Sample Examination #1
Closed book and notes. Calculators permitted
For full marks you must a) insert relevant diagrams, b) explain in a few words what
you are doing, c) present solutions neatly, d) indicate units where relevant,
e) carry 3 significant figures.
1.
The force F has a magnitude of 80 lb and acts within the volume shown.
Determine the magnitudes of the x, y, and z components of F.
(15 points)
Write down the direction cosines first, and use them to find the magnitudes. We have α,
β, but no γ yet, so start by finding γ.
cos2α + cos2β +cos2γ = 1
∴
(0.25 + 0.5 + cos2γ)
=1
∴
γ = 60° also
Now magnitudes are
Fx
Fy
Fz
= 80 cos 60° = 40 lb
= 80 cos 45° = 56.6 lb
= 80 cos 60° = 40 lb
2.
Determine the magnitude of the projected component of force F along the pole.
(15 points)
This is one of the uses of the dot product!
we need a unit vector along AO?
We need 2 vectors and only have 1: so
AO = [-3i + 2j -6k]
and its magnitude is
32 + 22 + 62
=
7
so a unit vector along AO is
=
1/7[-3i + 2j -6k]
and the dot product is F. AÔ
=
-20x-3/7 + 50x2/7 - 10x-6/7
=
60/7 + 100/7 + 60/7
=31.4 lb
3.
Determine the stretch in each of the two springs required to hold the 20 kg crate in
the equilibrium position shown. Each spring has an unstretched length of 2m and a
stiffness of k = 300N/m. (20 points)
FOC =
FOA =
FOB =
F
=
6i + 4j + 12k
62 + 42 + 122
- FOA j
- FOB i
- 196.2k
FOC
= 3/7 FOCi + 2/7 FOCj + 6/7 FOCk
and ΣF = 0
Equating i, j, and k components
3/7 FOC - FOB = 0
2/7 FOC - FOA = 0
6/7 FOC - 196.2 = 0 and solving, we have
FOC = 228.9 N
and FOB = 98.1 N and
Then using the spring formula, s = F/k:
sOB = 98.1/300 =
0.327m
sOA
= 65.4/300 =
0.218m
FOA = 65.4 N
4.
Two men exert forces of F = 80 lb and P = 50 lb on the ropes. Determine the
moment of each force about A. Will the pole rotate clockwise or counterclockwise?
(20 points)
P and F both can be resolved into vertical and horizontal components. The vertical
components pass through A and contribute zero moment. Therefore, we only consider
the horizontal components.
ΣMP = P . cos 45° . 18 = 636 ft.lb (acw)
ΣMF = F . 4/5 .12
= -768 ft.lb (cw)
ΣMTOTAL
= 636 - 768
= -132 ft.lb
= clockwise
(or, using r ^ F, for force F,
r = 0i + 12j + 0k
F = 80 4/5 i - 80 3/5 j + 0k
so
r^F =
i
j
k
0
12
0
64
-48 0
=
=
0i + 0j - 768k
768 lb.ft c.w.
etc. .. same approach for the other force, P, then find resultant to see if it is cw (ve) or acw (+ve)
Other questions that might have been asked
A.
Determine the coordinate direction angles of F1 and FR. (15 points)
Unit vector of F1 and FR??
uF1
=
4/5 i + 3/5 k =
0.8i + 0.6k
uFR
=
=
cos 45° sin 30° i + cos 45° cos 30° j + cos 45° j
0.3536i + 0.6124j + 0.707k
So the coordinate direction angles are:
F1
cos α = 0.8, ∴ α
=
cos β = 0 ∴ β
=
cos γ = 0.6 ∴ γ
=
FR
cos α = 0.3536
cos β = 0.6124
cos γ = 0.707
∴α =
∴β =
∴γ =
36.9°
90°
53.1°
69.3°
52.2°
45°
B.
Determine the moment created by the force F = [50i + 100j - 50k] N, acting at D,
about each of the joints at B and C. (20 points)
To find moments, use M = r ^ F where r = each r in turn.
rBD
= (0.75i + 0j - 0.3k) m
rCD
= (0.75i + 1.25j - 0.3k)
MB
=
i
j
0.75 0
50
100
MC
=
...................
k
-0.3
-50
= {30i + 22.5j + 75k} N.m
= {-32.5i + 22.5j + 12.5k} N.m