34
Forces and Newton’s 2 nd law

Forces and Newton’s 2nd law

  • Upload
    others

  • View
    4

  • Download
    0

Embed Size (px)

Citation preview

Page 1: Forces and Newton’s 2nd law

Forces and Newton’s 2nd law

Page 2: Forces and Newton’s 2nd law

Resultant vectors - as the single vector which has the same effect as the original vectors acting together

Vector addition 1. Tail to Head method

+ = 50N 40N 90N

=

Page 3: Forces and Newton’s 2nd law

Vector addition 1. Tail to Tail method …….also known as the parallelogram method

Page 4: Forces and Newton’s 2nd law

Worked Example

1

2

horizontal and vertical forces acting on the object

Page 5: Forces and Newton’s 2nd law

MEMORANDUMWorked Example

Page 6: Forces and Newton’s 2nd law

Practice Example 1

1

2

3

horizontal forces acting on the object

vertical forces acting on the object

Page 7: Forces and Newton’s 2nd law

MEMORANDUM 1

Rx = 300 + ( -180) = 120 N right ( taking right as +ve)

Ry = 170 + (-120) = 50 N up (taking up as +ve)

50 N R

120 N

Page 8: Forces and Newton’s 2nd law

Components of a Vector

• These are the individual vectors that would combine to create a resultant vector

• Two components – vertical and horizontal drawn tail to tail

Vertical component

Ry or RVC RESULTANT (R) (Opposite)

Horizontal component Rx or R

(adjacent)

Vertical component

sinθ =𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑒𝑢𝑠𝑒=

𝑅𝑦

𝑅

Ry = RSinθ

θ

Page 9: Forces and Newton’s 2nd law

A tow-truck tow a damaged car to thepanelbeaters. The tow cable applies a force of 500NOn the car at 300 to the horizonal. Calculate the vertical and horizontal components of the applied force

VERTICAL COMPONENT = Rsinθ

HORIZONTAL COMPONENT = RCosθ

WORKED EXAMPLE

Page 10: Forces and Newton’s 2nd law

Different kind of forces

Weight (Fg) – the gravitational force of the earth exerts on any object on, or near its surface

Fg = mg where g = 9,8 m.s-2

Normal (FN) – the perpendicular force exerted by a surface on an object in contact with it.

Frictional force (Ff) – the force that opposes the motion of an object and acts parallel to the surface with which the object is in contact

STATIC FRICTION (Ff(s)) KINETIC FRICTION (Ff(k))

Page 11: Forces and Newton’s 2nd law

Static friction (Ff(s))

Page 12: Forces and Newton’s 2nd law
Page 13: Forces and Newton’s 2nd law
Page 14: Forces and Newton’s 2nd law

Practice Example 2

Fg(x)

Page 15: Forces and Newton’s 2nd law

Memorandum 2

Page 16: Forces and Newton’s 2nd law

Kinetic Friction

Page 17: Forces and Newton’s 2nd law

Force and Free-Body diagramsbelow

Page 18: Forces and Newton’s 2nd law

Inclined Planes

horizontalFgCosθ

Fg(x)

Page 19: Forces and Newton’s 2nd law

The component of the weight

down the slope is its vertical

component Fg(y)

Fg Sinθ where θ is the angle

of incline

Fg(y)Fg(y) = Fg Sinθ

Fg(x) = Fg Cosθ

vertical

Fg(x)

Fg(y)

Ff

Page 20: Forces and Newton’s 2nd law

Practice Example 3

Page 21: Forces and Newton’s 2nd law

Memorandum 3a

Fx

horizontal

Fx

Fx

Page 22: Forces and Newton’s 2nd law

vertical

Memorandum 3b

Fy

Fy

Fy

vertical Fy

Page 23: Forces and Newton’s 2nd law

Newtons 2nd Law

• A net force will always cause the object to accelerate

• The net force is always the sum of all the forces acting in the plane of motion

and thus

Fnet = sum of all forces

Fnet = F1 + F2 + F3……….

Page 24: Forces and Newton’s 2nd law
Page 25: Forces and Newton’s 2nd law

Fnet = sum of all forces= Fapp + (-Ff) Fapp - Ff

Page 26: Forces and Newton’s 2nd law

Practice Example

Fnet = sum of all forces

= T + (-Ff)

+ve

T - Ff

Fnet = Fapp + (-T) + (-Ff)= Fapp - T - Ff

Fapp - T - Ff

Page 27: Forces and Newton’s 2nd law

Practice Example 4

Page 28: Forces and Newton’s 2nd law

Memorandum 4

Fnet = sum of all forces= Tx + (-Ff)= Tx - FfTx - Ff =

Page 29: Forces and Newton’s 2nd law

Practice Example 5

Page 30: Forces and Newton’s 2nd law

Memorandum 5

a)

b)

9,8 1470

1470 503 N

503103

0,67 m.s-2

0.67

Fg(y)

(Fg(y))

Fg(y)

Fg(y)

Fg(y)

We need to find the vertical component(Fy) which is

Page 31: Forces and Newton’s 2nd law

Practice Example 6

Practice Example 7

Calculate the tension in the cable when…….

a)

b)

c)

d)

Page 32: Forces and Newton’s 2nd law

Memorandum 6

9,8 49 000 N

49 000

149 000 N upwards

Fnet = sum of all forces = F + (-W )

Page 33: Forces and Newton’s 2nd law

Memorandum 7

9,8 7840 N downwards

7840 N upwards

78409040 N upwards

9,8 7840 N

7840 N upwards

Page 34: Forces and Newton’s 2nd law

7840

6240 N upwards