ENGR0135 - Statics and Mechanics of Materials 1 (2161)
Homework #2
Solution Set
1. Summing forces in the y-direction allows one to determine the magnitude of F2 :
X
Fy = 1000 sin 60◦ − 800 sin 37◦ − F2 sin 45◦ = 0
=⇒
F2 = 543.8689 N
Then, summing forces in the x-direction and using the value of F2 allows one to determine the magnitude of F1 :
X
Fx = F1 + F2 cos 45◦ − 1000 cos 60◦ − 800 cos 37◦ = 0
=⇒
F1 = 754.3350 N
Thus,
F1 = 754 N ,
F2 = 544 N
2. (a) Noting that the force F exerted on the block by the smooth surface is in the
direction normal to the surface, the free-body diagram for the block is
y
y
P
x
20◦
30◦
x
30◦
100 lb
F
The xy- and x0 y 0 -coordinate systems are for the two alternate solution approaches
below.
(b) Using the xy-coordinate system, the force equilibrium equations are:
X
Fx = P cos 50◦ − F sin 30◦ = 0
X
Fy = P sin 50◦ + F cos 30◦ − 100 = 0
Solving the two simultaneous equations for P and F ,
P = 53.2089 lb ,
F = 68.4040 lb
An alternative approach, that avoids having to solve simultaneous equations, is
to use the x0 y 0 -coordinate system. First summing forces in the x0 -direction,
X
Fx0 = P cos 20◦ − 100 sin 30◦ = 0
=⇒
P = 53.2089 lb
Then summing forces in the y 0 -direction,
X
Fy0 = F + P sin 20◦ − 100 cos 30◦ = 0
=⇒
Using either approach, the answer is
P = 53.2 lb ,
F = 68.4 lb
F = 68.4040 lb
3. The vectors and distances from D to A, B, and C are
p
rDA = 10i + 5j − 8k
rDA = (10)2 + (5)2 + (−8)2 = 13.7477 m
p
rDB = 5i − 5j − 8k
rDA = (5)2 + (−5)2 + (−8)2 = 10.6771 m
p
rDC = −7i − 2j − 8k
rDA = (−7)2 + (−2)2 + (−8)2 = 10.8167 m
It follows that the unit vectors in the directions of the cable forces are
rDA
= 0.7274i + 0.3637j − 0.5819k
rDA
rDB
=
= 0.4683i − 0.4683j − 0.7493k
rDB
rDC
=
= −0.6471i − 0.1849j − 0.7396k
rDC
eDA =
eDB
eDC
Noting that the net lift of the balloon is 10k (kN), the force equilibrium equation is
X
F = 10k + FDA eDA + FDB eDB + FDC eDC = 0
where FDA , FDB , and FDC are the forces exerted on the balloon by the three cables.
This gives three simultaneous equations to be solved for FDA , FDB , and FDC :
0.7274FDA + 0.4683FDB − 0.6471FDC = 0
0.3637FDA − 0.4683FDB − 0.1849FDC = 0
0.5819FDA + 0.7493FDB + 0.7396FDC = 10 kN
Solving these simultaneous equations gives
FDA = 5.73 kN ,
FDB = 1.48 kN ,
FDC = 7.51 kN
4. Note that the weight of the 500 kg mass is (500 kg)(9.81 m/s2 ) = 4905 N, let the tensions
in cables A and B be FA and FB , respectively, and consider the free-body diagram
below:
y
FB
45◦
FA
x
Fy = FB sin 45◦ − 4905 = 0
=⇒ FB = 6936.72 N
Fx = FB cos 45◦ − FA = 0
=⇒ FA = 4905 N
4905 N
The minimum allowable cable cross-sectional area, Amin , is given in terms of the cable
tension, F , and allowable normal stress, σall = 150 MPa, by
σ=
F
A
=⇒
Amin =
F
.
σall
The minimum allowable cross-sectional area is related to the minimum allowable diameter, dmin , by
π 2
Amin = dmin
.
