Mansoura National University Engineering Mechanics (1)

Faculty of Engineering Module Code: (BAS021)

AIE, MTE, BCE, BME, CCE Programs Time allowed :2 hours.
Level: Freshman (000) Irregular Full Mark:50 Marks
(4-page exam) & (5-Questions) Module Credit Hours (3)
Final Exam (14-6-2025) – Second Semester model (1)

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

Q1) [12 Marks] Determine the magnitude and
coordinate direction angles of the resultant force.

1) The Cartesian vector form of the force 𝑭𝟐

(a) 𝑭𝟐 = (−110𝒌) N, (b) 𝑭𝟐 = (−110𝒊) N
(c) 𝑭𝟐 = (110𝒌) N (d) 𝑭𝟐 = (−110𝒋) N

2) The Cartesian vector form of the force 𝑭1 is:

(a) 𝑭1 = (64𝒊 + 48𝒋 + 0𝒌) N, (b) 𝑭1 = (64𝒊 + 0𝒋 + 48𝒌) N
(c) 𝑭𝟏 = (64𝒊 + 0𝒋 − 48𝒌) N (d) 𝑭𝟏 = (48𝒊 + 0𝒋 + 64𝒌) N
3) The Cartesian vector form of the force 𝑭3 is:
(a) 𝑭3 = (42.43𝒊 + 73.48𝒋 − 84.85𝒌) N, (b) 𝑭3 = (73.48𝒊 + 42.43𝒋 − 84.85𝒌) N
(c) 𝑭3 = (42.43𝒊 + 73.48𝒋 + 84.85𝒌) N (d) 𝑭3 = (73.48𝒊 − 42.43𝒋 + 84.85𝒌) N
4) The Cartesian vector form of the resultant force 𝑹
(a) 𝑹 = (106.43𝒊 − 73.48𝒋 + 22.85𝒌) N, (b) 𝑹 = (22.85𝒊 + 73.48𝒋 + 106.43𝒌) N
(c) 𝑹 = (73.48𝒊 + 106.43𝒋 + 22.85𝒌) N (d) 𝑹 = (106.43𝒊 + 73.48𝒋 + 22.85𝒌) N
5) The magnitude of the resultant is
(a) |𝑹| = 333.35 N (b) |𝑹| = 139.35 N (c)|𝑹| = 231.33 N (d) |𝑹| = 131.335 N

6) The direction angles of the resultant are

(a) 𝜽𝒙 = 35.87𝑜 , 𝜽𝒚 = 55.98𝑜 , 𝜽𝒛 = 79.98𝑜

(b) 𝜽𝒙 = 79.98𝑜 , 𝜽𝒚 = 55.98𝑜 , 𝜽𝒛 = 35.87𝑜

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 (c) 𝜽𝒙 = 55.98𝑜 , 𝜽𝒚 = 35.87𝑜 , 𝜽𝒛 = 79.98𝑜

(d) 𝜽𝒙 = 35.87𝑜 , 𝜽𝒚 = 79.98𝑜 , 𝜽𝒛 = 55.98𝑜

7) The dot product between the force 𝑭1 and the force 𝑭3 , (𝑭1 ⊙ 𝑭3 ), is:
(a) 𝑭1 ⊙ 𝑭3 = −6788.23 , (b) 𝑭1 ⊙ 𝑭3 = 6788.23
(c) 𝑭1 ⊙ 𝑭3 = 7788.23 (d) 𝑭1 ⊙ 𝑭3 = −7788.23
8) The angle between the force 𝑭1 and the force 𝑭3 is:

(a) 𝜽 = 60𝑜 (b) 𝜽 = 50𝑜 (c) 𝜽 = 45𝑜 (d) 𝜽 = 127𝑜

Q2) [12 Marks] Determine the magnitudes

of forces F1, F2, and F3 necessary to hold the
force 𝑭 = {−20𝒊 − 10𝒋 − 20𝒌} kN in
equilibrium.

9) The Cartesian vector form of the force 𝑭𝟏 is:

√3 1 √3 √3 1 √3
(a) 𝑭1 = 𝐹1 { 𝒊 + 𝒋+ 𝒌} , (b) 𝑭1 = 𝐹1 { 𝒊 − 𝒋+ 𝒌}.
4 4 2 4 4 2

√3 1 √3 √3 1 √3
(c) 𝑭1 = 𝐹1 { 𝒊 − 𝒋+ 𝒌} . (d) 𝑭1 = 𝐹1 { 𝒊 − 𝒋− 𝒌} .
4 4 4 4 4 2

10) The Cartesian vector form of the force 𝑭2 is:

