These examples have been taken from Maths Quest Maths C for Queensland, Simpson & Campbell, Jacaranda Press.My sincere thanks to the authors and publishers.

First, construct triangle of forces using blue vector as hypotenuse.

(a) Horizontal component (green) = 25 cos 20 = 23.49 N

(b) Vertical component (pink) = 25 sin 20 = 8.55 N

20º

25N

23.49N

8.55N

20º

25N

23.49N

8.55N

(c) Friction acts horizontally opposite to the direction of motion. Let F be the friction vector.

F

As the cart is moving with constant velocity, its acceledration is zero and the horizontal forces balance.

F = 23.49 N

The frictional force is therefore 23.49N.

Note: The vertical forces would balance as well:

mg (not shown) would equal 8.55N. If g is 9.8, this enables you to calculate the mass she is pulling, i.e. 8.55/9.8 = 0.86kg. A light load!

(aerial view)

water resistance

R

T = 20000

T = 20000(a)

(b)The ship is not accelerating, therefore the magnitude of the resultant force (i.e. the force remaining after the resistance is subtracted from the tugboats’ engines driving force) will equal zero.

(c)

water resistance (friction)

R

T = 20000

T = 20000

As friction (water resistance) is horizontal, we need to resolve the two blue tugboat forces into horizontal and vertical components (as we did in Ex 1).

Each tugboat exerts a horizontal force of 20000 cos 10º = 19696 N

Total horizontal force of two tugboats = 2 x 19696 N = 39392 N

19696N

19696N

As the boat being towed is not accelerating, these forces must balance and so the friction must equal 39392N

There will be 3 forces –

• the gravitation force of the chair, mg vertically downwards

• the tension in the first rope, T1

• the tension in the second rope, T2

60º 30º

mg

= 8 x 9.8 = 78.4N

T1 T2

60º 30º

mg

= 8 x 9.8 = 78.4N

T1 T2

Resolve T1 into horizontal and vertical components

T1 cos 60

T1 sin 60

Resolve T2 into horizontal and vertical components

T2 cos 30

T2 sin 30NOTE the alternate (Z-rule) angles in this picture. 60º 30º

(b) Since the chair is not accelerating, horizontal forces balance and vertical forces balance

T1 cos 60 = T 2 cos 30

T1 sin 60 + T 2 sin 30 = mg

sin 30 = cos 60 = ½

sin 60 = cos 30 = 2

3

T1 cos 60 = T 2 cos 30

T1 sin 60 + T 2 sin 30 = mg

sin 30 = cos 60 = ½

sin 60 = cos 30 = 2

3

Substituting these values into the equations

2

3

221 TT

321 TT

4.7822

3 21 TT

8.1563 21 TT

substituting (H) into (V) we obtain

(H)

(V)

8.1563 22 TTNT 2.392 NT 9.671

(a) The forces will be the gravitation force on the speaker, mg, and the two (equal) Tension forces in the ropes

(b) Use the lengths given to calculate the angle θ (remember the triangle is isosceles). so we can split the triangle in half and use basic trig….

A B

6m

4m4m

θθ

mg = 50g

TT

4

3cos 1

4.41

41.4º(c) Resolving T vectors as before….

Vertical component of tension is T sin 41.4º, i.e. 0.66 T

(d) Right rope: T = 370.48 cos 41.4 i + 245 j = 278i + 245j

Left rope: T = – 370.48 cos 41.4 i + 245 j = – 278i + 245j.

Now since the vertical forces balance we have 2 x T sin 41.4º = mg

so T = 370.48 N

A B

mg = 50g

TT T sin 41.4ºT sin 41.4º

So the vertical component of the tension in each rope is 370.48 x sin 41.4º

= 245N

(e) is easiest done using Pythag….

From (d) we know the tensions are:

Right rope: T = 370.48 cos 41.4 i + 245 j = 278i + 245j

Left rope: T = – 370.48 cos 41.4 i + 245 j = – 278i + 245j.

