1.What is the weight of a 15 kg rock? 147 N 2.A skateboard (mass 12 kg ) accelerates from rest to...

1. What is the weight of a 15 kg rock?147 N

2. A skateboard (mass 12 kg ) accelerates from rest to 3.8 m/s over a distance of 9.4m. What is the skateboard ’s rate of acceleration? 

0.768 m/s2

3. What force was applied to the skateboard in to achieve this rate of acceleration?

9.22 N

4. An unbalanced force of 25 N is applied to a 12.5-kg mass. What should be the acceleration experienced by the mass?

2 m/s2

11/13 Forces Assignment Parts 2 & 3 due Friday

• NOW: Pick up Force Notes II. ON Force ntoes one, answer example J.

• Last Friday: A quiz over Fun with Forces & Fun with Weight.

• Monday: finished lab.

• Tuesday: reviewed lab, turned it in and took a quiz.

• If you were absent for any of these you need to make them up today after school or tomorrow morning.

• PM Test corrections & retakes Mon-Thur this week. You must show completed PM notes to be eligible for corrections & retakes. Retakes pm only. I have pm duty until 2:55

• There is an issue with the website. I will fix it today if possible

Ex J The brakes of a 1000-kg car exert 3000 N.a. How long will it take the car to come to a stop from a velocity of 30 m/s?

m = 1000 kgF = -3000 Nvi = 30 m/s

vf = 0 m/sm

Fa

1000

3000 = - 3 m/s2

a

vvt if

3

300

= 10 s

b. How far will the car travel during this time?

d = vit+ .5at2

= 30(10)+ .5 (-3)(10)2

= 150 m

Forces Part IIFriction

When surfaces are pressed together, we can identify

four forces

• Friction Force: FF

• Force opposing motion

• Measured in Newtons (N)

• Applied Force: FA

• The push or pull applied to the object

• Measured in Newtons (N)

• Fw Force of Weight or Gravity

• (Mass in kg) (Acceleration due to Gravity)

• Kg x 9.8 m/s2

• Measured in Newtons (N)

• Normal Force: FN

• Usually a 3rd Law reaction to gravity, that is equal and opposite of Force of Weight (Fw)

• Perpendicular to the surface.• Measured in Newtons (N)

• FN is NOT FNET

• Don’t confuse them just because they begin with N!

• A 50kg object is moving horizontally at a constant velocity. Is there acceleration? Is there a net force?

FW

FN

FAFF

Ex L No Friction

• Fnet = ma A 50 kg object experiences an applied force of 400 N,

what is the Fnet? What is the acceleration?:• Fnet = FA

• Fnet = Net Force, results in acceleration• a = Fnet /m = (400N )/ (50 kg) = 8 m/s2

Example L

FA

FW

FN

Friction

• Is a force that opposes motion.

Ex M Friction andConstant velocity. Is there a net

force?

• Fnet = 0 • FA+ -FF = ma or FA + - FF = Fnet

A 50 kg object moves at a constant velocity when acted upon by an applied force of 400 N. What is the FF? What is the Fnet? What is the acceleration?:

• FF = FA

• Fnet = FA + - FF = 0 N• If Fnet = 0 then acceleration = ?• 0

Example MFriction and constant velocity

FA

FW

FF

FN

How does friction affect net force?• Fnet = ma

• ma = FA+ FF

Where:• Fnet = Net Force • FF = Friction Force• FA = Applied Force• You are actually subtracting FF from FA, since Ff is

in the opposite direction• Friction will reduce the net force

• Do all surfaces provide the same amount of friction? How is this described?

FA

FW

FF

FN

Friction

What does it depend on?

• Depends on the surface of the materials.

• Depends on how tightly the surfaces are pressed together.

• FF = Force of Friction

Coefficient of Friction

• The coefficient of friction is a measure of how difficult it is to slide a material of one kind over another; the coefficient of friction applies to a pair of materials, and not simply to one object by itself

• Coefficient of Friction Reference Table - Engineer's Handbook

When Surfaces are Pressed Together

• Coefficent of Friction µ can be calculated

• It is a ratio of FF and FN

N

F

F

F

NF FF

Coefficient of friction

• What is it equal to?• What is the unit for Coefficient of friction?• If Coefficient of friction is small, what does

that mean about the FF?