(a) 𝑭2 = 𝐹2 (−√2𝒊 + 2𝒋 − 2𝒌)/2 ., (b) 𝑭2 = 𝐹2 (1𝒊 + 1𝒋 + 1𝒌)/√2 .
(c) 𝑭2 = 𝐹2 (−√2𝒊 + 1𝒋 + 1𝒌)/2 . (d) 𝑭2 = 𝐹2 (−1𝒊 − 1𝒋 + 2𝒌)/√2 .
11) The Cartesian vector form of the force 𝑭3 is:
(a) 𝑭3 = 𝐹3 (4𝒊 + 3𝒋 + 2𝒌)/3 ., (b) 𝑭3 = 𝐹3 (4𝒊 + 4𝒋 + 2𝒌)/6 .
(c) 𝑭3 = 𝐹3 (4𝒊 − 4𝒋 + 3𝒌)/6 . (d) 𝑭3 = 𝐹3 (4𝒊 + 4𝒋 − 2𝒌)/6.

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 (a) 16.78 KN, (b) 29.48 KN,
12) the magnitude of 𝑭1 is:
(c) 29.21 KN, (d) 26.78 KN.
(a) −29.48 KN, (b) 16.87 KN,
13) the magnitude of 𝑭2 is:
(c) 6.87 KN, (d) 18.75 KN.
(a) −29.48 KN, (b) 19.89 KN,
14) the magnitude of 𝑭3 is:
(c) 29.89 KN, (d) 18.75 KN.

Q3) [7.5 Marks] For the given system.

a) Replace the force system shown in
Figure by a resultant force and a
couple moment 𝑴𝑹𝒐 at point O.
b) Can we replace the resulting
system with a single resultant
force? Do it to a point 𝑃(𝑥, 𝑦, 0).
At point O,
15) The Cartesian vector form of the resultant force is
(a)𝑹 = −50𝒌 N (b) 𝑹 = −200𝒋 N (c) 𝑹 = 50𝒌 N (d) 𝑹 = 100𝒌 N
16) The Cartesian vector form of the Couple moment, 𝑴𝑪
(a) 𝑴𝑪 = −150𝒌 (N.m), (b) 𝑴𝑪 = −150𝒋 (N.m)
(c) 𝑴𝑪 = 100𝒌 (N.m) (d) 𝑴𝑪 = 100𝒊 (N.m)
17) The Cartesian vector form of the resultant moment at point O, 𝑴𝑹𝒐
(a) 𝑴𝑹𝒐 = −250𝒌 (N.m), (b) 𝑴𝑹𝒐 = −150𝒋 (N.m)
(c) 𝑴𝑹𝒐 = −100𝒋 (N.m) (d) 𝑴𝑹𝒐 = −250𝒋 (N.m)
18) Orthogonality 𝑴𝑹𝒐 ⊙ 𝑹 = ⋯

(a) −500 (b) 5000 (c) 500 (d) 0

19) The single resultant force 𝑹 pass by the point 𝑃(𝑥, 𝑦, 0)

(a) 𝑃(2, 2, 0) (b) 𝑃(0, −2,0) (c) 𝑃(0, 2, 0) (d) 𝑃(−2, 0, 0)

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 Q4) [7.5 Marks] Determine the
support reactions on the member
in the next Figure. The collar at A
is fixed to the member and can
slide vertically along the vertical
shaft.

Free body diagram (F. B. D.)

20) The reaction at the roller B is:

(a) 𝑩𝒙 → (b) 𝑩𝒚 ↑ (c) (𝑩𝒙 , 𝑩𝒚 ) →, ↑ (d) 𝑩𝒚 ↓

21) The reactions at collar A are:

(a) 𝑨𝒙 →, 𝑴 (b) 𝑨𝒚 ↑, 𝑴 (c) (𝑩𝒙 , 𝑩𝒚 ) →, ↑ (d) (𝑴𝒙 , 𝑴𝒚 )

22) The magnitude of reaction at the roller B is:

(a) 500 → (b) 300 ↑ (c) (300 ,500) →, ↑ (d) 500 ↑

23) ∑ 𝐹𝑥 = 0:

(a) 𝐴𝑥 = 300 → (b) 𝐴𝑥 = 300 ↑ (c) 𝐴𝑦 = 500 ↑ (d) 𝐴𝑦 = 300 ↑

24) The magnitude of the moment at the roller A is:

(a) 𝑀 = 658.55 ↻ (b) 𝑀 = 608.55 ↺ (c) 𝑀 = 608.55 ↻ (d) 𝑀 = 768.55 ↺

Solve the next question:

Q5) [11 Marks] The drum has a weight of 100 Ib and
rests on the floor for which the coefficient of static
friction is 𝜇𝑆 = 0.6. Determine the smallest magnitude
of the force P that will cause impending motion of the
drum.