So the magnitude of each tension is given by

22 245278 T

i.e. NT 371

(f)In this scenario, we are trying to find θ and ultimately the horizontal distance between A and B, given T = 4000 (in each rope)

θ

mg = 50 x 9.8 = 490N

4000sin θ4000sin θ

8000 sin θ = 490

5.38000

490sin 1

Balancing vertical forces,

x = 4 cos 3.5

x

4m

x = 3.99

So length AB = 2x

= 7.98 or 8 metres

(a)

F =

120N

mg = 40 x 9.8 = 392N

N

Friction F

F =

120N

N

Friction R

-120cos40ºj

– 120sin 40ºi

40º

(b)

Decomposing F into horizontal and vertical components, we get

F = – 120 sin 40 i – 120 cos 40 j

F = – 77.13 i – 91.93 j

F =

120N

N

Friction Fr

-120cos40ºj

120sin 40ºi

40º

(c)

We also need to place mg on the diagram

mg

R = (Friction – 120 sin 40)i + (N – 120 cos40 – mg)j

As the mower moves at constant velocity, R = 0 and so

(Friction – 120 sin 40 )i + (N – 120 cos40 – mg)j = 0

(Friction – 77.13)i + (N – 91.93 – 392)j = 0

i.e., (Friction – 77.13)i + (N – 483.9)j = 0

(d)

Equating horizontal forces, Friction = 77.13N

From (c), (Friction – 77.13)i + (N – 483.9)j = 0

(e)

Equating vertical forces, Normal = 483.9N

W = mg

mg cos 30º

mg sin 30º

When dealing with inclined planes it is best to resolve parallel and perpendicular to the incline

Note the angle at A will always equal the angle BCD (in this case 30º)

30º

A B

C

D

30º

Note – smooth plane therefore no friction!

NT

(a)

(b) As the mass is not moving, the resultant force R must equal 0, i.e. R = 0i + 0j

(c) Using

i

jas reference axes,

W = – mg sin 30 i – mg cos 30 j

W = (– 1.5 x 9.8 x 0.5) i – (1.5 x 9.8 x 0.866) j

W = – 7.35 i – 12.73 j

(d)

T = mg sin 30 i

T = 7.35i with magnitude 7.35 N

As object is not moving (or accelerating), the forces balance. Balancing the (green) forces parallel to the plane, we get

(e) As object is not moving (or accelerating), the forces balance. Balancing the (pink) forces perpendicular to the plane, we get

N = mg cos 30 j

N = 12.73 j with a magnitude of 12.73N

30º

W

H

N

The forces H and W need to be resolved into components that are parallel and perpendicular to the plane.

Remember to make H and W the hypotenuse of their respective triangles. The component vectors will then be the opposite and adjacent sides of the triangles, and will go in a direction so the hypotenuse is the resultant of the two components.

(a)

30º

W

H

N

First, we’ll do H

H cos 30

H sin

30

Second, we’ll do W

W sin 30

W cos 30

30

30

(b) W = – Wsin 30 i – W cos 30 j

= – 1.5 x 9.8 x 0.5 i – 1.5 x 9.8 x 0.866 j

= – 7.35 i – 12.73 j

(c) H = H cos 30 i – H sin 30 j

Balancing the forces parallel to the plane, we have

W sin 30 = H cos 30

i.e., 7.35 = H x 0.866

thus H = 8.49

To get H in i-j form we use the H triangle in the diagram.

H = H cos 30 i – H sin 30 j

i.e. H = 7.3i – 4.3j

(d) Balancing the (pink) forces perpendicular to the plane, we get

N = W cos 30 + H sin 30

N = mg cos 30 + H sin 30

N = 1.5 x 9.8 x 0.866 + 8.49 x 0.5

N = 16.98

N = 16.98j

From (c) , W sin 30 = H cos 30(e)

2

3

2..

HWei

3HW

F

N

mg

Fric

F = μN

N = mg = 392

F = Fric = 300

300 = μ x 392

μ = 0.77

40º

Fric

2g

2gsin40

40º

2gcos40

Res

N

Resolving horizontally, Res = 2g sin 40 – Fric …(1)

Resolving vertically, N = 2g cos 40 ……………….(2)

Fric = μN

= 0.2 x 2g cos 40 (substituting 2g cos 40 for N)

= 3

Res = 2g sin 40 – 3 (from (1)

= 9.6 N