Unit_7_Statics_Dynamics_Vectors_2 (Extension)
Answer the following questions showing the working and answers in full on
file paper. You will get credit for both. NB: some of these questions are
harder than AS and can be done for fun!
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
tan = opp/adj
a
b
c


SinA SinB SinC
c2 = a2 + b2 - 2ab cos C
sin = opp/hyp
cos = adj/hyp
A vector quantity has magnitude and direction. Good examples are forces and
velocities. They can be represented by a line and arrow to show the direction
and an angle from the horizontal or vertical to represent the direction and
magnitude. We can draw scale diagrams (see below) to interpret problems or
use trigonometry formulae to solve them (see above).
1) What vector can you use to replace these 3
vectors? In the diagram we see 3 vectors with
their associated magnitudes and angles. In
order to add these, we always must connect
vectors 'head to tail' and the resultant vector
(which represents the vector sum) is drawn
from the tail of the first vector to the head of
the last vector. Do this problem by drawn out
method. (2 marks)
HELP!!!
Vectors can also be solved by
making a parallagram of forces and
transposing vectors to make triangles which
can be solved using trigonometry (as shown
above).
2) Use the parallelogram method to resolve the forces acting on this object
placed. (Hint employ both cosine & sine rule). (4 marks)
3) Two forces of magnitude 10.0N and F Newton’s produce a resultant of
magnitude 30.0N in the direction OA. Find the direction and magnitude of F.
(2 marks) (Hint use Pythagoras)
4) The graph shows a part completed
vector diagram. You task is to find out
the vector R by mathematical
analysis. The vectors A & B are
shown in both coordinate and bearing
formats. Show working for both a
mathematical method (2 marks) and
drawn out scaled method. (2 marks)
Hint: this is not as hard as it might
initially look!
A simple method of resolving vectors in directions is by dissembling the 2
vectors into horizontal and vertical components then summating them to make
a new triangle of vectors.
5) Use this method to find the resultant forces in diagrams a & b using
mathematical methods (4 marks)
6) Forces of 60.0N and F Newton’s act at point O. Find the magnitude and
direction of F if the resultant force is of magnitude 30.0N along OX (2 marks)
7) Forces of 60.0N and F Newton’s act at point O. Find the magnitude and
direction of F if the resultant force is of magnitude 30.0N along OX (2 marks)
Answers
1) In this example, using a ruler and protractor,
we are able to get a resultant vector of about
magnitude 11 and an angle of 102°.
2)
3)
Finding the components of vectors for
vector addition involves forming a right
triangle from each vector and using the
standard triangle trigonometry.
4)
The vector sum can be found by
combining these components and
converting to polar form.
After finding the components for the
vectors A and B, and combining them
to find the components of the resultant
vector R, the result can be put in polar
form by
5) a & b
a) for 10N at 30 -> RSin = 5N RCos = 8.66N, for 15N at 0 -> RSin = 0N
RCos = 15N. Hence -> H -> 23.66N, Sum V = 5N , Make new triangle -?
Opp/adj = tan ->  = 11.93, Pythag -> 584.79 = 24.18.
b) Same method but take away as angle is pointed away;
for 40N at 45 -> RSin = 28.284N (up) RCos = 28.284N (left), for 50N at 0 ->
RSin = 0N RCos = 50N. Hence -> H -> 21.716N (right), Sum V = 28.284N ,
Make new triangle -? Opp/adj = tan ->  = 52.48, Pythag -> R = 35.66
6) Use standard trig F2 = (60N) 2 + (30N) 2 => F = 67.08N, then to tan = opp / adj
= tan-1(30/60) = 26.565 Then  = 90 – 26.565 = 63.43
7) Resolve vectors separately. F should actually be down at 90 51.962N.