force on an object due to gravity. Unit of Force: U. S. Customary: Pound (lb)

• S. I. : Newton (N), 1 N = 0.22 lb (on Earth)

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object.

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object.

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object. − → F1

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object. − → F1 − → F2 − → F3

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object. − → F1 − → F2 − → F3 Superposition - The net result of two or more forces is given by the vector sum. Σ− → F = − → F1 + − → F2 + − → F3 . . .

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object. − → F1 − → F2 − → F3

⇒

Superposition - The net result of two or more forces is given by the vector sum. Σ− → F = − → F1 + − → F2 + − → F3 . . . Σ− → F

Superposition

Forces 17th June 2014

Usually there is more than one force acting on an object. − → F1 − → F2 − → F3

⇒

Superposition - The net result of two or more forces is given by the vector sum. Σ− → F = − → F1 + − → F2 + − → F3 . . . Σ− → F One single force. Applied at the center to avoid rotation.