With all Best Wishes

Examiner: Assoc. Prof. Dr. Galal I. El-Baghdady.
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 Mansoura National University Engineering Mechanics (1)
Faculty of Engineering Module Code: (BAS021)
AIE, MTE, BCE, BME, CCE Programs Time allowed :2 hours.
Level: Freshman (000) Irregular Full Mark:50 Marks
(4-page exam) & (5-Questions) Module Credit Hours (3)
Final Exam (14-6-2025) – Second Semester model (2)

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

Q1) [12 Marks] Determine the magnitude and
coordinate direction angles of the resultant force.

1) The Cartesian vector form of the force 𝑭𝟐

(a) 𝑭𝟐 = (−110𝒊) N, (b) 𝑭𝟐 = (−110𝒌) N
(c) 𝑭𝟐 = (110𝒋) N (d) 𝑭𝟐 = (−110𝒋) N

2) The Cartesian vector form of the force 𝑭1 is:

(a) 𝑭1 = (64𝒊 + 0𝒋 + 48𝒌) N, (b) 𝑭1 = (48𝒊 + 48𝒋 + 64𝒌) N
(c) 𝑭𝟏 = (64𝒊 + 0𝒋 − 48𝒌) N (d) 𝑭𝟏 = (48𝒊 + 20𝒋 + 64𝒌) N
3) The Cartesian vector form of the force 𝑭3 is:
(a) 𝑭3 = (42.43𝒊 + 73.48𝒋 − 84.85𝒌) N, (b) 𝑭3 = (73.48𝒊 + 42.43𝒋 − 84.85𝒌) N
(c) 𝑭3 = (73.48𝒊 − 42.43𝒋 − 84.85𝒌) N (d) 𝑭3 = (42.43𝒊 + 73.48𝒋 + 84.85𝒌) N
4) The Cartesian vector form of the resultant force 𝑹
(a) 𝑹 = (106.43𝒊 + 73.48𝒋 − 22.85𝒌) N, (b) 𝑹 = (22.85𝒊 + 73.48𝒋 + 106.43𝒌) N
(c) 𝑹 = (106.43𝒊 + 73.48𝒋 + 22.85𝒌) N (d) 𝑹 = (73.48𝒊 + 106.43𝒋 + 22.85𝒌) N
5) The magnitude of the resultant is
(a) |𝑹| = 233.35 N (b) |𝑹| = 131.335 N (c)|𝑹| = 231.33 N (d) |𝑹| = 161.35 N

6) The direction angles of the resultant are

(a) 𝜽𝒙 = 35.87𝑜 , 𝜽𝒚 = 79.98𝑜 , 𝜽𝒛 = 55.98𝑜

(b) 𝜽𝒙 = 55.98𝑜 , 𝜽𝒚 = 35.87𝑜 , 𝜽𝒛 = 79.98𝑜

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 (c) 𝜽𝒙 = 79.98𝑜 , 𝜽𝒚 = 55.98𝑜 , 𝜽𝒛 = 35.87𝑜

(d) 𝜽𝒙 = 35.87𝑜 , 𝜽𝒚 = 55.98𝑜 , 𝜽𝒛 = 79.98𝑜

7) The dot product between the force 𝑭1 and the force 𝑭3 , (𝑭1 ⊙ 𝑭3 ), is:
(a) 𝑭1 ⊙ 𝑭3 = 6788.23 , (b) 𝑭1 ⊙ 𝑭3 = 7788.23
(c) 𝑭1 ⊙ 𝑭3 = −7788.23 (d) 𝑭1 ⊙ 𝑭3 = −6788.23
8) The angle between the force 𝑭1 and the force 𝑭3 is:

(a) 𝜽 = 50𝑜 (b) 𝜽 = 45𝑜 (c) 𝜽 = 145𝑜 (d) 𝜽 = 60𝑜

Q2) [12 Marks] Determine the magnitudes

of forces F1, F2, and F3 necessary to hold the
force 𝑭 = {−20𝒊 − 10𝒋 − 20𝒌} kN in
equilibrium.

9) The Cartesian vector form of the force 𝐹1 is:

√3 1 √3 √3 1 √3
(a) 𝑭1 = 𝐹1 { 𝒊 − 𝒋+ 𝒌} , (b) 𝑭1 = 𝐹1 { 𝒊 + 𝒋+ 𝒌}.
4 4 2 4 4 2

√3 1 √3 √3 1 √3
(c) 𝑭1 = 𝐹1 { 𝒊 − 𝒋+ 𝒌} . (d) 𝑭1 = 𝐹1 { 𝒊 − 𝒋− 𝒌} .
4 4 4 4 4 2

10) The Cartesian vector form of the force 𝑭2 is:

