Model : C
Mansoura University
Faculty of Engineering
Eng. Math. & Physics Dept.
Engineering Mechanics I
Midterm Exam (2011-2012)
Time: One hour
Instructors: Prof. Dr. Bishri Abdel-Mo'emen, Assoc. Prof. Ayman Ashour, and Dr. Waleed Albeshbeshy
Student Name: ……………………………………………………………………………. :‫اإلسم‬
Section No:
……………………
F= 10 kN
.
Question 1: ( 5 Points )
Resolve the 10 kN force F in Fig. (P1) into components acting along the u and v axes
and determine the magnitudes of these components. Take θ=95o and β=25o.
Fu
95
u
Fu= -10 cos 25o
= -9.063 N
F= 10 kN
60
β
Fv
u
25
θ
Fv = 10 sin 25o
= 4.226 N
v
θ
v
𝐹𝑢
10
=
sin 25 sin 95
𝐹𝑣
10
=
sin 60 sin 95
𝐹𝑣 =
10
sin 60
sin 95
= 8.693 𝑘𝑁
𝐹𝑢 =
10
sin 95
sin 25 = 4.242 𝑘𝑁
(in the negative u direction
y
F1= 500 N
Question 2: ( 5 Points )
I If the magnitude of the resultant force acting on the eyebolt is 1000 N and is
directed along the positive x axis, determine the magnitude of F2 and the angle θ.
30
o
x
θ
5
Given: Rx= 1000 N
Ry= 0
Rx= 500 cos 30o + F2 cosθ - 350 (3/5)
1000 = 433 + F2 cosθ - 210
F2 cos θ = 777 N
………………………………… (1)
4
3
F3= 350 N
Ry= 500 sin 30o - F2 sinθ - 350 (4/5)
0= 250 - F2 sinθ - 280
F2 sin θ = -30
……………………………….. (2)
Dividing (2) by (1),
5
tan θ = (-30)/(777)= -0.039
4
∴ 𝜃 = −2.21𝑜 ( the negative means that F2 in the
3
first quadrant )
Substituting in either (1) or (2)
F3= 350 N
∴ 𝐹2 = 777.6 𝑁
F2
y
F1= 500 N
o
30
x
θ
F2
B
Question 3: ( 5 Points )
In order to raise the lamp post from the position shown, the force F on the cable must
create a counterclockwise moment of 2100 N.m about point A. Determine the
magnitude of F that must be applied to the cable
F
3m
o
65
C
From Triangle ABC: The cosine law gives
𝐶𝐵 = 32 + 1.52 − 2 1.5 3 𝑐𝑜𝑠115
= 15.054 = 3.88
From the sine law:
1.5
3.88
=
𝑠𝑖𝑛𝛽 𝑠𝑖𝑛115
B
∴ 𝑠𝑖𝑛𝛽 = 0.35
F
β
The moment about A:
MA = 𝐹 𝑠𝑖𝑛𝛽 3
∴ 2100 = 𝐹 × 0.35 × 3
∴ F =2000 N
3m
α
C
o
115
1.5 m A
Question 4: ( 5 Points )
As an airplane's brake are applied, the nose wheel exerts two forces on the end
of the landing gear as shown. Determine the horizontal and vertical
components of reaction at the pin B and the force in strut AC.
C
B
o
30
30 cm
A
o
20
50 cm
From the FBD:
a=0.8 tan 20o= 0.291 m
b= 0.3 tan 20o=0.109 m
Equilibrium conditions:
𝑀𝐵 = 0;
−6 × 𝑎 + 3 × 0.8 − 𝐹𝑠𝑖𝑛50 × 0.3 + 𝐹𝑐𝑜𝑠50 × 𝑏 = 0
−6 × 0.291 + 3 × 0.8 − 𝐹𝑠𝑖𝑛50 × 0.3 + 𝐹𝑐𝑜𝑠50 × 0.109 = 0
∴ 0.16 𝐹 = 0.653
∴
𝐹 = 4.081 𝑘𝑁
A
1.5 m
3 kN
O
6 kN
a
Bx
By
b
B
C
o
30
A
F
Equilibrium conditions:
o
20
𝐹𝑥 = 0;
−𝐵𝑥 − 𝐹𝑠𝑖𝑛50 + 3 = 0
∴
𝐵𝑥 = 3 − 4.081 𝑠𝑖𝑛50 = −0.126 𝑘𝑁 (opposite to assumed direction)
3 kN
O
Equilibrium conditions:
𝐹𝑦 = 0;
𝐵𝑦 − 𝐹𝑐𝑜𝑠50 + 6 = 0
6 kN
∴
𝐵𝑦 = −6 + 4.081 𝑐𝑜𝑠50 = −3.377 𝑘𝑁 (opposite to assumed direction )
0.3 m
0.5 m