ENS161
1st Chapter Exam
Solutions
by:
Engr. Steve Anthony N. Lim
CE Department Faculty
October 5, 2022
1
Problem #1
Determine the angle θ in the figure shown such that the horizontal component of the force exerted by the spring on
the pin P is maximum. The free length of the spring is r.
Figure 1: Problem #1
Figure 2: Problem #1 (FBD)
In order to solve the value of θ for which the horizontal component of the spring force FSx is maximum, we have to
use concepts of minima and maxima through differentiation. As such, the first goal is to define FSx in terms of θ.
From Fig.2, it can be seen that
√
L = r 5 − 4 cos θ
and
x = r sin θ
such that
cos ϕ =
sin θ
x
=√
L
5 − 4 cos θ
and so the force on the spring must then be
FS = k (L − L0 ) = kr
√
5 − 4 cos θ − 1
and its horizontal component is
FSx
sin θ
= FS cos ϕ = kr sin θ − √
5 − 4 cos θ
solving for the derivative of FSx
−2 cos2 θ + 5 cos θ − 2
dFSx
= kr cos θ −
3/2
dθ
(5 − 4 cos θ)
!
equating this to zero to calculate the value of θ for a maximum value of FSx
3/2
−2 cos2 θ + 5 cos θ − 2 = cos θ(5 − 4 cos θ)
to simplify the solution, let y = cos θ, such that
3/2
−2y 2 + 5y − 2 = y(5 − 4y)
4y 4 − 20y 3 + 33y 2 − 20y + 4 = 125y 2 − 300y 3 + 240y 4 − 64y 5
2
so finally
64y 5 − 236y 4 + 280y 3 − 92y 2 − 20y + 4 = 0
solving this equation will yield (other values of y are not valid)
y1 = +0.142
y2 = −0.225
consequently, we can solve θ by using the earlier representation y = cos θ
θ1 = 81.81◦
θ2 = 103.01◦
and the corresponding values of FSx are
FSx1 = 0.519kr
FSx2 = 0.573kr
therefore, FSx is maximum when
θ = 103.01◦
3
2
Problem #2
Two cables tied together at C is loaded as shown. If the cables can only carry a maximum load of 800 N, determine
the maximum possible value of the force P and the corresponding value of θ. Note that 0 ≤ θ ≤ 90◦ .
Figure 3: Problem #2
Figure 4: Problem #2 (FBD)
To solve this problem quickly, we can simply assume both cables reach their maximum capacity of 800 N. We can
then verify if this assumption is valid by checking if the value of θ is still within the range as indicated in the problem.
So, to start
TCA = 800 N
TCB = 800 N
P
P
Fy = 0
P sin θ = 800 (sin 35◦ + sin 50◦ )
(1)
P cos θ = 800 (cos 35◦ − cos 50◦ )
(2)
Fx = 0
dividing Eq.1 by Eq.2
tan θ =
sin 35◦ + sin 50◦
cos 35◦ − cos 50◦
which yield
θ = 82.5◦
since this value is within 0 ≤ θ ≤ 90◦ our initial assumption is correct, and we can now solve P
P = 1081 N
4
3
Problem #3
The boom OA carries a load P = 137 N and is supported by two cable as shown. If the cable AB can only sustain a
tension of 183 N, determine the maximum possible length of the boom OA so that the resultant of the tensile forces
and the load is directed along the x-axis.
Figure 5: Problem #3
Figure 6: Problem #3 (FBD)
From the figure and letting x = 10d to remove the decimals in our equations
TAB = − √
TAC = − √
183x
5307
4392
·i+ √
·j+ √
·k
x2 + 1417
x2 + 1417
x2 + 1417
T2 x
25T2
36T2
·i+ √
·j− √
·k
x2 + 1921
x2 + 1921
x2 + 1921
and
P = −137 · j
because
the problem states that the resultant of these three forces is directed along the x-axis, we can say that
P
Fz = 0
√
and
P
T2
122
=√
x2 + 1921
x2 + 1417
(3)
Fy = 0
√
25T2
5307
+√
= 137
2
+ 1417
x + 1921
x2
substituting Eq.3 into Eq.4
8357
√
= 137
2
x + 1417
p
x2 + 1417 = 61
x = 48
and so:
d = 4.8 m
5
(4)
4
Problem #4
A homogenous circular disc with a mass of 12 kg and a radius of 7 cm is suspended using three cables and remains
in a horizontal position as shown. If the tension developed in the cable AB is 34.72 N, determine the height y.
Figure 7: Problem #4
Figure 8: Problem #4 (FBD)
Because the original figure for this problem did not include a coordinate system axis, the FBD drawn provides one.