(a) 𝑭2 = 𝐹2 (−√2𝒊 + 2𝒋 − 2𝒌)/2 ., (b) 𝑭2 = 𝐹2 (1𝒊 + 1𝒋 + 1𝒌)/√2 .
(c) 𝑭2 = 𝐹2 (−√2𝒊 − 1𝒋 − 1𝒌)/2 . (d) 𝑭2 = 𝐹2 (−√2𝒊 + 1𝒋 + 1𝒌)/2 .
11) The Cartesian vector form of the force 𝑭3 is:
(a) 𝑭3 = 𝐹3 (4𝒊 + 3𝒋 + 2𝒌)/3 . (b) 𝑭3 = 𝐹3 (4𝒊 + 4𝒋 − 2𝒌)/6 .
(c) 𝑭3 = 𝐹3 (4𝒊 − 4𝒋 + 3𝒌)/6 . (d) 𝑭3 = 𝐹3 (4𝒊 + 4𝒋 + 2𝒌)/6.

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 (a) 26.78 KN, (b) 29.48 KN,
12) the magnitude of 𝑭1 is:
(c) 29.21 KN, (d) 18.75 KN.
(a) −29.48 KN, (b) 6.87 KN,
13) the magnitude of 𝑭2 is:
(c) 29.21 KN, (d) 18.75 KN.
(a) −29.48 KN, (b) 29.48 KN,
14) the magnitude of 𝑭3 is:
(c) 19.89 KN, (d) 18.75 KN.

Q3) [7.5 Marks] Determine the

support reactions on the member
in the next Figure. The collar at A
is fixed to the member and can
slide vertically along the vertical
shaft.

Free body diagram (F. B. D.)

15) The reaction at the roller B is:

(a) 𝑩𝒙 ← (b) 𝑩𝒚 ↓ (c) (𝑩𝒙 , 𝑩𝒚 ) →, ↑ (d) 𝑩𝒚 ↑

16) The reactions at collar A are:

(a) (𝑴𝒙 , 𝑴𝒚 ) (b) (𝑩𝒙 , 𝑩𝒚 ) →, ↑ (c) 𝑨𝒚 ↑, 𝑴 (d) 𝑨𝒙 →, 𝑴

17) The magnitude of reaction at the roller B is:

(a) 500 ↑ (b) (500 ,300) →, ↑ (c) 300 ↑ (d) 500 ↓

18) ∑ 𝐹𝑥 = 0:

(a) 𝐴𝑥 = 300 ← (b) 𝐴𝑥 = 300 → (c) 𝐴𝑦 = 500 ↓ (d) 𝐴𝑦 = 300 ↑

19) The magnitude of the moment at the roller A is:

(a) 𝑀 = 658.55 ↻ (b) 𝑀 = 608.55 ↺ (c) 𝑀 = 608.55 ↻ (d) 𝑀 = 768.55 ↺

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Q4) [7.5 Marks] For the given system.
a) Replace the force system shown in
Figure by a resultant force and a
couple moment 𝑴𝑹𝒐 at point O.
b) Can we replace the resulting
system with a single resultant
force? Do it to a point 𝑃(𝑥, 𝑦, 0).
At point O,
20) The Cartesian vector form of the resultant force is
(a)𝑹 = 50𝒌 N (b) 𝑹 = −200𝒋 N (c) 𝑹 = −50𝒌 N (d) 𝑹 = 100𝒌 N
21) The Cartesian vector form of the Couple moment, 𝑴𝑪
(a) 𝑴𝑪 = −150𝒋 (N.m), (b) 𝑴𝑪 = 150𝒋 (N.m)
(c) 𝑴𝑪 = −100𝒌 (N.m) (d) 𝑴𝑪 = −100𝒊 (N.m)
22) The Cartesian vector form of the resultant moment at point O, 𝑴𝑹𝒐
(a) 𝑴𝑹𝒐 = −100𝒊 (N.m), (b) 𝑴𝑹𝒐 = −100𝒋 (N.m)
(c) 𝑴𝑹𝒐 = −150𝒋 (N.m) (d) 𝑴𝑹𝒐 = −250𝒋 (N.m)
23) Orthogonality 𝑴𝑹𝒐 ⊙ 𝑹 = ⋯

(a) 0 (b) 500 (c) 5000 (d) −500

24) The single resultant force 𝑹 pass by the point 𝑃(𝑥, 𝑦, 0)

(a) 𝑃(−2, 2, 0) (b) 𝑃(0, −2,0) (c) 𝑃(−2, 0, 0) (d) 𝑃(2, 0, 0)

Solve the next question:

Q5) [11 Marks] The drum has a weight of 100 Ib and
rests on the floor for which the coefficient of static
friction is 𝜇𝑆 = 0.6. Determine the smallest magnitude
of the force P that will cause impending motion of the
drum.

With all Best Wishes

Examiner: Assoc. Prof. Dr. Galal I. El-Baghdady.
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