“ Friendship is like peeing on yourself; everyone can see it, but only you get the warm feeling it brings.” Funnyquotes.com Course web page 3 Cell phones put away when class begins please.

Announcements Reading: Chapter 2, sections 1-3; Chapter 3 section 1; and Chapter 4, all. HW2 is on WileyPlus now and due Monday by 5pm. Sample problems also on WileyPlus. Test 1 is on Wednesday. A formula sheet is posted on the course web page.

Chapter 4: Forces and Newton's Laws of Motion

Forces we will use this semester: 1) Gravity 2) Normal: Component of gravity perpendicular to the surface: N=F w 3) Friction 4) Tension 5) Spring (Chapter 10, so later)

N= -F w N= -F w -F 1 N= -F w F1F1 From last time: The Normal Force

Friction F f =  N (  is always given) → N = -F w → F w = mg Static: the object doesn't move. Kinetic: the object is moving along the surface.

Example we did last time: +Y +X -X

+Y +X The key was choosing axes to make it easiest. -X

Catalog of Forces Tension – force exerted through a rope (etc.) by pulling on an object with the rope. Tension problems are just vectors!

Tension example: A box is hanging from two ropes. If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? Think of the forces just as vectors and solve. Nothing more than that.

Tension example: A box is hanging from two ropes. If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? Choose axes and solve for components. Remember gravity! Firstly, the box is going nowhere, so the forces must completely cancel in all directions.

Tension example: If the mass of the box is 23kg, what is the tension in ropes 1 and 2? Choose axes +X +Y 71 12

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 Choose axes: put angles from a common point. I'll choose the +X direction. For 1: =109 o +X +Y

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 Look at the forces in X and Y directions: Y: T 1Y +T 2Y = -F g X: T 1X = -T 2X +X +Y We now have 2 equations and 2 things we don't know (T 1 and T 2 ). Now it's just solving equations, which you know how to do.

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 Look at the forces in X and Y directions: Y: T 1Y +T 2Y = -F g X: T 1X = -T 2X +X +Y I'll solve for T 1 in relation to T 2 Using the X equation. T 1 cos(109) = -T 2 cos(12) T 1 = 3.0T 2

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 Look at the forces in X and Y directions: Y: T 1Y +T 2Y = -F g X: T 1X = -T 2X +X +Y Substitute this into our Y equation. Now down to 1 unknown T 1 = 3.0T 2 T 1 sin(109)+T 2 sin(12) = -mg 3.0T 2 sin(109)+T 2 sin(1w) = -(23)(-9.8)

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 Look at the forces in X and Y directions: Y: T 1Y +T 2Y = -F g X: T 1X = -T 2X +X +Y Substitute this into our first equation. Now down to 1 unknown T 1 = 3.0T 2 3.0T 2 sin(109)+T 2 sin(12) = -(23)(-9.8) 2.84T T 2 = T 2 =225.4

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 Look at the forces in X and Y directions: Y: T 1Y +T 2Y = -F g X: T 1X = -T 2X +X +Y Substitute this into our first equation. Now down to 1 unknown T 1 = 3.0T 2 3.0T 2 sin(109)+T 2 sin(12) = -(23)(-9.8) T 2 = 73.9N

Tension example: If the mass of the box is 23kg, what are the tensions in ropes 1 and 2? 2 1 +X +Y Solve this back for T 1 T 2 = 73.9N T 1 = 3.0T 2 = 3(73.9) = 221.7N

The advantage of pulleys. In the image on the left, a pull on the rope translates into a tension force on the block. 1 pull = 1 F T

The advantage of pulleys. In the image on the right, a pull on the rope translates into two tension forces on the block. 1 pull = 2 F T Note that the block moves ½ as far as the pull on the rope.

The advantage of pulleys. This can be continued for more pulleys. The force on the rope converts to the number of ropes on the weight end. So 1F pull → #F T

The advantage of pulleys. The force on the rope converts to the number of ropes on the weight end. So 1F pull → #F T If the Weight is 200N, how much is F pull for each case if the weight does not move?