So that, from the figures
TBA = TBA cos θ cos 40◦ i + TBA sin θj − TBA cos θ sin 40◦ k
TCA = TCA cos θ cos 40◦ i + TCA sin θj + TCA cos θ sin 40◦ k
TDA = TDA cos θi + TDA sin θj
W = −12(9.81)j
P
P
P
Fx = 0
TBA cos θ cos 40◦ + TCA cos θ cos 40◦ − TDA cos θ = 0
(5)
TBA sin θ + TCA sin θ + TDA sin θ = 12(9.81)
(6)
−TBA cos θ sin 40◦ + TCA cos θ sin 40◦ = 0
(7)
Fy = 0
Fz = 0
from Eq.7, we can say that
TBA = TCA
using this, we can restate Eq.6 into
12(9.81) − 2TBA sin θ
sin θ
= TCA
TDA =
substituting Eq.8 into Eq.5 and still using TBA
2TBA cos θ cos 40◦ −
12(9.81)
+ 2TBA cos θ = 0
tan θ
knowing that TBA = 34.72 N
6
(8)
θ = sin
−1
12(9.81)
2(34.72)(cos 40◦ + 1)
θ = 73.73◦
and so, from the figure
y = 7 tan θ
y = 23.977 ≈ 24 cm
7
ENS161
2nd Chapter Exam
Solutions
by:
Engr. Steve Anthony N. Lim
CE Department Faculty
November 2, 2022
ENS161 - Statics of Rigid Bodies
2nd Chapter Exam (11-02-22)
Instructions
Answer the problems shown completely (numerical solutions and ancillary diagrams included).
Box the final answer/s.
1. Three (3) forces are applied tangentially to a circular plate as shown. Determine the resultant moment of these forces
about the center of the plate.
2. Solve for the magnitude of the smallest force P that can be applied at point A so that the three (3) forces shown can
be replaced by a single force whose line of action passes through point E. The shape shown is a regular heptagon and
the 15 kN forces are directed vertically.
3. The rod shown supports a cylinder of mass 50 kg and is pinned at its end at A. It is also subjected to a concentrated
(couple) moment of 600 N · m. This rod is connected to a spring at B that is attached to a roller so that it remains in
a vertical position at all times. If the spring has a free length of 1.0 m when θ = 0, determine the corresponding value
of this angle for equilibrium.
4. Determine the range of values of θ for which the semicircular rod shown can be maintained in equilibrium.
Figure 1: Problem #1
Figure 2: Problem #2
Figure 3: Problem #3
Figure 4: Problem #4
“Give me lever long enough and a fulcrum on which to place it, and I shall move the world.” - Archimedes
1
Problem #1
Three (3) forces are applied tangentially to a circular plate as shown. Determine the resultant moment of these forces
about the center of the plate.
Figure 1: Problem #1
Figure 2: Problem #1 (FBD)
Solution:
To solve the moment at the center of the circular plate, its radius needs to be solved. Using the general equation for
circles on a plane:
x2 + y 2 + Dx + Cy + E = 0
at the point (4, 2) the general equation becomes
4C + 2D + E = −20
(1)
at the point (−1, 1) the general equation becomes
−C + D + E = −2
(2)
at the point (3, −3) the general equation becomes
3C − 3D + E = −18
simultaneously solving equations (1), (2), and (3) will yield
C=−
11
3
1
3
E = −6
D=
so the general equation of the circle is
11x y
+ −6=0
3
3
rearranging this equation to the center-radius form will yield
x2 + y 2 −
x−
11
6
2
2
1
169
+ y+
=
6
18
2
(3)
which that the radius of the circle is
r
r=
√
169
13 2
=
= 3.06 m
18
6
so the moment at the center is
(⟳ +)ΣMD = 30r + 40r − 50r
= 20r
= 61.3 kN · m
3
2
Problem #2
Solve for the magnitude of the smallest force P that can be applied at point A so that the three (3) forces shown can
be replaced by a single force whose line of action passes through point E. The shape shown is a regular heptagon
and the 15 kN forces are directed vertically.
Figure 3: Problem #2
Figure 4: Problem #2 (FBD)
Solution:
From the figure shown, in order for the magnitude of the force P to be as small as possible, its line of action needs
to be perpendicular with the line drawn from A to E. Interpretation of the problem also states that ΣME = 0.
(⟳ +)ΣME = 15d − P d = 0
such that
P = 15 kN
4
3
Problem #3
The rod shown supports a cylinder of mass 50 kg and is pinned at its end at A. It is also subjected to a concentrated
(couple) moment of 600 N · m. This rod is connected to a spring at B that is attached to a roller so that it remains
in a vertical position at all times. If the spring has a free length of 1.0 m when θ = 0, determine the corresponding
value of this angle for equilibrium.
Figure 5: Problem #3
Figure 6: Problem #3 (FBD)
Solution:
Defining, firstly, the force on the spring in terms of θ
Fs = kx
= 600(1 + 3 sin θ − 1)
= 1800 sin θ
such that (⟳ +)ΣMA = 0
600 + 1.5W cos θ − 3Fs cos θ = 0
600 + 1.5(50 · 9.81) cos θ − 3(1800 sin θ) cos θ = 0
600 + 735.75 cos θ − 5400 sin θ cos θ = 0
solving this equation for θ will yield
θ1 = 82.5◦
θ2 = 14.5◦
5
4
Problem #4
Determine the range of values of θ for which the semicircular rod shown can be maintained in equilibrium.
Figure 7: Problem #4
Figure 8: Problem #4 (Position 1)
Figure 9: Problem #4 (Position 2)
6
Solution:
As shown in the figures above, the lower limit of the range for the angle θ is at Position 1, and the upper limit is at
Position 2. So, from Fig. 8 we have
R tan 50◦
2R
tan 50◦
−1
θL = tan
2
tan θL =
θL = 30.8◦
from Fig. 9
R tan 30◦
2R
tan 30◦
◦
−1
180 − θU = tan
2
tan (180◦ − θU ) =
θU = 163.9◦
so the range of possible values of θ is
30.8◦ ≤ θ ≤ 163.9◦
7