Mansoura National University Engineering Mechanics (1)

Faculty of Engineering Module Code: (BAS021)

AIE, MTE, BCE, BME, CCE Programs Time allowed :2 hour.
Level: Freshman (000) Full Mark:50 Marks
(4-page exam) & (5-Questions) Module Credit Hours (3)
Final Exam (3-9-2024) – Summer Semester Model (1)

Solve, then choose the nearest correct answer to the following:

Q1) [10 Marks] Determine the magnitude and
coordinate direction angles of the resultant of
the two forces shown. Let 𝐹1 = 20𝐾𝑁 and
𝐹1 = 10𝐾𝑁.

1) The Cartesian vector form of the resultant

force 𝑭𝟏 is

(a) 𝑭𝟏 = (8.66𝒊 − 5𝒋 + 17.32𝒌) KN, (b) 𝑭𝟏 = (−8.66𝒊 − 5𝒋 + 17.32𝒌) KN

(c) 𝑭𝟏 = (−8.66𝒊 + 5𝒋 + 17.32𝒌) KN (d) 𝑭𝟏 = (8.66𝒊 − 5𝒋 − 17.32𝒌) KN
2) The Cartesian vector form of the resultant force 𝑭𝟐 is
(a) 𝑭𝟐 = (7.07𝒊 + 5𝒋 + 5𝒌) KN, (b) 𝑭𝟐 = (−7.07𝒊 + 5𝒋 + 5𝒌) KN
(c) 𝑭𝟐 = (−7.07𝒊 + 5𝒋 − 5𝒌) KN (d) 𝑭𝟐 = (−7.07𝒊 − 5𝒋 + 5𝒌) KN
3) The Cartesian vector form of the resultant force 𝑹 is
(a) 𝑹 = (1.59𝒊 + 10𝒋 − 22.32𝒌) Ib, (b) 𝑹 = (1.59𝒊 + 10𝒋 + 22.32𝒌) Ib
(c) 𝑹 = (1.59𝒊 + 0𝒋 + 22.32𝒌) Ib (d) 𝑹 = (1.59𝒊 − 10𝒋 + 22.32𝒌) Ib
4) The magnitude of the resultant is
(a) |𝑹| = 22.377 KN (b) |𝑹| = 20.37 KN (c)|𝑹| = 16.45 KN (d) |𝑹| = 21.37 KN
5) The direction angle in y-direction of the resultant is …
(a) 𝜽𝒚 = 75.1𝑜 (b) 𝜽𝒚 = 4.1𝑜 (c)𝜽𝒚 = 85.9𝑜 (d) 𝜽𝒚 = 90.0𝑜

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Q2) [10 Marks] A 90 N load is suspended
from the hook as shown in Figure. The load
is supported by two cables and a spring
having a stiffness k = 500 N/m. Cable AD lies
in the x-y plane, and cable AC lies in the x-z
plane.

6) The Cartesian vector form of the force in the spring 𝑭

(a) 𝑭 = (𝐹𝒊 + 𝟎𝒋 − 0𝒌) N, (b) 𝑭 = (0𝒊 − 𝐹𝒋 − 0𝒌) N
(c) 𝑭 = (0𝒊 + 0𝒋 − 𝐹𝒌) N (d) 𝑭 = (0𝒊 + 𝐹𝒋 + 0𝒌) N
7) The Cartesian vector form of the force in the cable AC is
(a) 𝑻𝐴𝐶 = 𝑇𝐴𝐶 (−0.8𝒊 + 𝟎𝒋 + 0.6𝒌) N, (b) 𝑻𝐴𝐶 = 𝑇𝐴𝐶 (0.8𝒊 + 0.6𝒋 + 0.6𝒌) N
(c) 𝑻𝐴𝐶 = 𝑇𝐴𝐶 (−0.8𝒊 + 𝟎𝒋 − 0.6𝒌) N (d) 𝑻𝐴𝐶 = 𝑇𝐴𝐶 (−0.6𝒊 + 0𝒋 − 0.8𝒌) N

8) The tension in Cable AC (a) −240 N, (b) 240 N,

(c) 230 N, (d) 150 N.
9) The force in the spring (a) 𝐹 = −240 N, (b) 𝐹 = 240 N,
(c) 𝐹 = 208 N, (d) 𝐹 = 150 N.
10) The stretch of the spring ∆𝑙 (a) 0.416 m, (b) 0.516 m,
(c) 0.465 m, (d) 0.614 𝑚.

