1. I NCLINED P LANES I LOVE these!

2. W HAT E XACTLY WILL WE BE LOOKING AT ? Visualize an object placed on a ramp. In physics we refer to the ramp as an “ inclined plane.” o We are still NEGLECTING friction so, if you put an object on an incline…what will it do? o Yup! It will slide down the incline. But, WHY will it slide down the incline?

3. WHY WILL IT SLIDE DOWN THE INCLINE ? Seems simple… It moves downhill because of gravity. If there is no friction (to resist the motion), then gravity is “allowed” to accelerate it down the incline. The force of gravity IS why it is pulled downhill…but there is more to it. o What is the direction of the gravitational force on the box?

4. L OOK AT THE D IRECTION OF G RAVITY … The force of gravity (i.e. weight) ALWAYS acts directly down (toward the center of Earth). Hang on…the box doesn’t move straight down – it moves down the hill. The concept of net force (F net ) indicates that there must be a force acting parallel to the motion, that causes it to accelerate down the incline. o What is this parallel force? F g = mg motion

5. W HAT IS THIS PARALLEL FORCE ? Let’s say that the incline makes an angle of  with the ground. The box pushes down on the incline and the incline pushes back on the box. The support force that the incline supplies to the box is the normal force. Remember that the normal force is always perpendicular to the surface. o We will use the normal force to lay out the rest of the picture. F g = mg motion  FNFN Do NOT draw the normal force vertically upward…it must be perpendicular to the surface!

6. L ET THE NORMAL FORCE HELP YOU … Extend the normal force in the opposite direction. This force is the component of the weight that is perpendicular to the surface. o This still isn’t parallel to the motion…but we’re getting there... F g = mg motion  FNFN FF

7. W HAT IS THE FORCE DOWN THE HILL ? Draw a line PARALLEL to the incline This force is the component of the weight that is parallel to the surface. The “parallel force” is what causes the pull down the hill! o We use F p to find F net … o So how do we find F p ? F g = mg motion  FNFN FpFp FF

8. F OCUS ON THE TRIANGLES. Notice that the inclined plane itself is a right triangle. Notice that the triangle made up of F , F p and mg is also a right triangle. These two triangles are similar triangles. o How does it help us that these two triangles are similar? F g = mg motion  FNFN FpFp FF

9. H OW DOES IT HELP US THAT THESE TWO TRIANGLES ARE SIMILAR ? The base angle of the inclined plane (  ) is the same as the angle between F  and mg. You also want to pay attention to where the right angle is. o You now have the picture you need to analyze the situation. F g = mg motion  FNFN FpFp FF 

10. W HAT DO WE DO NOW ? Let’s do a problem! A block is released (from rest) from the top of an incline that makes an angle of 30  with the ground. Neglecting friction, what is the acceleration of the block down the incline? F g = mg motion 30 FNFN o Visualize the forces acting on the box as well as the direction of the acceleration.

11. W HAT DO WE DO NOW ? If the block is going to move down the incline there must be a force directed PARALLEL to that motion. We now know that we need to break the weight (mg) up into parallel and perpendicular components. F g = mg motion 30 FNFN F  = mgcos30  F p = mgsin30  30 o The parallel component (F p = mgsin30  ) is (part of) the net force.

12. W HAT DO WE DO NOW ? F net = ma The net force includes all forces (or components of forces that are parallel to the motion) F p = ma mgsin30  = ma To solve for the acceleration divide both sides by mass. mgsin30  = a m gsin30  = a (9.8)sin30  = a = 4.9m/s 2 F g = mg motion 30 FNFN F  = mgcos30  F p = mgsin30  30 That’s it… … KNOW HOW TO SET UP THE DIAGRAM! … UNDERSTAND HOW TO WORK WITH F net !

13. W HAT IF WE PUSH THE BOX ? A 5-kg block is pushed up an incline via a force of 30N as shown. If the incline makes an angle of 30  with the ground and friction is neglected, what is the acceleration of the block? motion 30 o Visualize the forces acting on the box as well as the direction of the acceleration. F = 30N

14. W HAT IF WE PUSH THE BOX ? Don’t forget about the weight (straight down) and the normal force (perpendicular to the surface and upward). Don’t forget to break the weight (i.e. force of gravity) into components that are parallel to and perpendicular to the incline. motion 30 o Now you are ready. F = 30N F g = mg FNFN F  = mgcos30  F p = mgsin30 

15. W HAT IF WE PUSH THE BOX ? F net = ma The net force includes all forces (or components of forces that are parallel to the motion) F - F p = ma F is positive b/c it is in the same direction as the motion and F p is negative b/c it is in the opposite direction of the motion. F - mgsin30  = ma To solve for the acceleration divide both sides by mass. F - mgsin30  = a m 30 – (5)(9.8)sin 30  = a = 1.1m/s 2 5 F p = mgsin30  motion 30 F = 30N F g = mg FNFN F  = mgcos30  That’s it… … KNOW HOW TO SET UP THE DIAGRAM! … UNDERSTAND HOW TO WORK WITH F net !

16. G OT IT ? Try it on your own… …BUT… …remember you have lots of support. 1) We will do this in class together. 2) Use Mrs. McGrath’s pencast for help 3) See the notes on the back of handout 4-5 for help 4) ASK QUESTIONS!