The advantage of pulleys. The force on the rope converts to the number of ropes on the weight end. So 1F pull → #F T If the Weight is 200N, how much is F pull for each case if the weight does not move? For 1 pulley, 1 rope, F pull =200N (just like the weight). For 2 pulleys, 2 ropes, F pull =100N

Sample Problem A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? Where do we start?

Sample Problem A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? Where do we start? Define our axes and examine the forces.

+X 4T +Y FkFk FwFw  FNFN A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? Define our axes: We swiveled our gravity vector 30 o clockwise, from the +X axis this is o. 90 o =-120 o

+X 4T +Y FkFk FwFw  FNFN A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? Examine the forces. What is the net acceleration in the Y direction? =-120 o

+X 4T +Y FkFk FwFw  FNFN A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? Define our axes and examine the forces. What is the net acceleration in the Y direction? 0 the forces must balance. =-120 o

+X 4T +Y FkFk FwFw  FNFN A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? F N = -F Wy F N = -mgsin(-120 o ) F N = 42.4N Find the normal force. =-120 o

A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? What about in the X direction? Here we have a net acceleration- the block will move along the plane. So F netX =ma x 4T+F k +F wx =ma x F N = 42.4N =-120 o

A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? What about in the X direction? Here we have a net acceleration- the block will move along the plane. So F netX =ma x 4T+F k +F wx =ma x 4(30)+  k F N +mgcos(-120 o )=ma x (42.4) =5a x F N = 42.4N =-120 o

A force of 30N is pulling on the rope. If  is 30 o and  k is 0.2, what is the acceleration of the block if it has m=5kg? What about in the X direction? Here we have a net acceleration- the block will move along the plane. So F netX =ma x 4T+F k +F wx =ma x 4(30)+  k F N +mgcos(110 o )=ma x (42.4)-24.5=5a x 87.0 = 5a x a x =17.4 m/s 2 F N = 42.4N =-120 o

F Notes on transmitted forces. Frictionless With friction

4.7 The Gravitational Force

Every object with mass exerts an attractive force on every other object with mass. Newton’s Law of Universal Gravitation

Universal gravitational constant

We only use this for large objects, otherwise we use F=mg, which is weight: F w We use:But gravity is really:So we are abbreviating

In this formula, we are only interested in how F changes when the masses or distance changes. If m 1 were to become 3m 1 how would F change? A) increase by 3. B) decrease by 3. C) increase by 9. D) decrease by 9.

In this formula, we are only interested in how F changes when the masses or distance changes. If m 1 were to become 3m 1 how would F change? A) increase by 3.

In this formula, we are only interested in how F changes when the masses or distance changes. If r were to become 3r how would F change? A) increase by 3. B) decrease by 3. C) increase by 9. D) decrease by 9.

In this formula, we are only interested in how F changes when the masses or distance changes. If r were to become 3r how would F change? D) decrease by 9.

Equilibrium State of a body in which there is no change in its state of motion => either at rest or moving at constant motion (speed and direction).

Equilibrium If there is no acceleration, there is no NET force (though forces may be applied). If a=0, then  F = 0. This may be for the entire object, or just for one direction (as in sitting on a surface, like we have done). For these problems, sum all the forces and set them equal to zero.

ALWAYS DRAW A PICTURE OF ALL FORCES ACTING ON AN OBJECT WITH LABELS AND DIRECTION. This is a FREE-BODY DIAGRAM. PAY ATTENTION TO IT!!!!!

Free-Body Diagrams 1)Identify all forces acting on the body. 2)Draw a coordinate system. 3)Represent the object as a dot. 4)Draw vectors representing each of the forces. 5)Resolve all forces into parallel and perpendicular components relative to the surface. 6)Draw and label the net force.

A box is pulled along a surface with a coefficient of kinetic friction of

  N N FfFf FfFf FWFW FWFW Rotated so motion is along the surface.

TTT F W1 F W2