Q3) [14 Marks] Replace the two forces

F1 and F2 and the couple moment M
shown in figure by a resultant force and
couple at the origin. Then reduce this
system to a single force and calculate the
coordinates of the point P in the xy plane
through which the resultant force acts.

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11) The Cartesian vector form of the a) (10𝒌) Ib b) (10𝒋) Ib
force 𝑭𝟏 , is ….. c) (−10𝒌) Ib d) (−50𝒌)Ib
12) The Cartesian coordinate of the a) (4, −3, 0) ft b) (−4, −3, 0) ft
point at 𝑭𝟐 , is ….. c) (−4,3, 0) ft d) (−4,0, 0) ft
13) The Cartesian vector form of the a) (10𝒌) Ib b) (50𝒋) Ib
force 𝑭𝟐 , is ….. c) (−10𝒌) Ib d) (−50𝒌)Ib
14) The Cartesian vector form of the a) (40𝒊 − 30𝒋) b) (−40𝒊 + 30𝒋)
couple moment 𝑴, is ….. c) (40𝒊 + 30𝒋) d) (−40𝒊 − 30𝒋)
15) The resultant force at point O, 𝑭𝑹 , is a) (40𝒌) Ib b) (10𝒋) Ib
….. c) (−40𝒌) Ib d) (−50𝒌)Ib
16) The resultant moment about point a) −(190𝒊 + 170𝒋) b) (190𝒊 − 170𝒋)
O, 𝑴𝑹 , is ….. c) (190𝒊 + 170𝒋) d) (−190𝒊 + 170𝒋)
17) This system can reduced to a single a) (−4,4, 0) ft b) (−4.25,4.75,0)ft
force at point P with coordinates ….. c) (−4,4, 0) ft) d) (4.25,4.75,0) ft

Q4) [8 Marks] The pipe assembly

shown in Figure has a built-in (fixed)
support and is subjected to two forces
and a couple moment. What are the
reactions at the support?

18) The magnitude of the horizontal a) 150 N b) 86.6 N

component at A is c) 73.2 N d) 0 N
19) The magnitude of the vertical a) 150 N b) 86.6 N
component at A is c) 73.2 N d) 0 N
20) The magnitude of the total force at a) 150 N b) 86.6 N
A is c) 173.2 N d) 0 N
21) The magnitude of the moment M A a) 75.8 N.m b) 85.6 N.m
is c) 88.5 N. m d) 73.2 N.m

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Q5) [8 Marks] A 100 lb force acts as shown on
a 300 lb block placed on an inclined plane, as
shown in Figure (a). The coefficients of friction
between the block and the plane are  s = 0.25
and k = 0.2. Determine whether the block is in
equilibrium and find the value of the friction
force.

a) 240 Ib b) 100 Ib
22) The magnitude of the Normal force is
c) 120 Ib d) 200.4 Ib
a) 40 Ib b) -80 Ib
23) The friction force F is
c) 48 Ib d) 80 Ib
a) 40 Ib b) 60 Ib
24) The Maximum friction force 𝐹𝑠 is
c) 48 Ib d) -66 Ib
25) The magnitude of the actual friction a) 40 Ib b) -80 Ib
force will be c) 48 Ib d) 80 Ib

With all Best Wishes

Examiner: Dr. Galal I. El-Baghdady.

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 Mansoura National University
Engineering Mechanics (1)
Faculty of Engineering
Module Code: (BAS 021)
AIE, MTE, BCE, BME, CCE
Time allowed: 2 Hours.
Programs
Full Mark: 50 Marks
Level: Freshman (000)
Module Credit Hours (3)
(4 pages exam)
Final Exam - (3-9-2024) – Summer Semester
Model Answer

Model (1) Answer Model (2) Answer

Q. Number Correct answer Q. Number Correct answer

1 A 1 B

2 B 2 C

3 C 3 B

4 A 4 C

5 D 5 C

6 D 6 D

7 A 7 B

8 D 8 B

9 C 9 B

10 A 10 C

11 A 11 C

12 C 12 A

13 D 13 B

14 B 14 C

15 C 15 A

16 A 16 C

17 B 17 D

18 B 18 C
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19 A 19 A

20 C 20 B

21 D 21 D

22 A 22 B

23 B 23 C

24 B 24 D

25 C 25 D

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