• If Coefficient of friction is large, what does that mean about the FF?

• The higher the coefficient of friction, the more difficult to slide

I call these FAWN problems

Example N• A 400 N force is applied to a 50kg

object. Calculate the acceleration of the object if = 0.3.

FA

Fg

FF

FN

Example Cont’d.

• FA = 400 N

• Fg = m·g (50kg)(9.8m/s2) Fg = 490 N

• FN = 490 N also

• FF = (0.3)(490N) = 147 N

Example cont’d.

FA = 400N

Fg = 490N

FF = 147N

FN = 490N

To solve for Acceleration must calculate Net Force

• FNET = ma= 400N – 147N

• ma = 253N

FNET = ma = FA+ -FF

Now use Net Force and mass of object in F=ma formula

a = F/m

a = 253N/50kg

a =5.06 m/s2

Ex OPutting it all together

A 50 kg object accelerates horizontally at 0.2 m/s2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the Fnet? What is the FF ? What is the FA ?How far does it travel in 5 seconds?:

• Fnet = (50 kg) (0.2 m/s2) = 10 N• FF = μ FN

• FN = (50kg)(9.8 m/s2) = 490 N• FF = (0.01) (490 N) =4.9N• FA = Fnet –

-FF

• FA = 10 N + 4.9 N = 14.9 N

How does friction affect net force? Putting it all together

A 50 kg object accelerates horizontally at 0.2 m/s2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the Fnet? What is the FF ? What is the FA ?How far does it travel in 5 seconds?:

• Fnet = (50 kg) (0.2 m/s2) = 10 N• Set up DVVAT• d = ?• vi = 0 m/s• vf = ?• a =0.2 m/s2

• d = (.5)(.2)(5)2

• d = 2.5m

As the coefficient of friction decreases and the object and the applied force remain the same,

What happens to:

• FA

• FW

• FN

• FNET

TENSION aka FT

• is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.

• It is the opposite of compression. It is a “response force”• That is to say, if one pulls on the rope, the rope fights

back by resisting being stretched• Ropes, strings, and cables can only pull. They cannot push

because they bend.• is measured in newtons • is always measured parallel to the string on which it

applies.

• What does the rope provide?

• A lift (vertical force) and a pull (horizontal force)

• If there was no angle, would there be any vertical force?

• No

• If the angle was at 90°, how would that affect the force components?

• Force would only be in the vertical plane

• How would you calculate the horizontal and vertical force components if the angle of the rope with the floor was 57° and the Force of tension (FA) in the rope was 400 N?

Ex P. This crate is be pulled with a rope across a friction based horizontal surface at a constant velocity. The rope exerts a tension of 400 N at an angle of57°. What is the

coefficient of friction?

50 kg

57°

A box is pulled into motion with a rope across a horizontal surface. The rope makes an angle of 57° to the floor. The Force of tension

(FA) in the rope is 400N

50 kg 57º

FAX

FAY

400 N

A. Determine FAX or FHoriz

= cos (57) (400N)= 217.86 N

B. Determine FAY or FVert

= sin (57) (400N)= 335.47 N

I work in a circle• Determine FW

• Determine FAX

• Determine FAY

• Determine FN

• The FAY supplies part of the upward force. The total upward force is FAY + FN and together these are equal but opposite the FW. FN = FW – FAY

• Determine FF

• Determine coefficient of friction

50 kg57°

FAX 217.86 N

FAY 335.47 N

400 N

Determine Fw

= (50 kg)(9.8 m/s2)= 490 N

Determine FN

= FW- FAY

= 490 N – 335.47 N= 154.53 N

50 kg57°

FAX 217.86 N

FAY 335.47 N

400 N

Determine FF

FF= Fax (Constant Velocity)FF= 217.86N

Determine µ= FF/ FFN

= 217.86N/ 154.53 N= 1.41

The End