4
Cable A
4905 N
π 2
dmin = Amin =
= 3.27 × 10−5 m2
6
2
4
150 × 10 N/m
dmin = 6.4525 × 10−3 m = 6.45 mm
Cable B
6936.72 N
π 2
dmin = Amin =
= 4.6245 × 10−5 m2
4
150 × 106 N/m2
dmin = 7.6734 × 10−3 m = 7.67 mm
5. Note: In a problem, like this one, where an allowable normal stress is given without mentioning tension or compression, the allowable normal stress is assumed to be
the same in both tension and compression. This may not be entirely clear from the
textbook.
Use appropriate free-body diagrams, as shown below, to determine the axial force, and
subsequently the minimum allowable cross-sectional area, for each pipe section.
Segment AB
5 kip
30 kip
PAB
Fx = −PAB − 30 + 5 + 15 = 0
AAB =
D
C
B
15 kip
PAB = −10 kip
=⇒
PAB
−10 kip
=
= 0.417 in2
σall
−24 kip/in2
Segment BC
5 kip
PBC
D
C
Fx = −PBC + 5 + 15 = 0
ABC =
15 kip
=⇒
PBC = 20 kip
PBC
20 kip
=
= 0.833 in2
σall
24 kip/in2
Segment CD
15 kip
PCD
D
Fx = −PCD + 15 = 0
ACD =
=⇒
PCD = 15 kip
PCD
15 kip
2
=
2 = 0.625 in
σall
24 kip/in
6. Using a free-body diagram of the solid aluminum rod to find the shear force V :
P
V
L
Fx = P − V = 0
=⇒
V = P = 100 lb
The interface area that the shear force is distributed over is
A = (πd)L = π(0.75 in)(0.75 in) = 1.7671 in2
Thus, the average shear stress is
τ=
100 lb
V
=
= 56.6 psi
A
1.7671 in2
7. Note first that the cross-sectional area of the bar is A = (0.125 m)t and the angle of
inclination of the weld is θ = 90◦ − 60◦ = 30◦ . There will be a minimum thickness
based on the allowable normal stress and a minimum thickness based on the allowable
shear stress. The answer is the larger of the two, so that neither allowable stress is
exceeded.
Normal Stress
P
P
cos2 θ
=⇒
Amin =
cos2 θ
A
σall
750 × 103 N
=
cos2 30◦ = 7.0312 × 10−3 m2
80 × 106 N/m2
Amin
=
= 5.625 × 10−2 m
0.125 m
σn =
Amin
tmin
Shear Stress
P
P
sin θ cos θ
=⇒
Amin =
sin θ cos θ
A
τall
750 × 103 N
sin 30◦ cos 30◦ = 6.4952 × 10−3 m2
=
50 × 106 N/m2
Amin
=
= 5.196 × 10−2 m
0.125 m
τn =
Amin
tmin
Answer:
tmin = 56.3 mm
8. Note first that the angle of inclination of the joint is θ = 90◦ − φ and the range of
inclinations to be considered is 0 ≤ θ ≤ 45◦ . The cross-sectional area of the member
is A = (2 in)(3 in) = 6 in2 .
Normal Stress: Given the equation for the normal stress on the plane of the joint,
σn =
P
cos2 θ
A
the maximum safe load based on the allowable tensile stress in the glue is
Pσ =
(50 psi)(6 in2 )
300 lb
σall A
=
=
2
2
cos θ
cos θ
cos2 θ
Shear Stress: Given the equation for the shear stress on the plane of the joint,
τn =
P
sin θ cos θ
A
the maximum safe load based on the allowable shear stress in the glue is
Pτ =
τall A
(35 psi)(6 in2 )
210 lb
=
=
sin θ cos θ
sin θ cos θ
sin θ cos θ
At any given inclination level, θ, the maximum safe load is the smaller of Pσ and Pτ .
Plotting these results as a function of θ for 0 ≤ θ ≤ 45◦ ,
Pall (lb)
Pσ
600
Popt
420
300
Pτ
θopt 45◦ θ
(a) The optimum angle is where the two plots cross, Pσ = Pτ ,
210 lb
300 lb
=
2
cos θopt
sin θopt cos θopt
=⇒
tan θopt =
210
300
=⇒
Thus,
φopt = 90◦ − θopt = 55.0◦
(b) The maximum safe load at this optimum angle is
Popt =
300 lb
210 lb
=
= 447 lb
2
cos θopt
sin θopt cos θopt
θopt = 35.0◦