Maths Quest C Year 11 for Queensland
WorkSHEET 8.1
1
Chapter 8 Vector applications WorkSHEET 8.1
Vector applications
For the three forces
F  2 ~i  j, ~F  3 ~i  2 j and ~F   ~i  3 j ,
~
1
~
2
~
3
1
Name: _________________________
(a)
R
 ~F  ~F  ~F
~
1
2
2
3
 2 ~i  j  3 ~i  2 j  i  3 j
~
~
find:
(a) the resultant force R~
~
~
~
 4 i  4j
~
~
(b) the magnitude of the resultant force.
(b)
R  42  42
~
 32
4 2
2
3
, B~ and C~ have
Three coplanar forces, A
~
magnitudes of 80 N, 60 N and 90 N
respectively. Find the angle between B~ and C~
if the resultant force is zero.
or
Using the cosine rule

80 2  90 2  60 2  29060 cos 180   

6400  8100  3600  10800 cos 180   
 11700  10800 cos 
10800 cos   5300
 5300
cos  
10800
 0.4907
  119.4
The angle between forces B~ and C~ is 119.4


Maths Quest C Year 11 for Queensland
3
Chapter 8 Vector applications WorkSHEET 8.1
A boy pulls a cart along by an attached rope.
The cart is on a horizontal path and the rope
makes an angle of 30 to the horizontal. If the
rope exerts a force of magnitude 20 N on the
cart, find:
(a) the horizontal component of the force
exerted on the cart by the rope
(b) the vertical component of the force
exerted on the cart by the rope.
2
(a)
Horizontal component  20 cos 30
3
2
 10 3 N
 20 
Vertical component  20 sin 30
 20  0.5
 10 N
(b)
4
2
A 2-kg mass is held on a smooth inclined
plane, angled at 45 to the horizontal, by a
string which is parallel to the plane.
Draw a vector force diagram showing the
forces acting on the mass. Find the magnitude
of the resultant force, R~ .
2
Since the mass is not moving, the forces are in
equilibrium.
R~ = 0
5
For the situation described in the previous
question, use ~i as a unit vector up the plane and
3
j is at right angles to ~i
~
j as a unit vector perpendicular to the plane to
~
find:
(a) the weight vector W~ using ~i  j notation.
~
(b)
(a)
W  2 g sin 45i  2 g cos 45 j
~
~
~
  2gi  2g j
the tension force in the string,
T~ using ~i  j notation.
~
~
~
(c)
the magnitude of the normal contact
force, N.
(b)

~ 
~
Since R
0
~
T  2g  0
T 
2g
T 
2g i
~
(c)
~
R  T  2g i  N  2g j
~
N  2g  0
N  2 g newton.
Maths Quest C Year 11 for Queensland
6
Chapter 8 Vector applications WorkSHEET 8.1
3
A 2.4-kg mass rests on a plane inclined at 24
such that it is just about to slide.
5
The plane is then inclined at an angle of 36 .
Determine the resultant force on the mass.
R
 F  2.4 g sin 24 ~i
~
  N  2.4 g cos 24 j
~
But R
0
~
F  2.4 g sin 24
F
F
N  2.4 g cos 24
N






0
2.4 g sin 24
9.57 N
0
2.4 g cos 24
21.49 N
F
Therefore  
N
9.57

 0.445
21.49
Now R~  F1  2.4 g sin 36 ~i
  N 1  2.4 g cos 36 j
~
N 1  2.4 g cos 36  0
N 1  2.4 g cos 36
N 1  19.03 N
F1  N 1 
F1  19.03  0.44
Resultant force  F1  2.4 g cos 36 
 10.56 N
Maths Quest C Year 11 for Queensland
7
Chapter 8 Vector applications WorkSHEET 8.1
Two masses, m1  1 kg and m2  2 kg ,
connected by a light, inextensible string, are
pulled by a horizontal force, A
.
~
4
7
(a)
The force is applied to m1 and the connected
masses slide across a horizontal table at
constant speed. If the coefficient of friction is
0.25, determine:
(a) the magnitude of the tension in the string
(b)
the magnitude of A
.
~
For m1 :
R
  A  T  F1  ~i   N1  1g  j  0
~
~
(constant speed  a  0)
N1  1g  0
N1  g
F1  N1 
 g  0.25
g

4
A  T  F1  0
g
AT   0
(1)
4
For m 2 :
R
 T  F 2  ~i   N 2  2 g  j  0
~
~
N 2  2g  0
N 2  2g
F2  N 2 
 2 g  0.25
g

2
T  F2  0
g
T  0
2
g
T 
2
 4.9 N
(b)
substitute T into equation (1)
g
AT  0
4
A  4.9  2.45  0
A  7.35 N
Maths Quest C Year 11 for Queensland
8
Chapter 8 Vector applications WorkSHEET 8.1
5
A horizontal force of 16 N acts on a 1.5-kg
block resulting in uniform motion across a flat
floor. Find the coefficient of friction.
4
R~  16  F  ~i  N  1.5 g  j
~
Since motion is uniform (that is, velocity is
constant),
R~  0
16  F
F
and N  1.5 g
N
0
 11
0
 1 .5 g
F
M 
N
16
M 
1.5 g
 1.09
The coefficient of friction is approximately 1.09
9
The following three forces are in equilibrium
F  2 ~i  j, ~F  3 ~i  2 j and ~F .
~
1
~
2
~
3
2
F  ~F  ~F  0
~1
2
Calculate F3
3
 ~F   ~F   ~F
3
~
1
2
 (2 ~i  j )  (3 ~i  2 j )
~
~
 5 ~i  j
~
10
Calculate the resultant of the two forces shown
below.
2
2
R  15 2  25 2  2  15  25 cos 80 
R  26.8
25  sin 80 
 26.8 N, 16.7  S of E
26.8
  66.7
sin  