1) The document discusses resolving forces into components that are parallel and perpendicular to inclined planes. It shows examples of resolving weights and tensions into these components. 2) Key concepts covered include resolving forces, balancing horizontal and vertical forces, and calculating tensions, normal forces, and coefficients of friction. 3) Several multi-step examples are worked through that demonstrate resolving forces for objects on inclined planes and calculating unknown values like tensions and coefficients of friction.

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

2. 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

3. • 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!

4. (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.

5. (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

6. 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

7. 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

8. 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
2
21 TT
= 321 TT =
4.78
22
3 21 =+
TT
8.1563 21 =+TT
substituting (H) into (V) we obtain
(H)
(V)
8.1563 22 =+TT
NT 2.392 =
NT 9.671 =

10. (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
3
cos 1
θ

4.41=θ

11. 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

12. (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=

13. (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

5.3
8000
490
sin 1
=





= −
θ
Balancing vertical forces,
x = 4 cos 3.5
x
4m
x = 3.99
So length AB = 2x
= 7.98 or 8 metres

15. (a)
F
=120N
mg = 40 x 9.8 =
392N
N
Friction F
•

16. 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

17. 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

18. (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

20. W = mg
mgcos30º
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!
N
T
(a)
(b) As the mass is not moving, the resultant force R
must equal 0, i.e. R = 0i + 0j

21. (c) Using
i
j
as 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

22. (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

24. 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)

25. 30º
•
W
H
N
First, we’ll do H
H cos 30
Hsin30
Second, we’ll do W
W
sin 30
W
cos30
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

26. (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

27. (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
..
HW
ei =
3HW =∴

29. F
N
mg
Fric
F = μN
N = mg = 392
F = Fric = 300
300 = μ x 392
μ = 0.77

31. 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