Below the 1000+ Questions with detail solution on Force Analysis of Engineering Mechanics subject, which were asked in various competitive exam of Civil Engineer for various posts.
Question 1.
The load shared by the member BC of the structure shown in figure below is
Previously Asked in : TNPSC AE 2012 ||
2√3 KN
3√2 KN
4 KN
3 KN
Answer (Detailed Solution Below)
Option 4 KN is correct
Detailed Explanation :
As shown above figure the value of TBC force is
TBC sin30° = – 2 KN
TBC = -2 / sin30° = 4 KN (Compressive)
Question 2.
A force p of 50 N and another force Q of unknown magnitude act at 90° to each other. They are balanced by a force of 130 N. The magnitude of Q is
Previously Asked in : ||
60 N
80 N
100 N
120 N
Answer (Detailed Solution Below)
Option 120 N is correct
Detailed Explanation :
Given that : P = 50N ; θ = 90° ; Resultant force (R) = 130N.
we known that ;
Resultant force (R) = √ (P2 +Q2 + 2PQcosθ)
Put the given value in above equation;
130 = √ 502 +Q2 + 2 x 50 x Q x cos90°)
On Squaring both side; Also we know that cos90° = 0
⇒ 1302 = 502 + Q2
⇒ 16900 = 2500 + Q2
⇒ Q2 = 16900 – 2500
⇒ Q2 = 14400
⇒ Q = √ (14400)
Q = 120 N
Hence 120 N option is correct
Question 3.
If the resultant of two forces has the same magnitude as either of the force, then the angle between the two forces is
Previously Asked in : ESE PRE 1998 || UP TGT Mathematics 2016 || UPPSC Polytechnic Lecturer Civil 22 Dec 2021
30°
45°
60°
120°
Answer (Detailed Solution Below)
Option 120° is correct
Detailed Explanation :
Concept:
When two forces ‘P’ and ‘Q’ have the same magnitude as their resultant ‘R’, the relationship can be expressed by Parallelogram law of vector addition,. The magnitude of the resultant force is given by:
Resultant force (R) = √ (P2 +Q2 + 2PQcosθ)
Calculation:
Let us consider two forces P” and ‘Q’.
Given that; R = P = Q
Let the angle between P and Q is θ
Using parallelogram law;
R = √ (P2 +Q2 + 2PQcosθ)
R = √ (R2 +R2 + 2R.R.cosθ)
R = √ (2R2 (1+ cosθ)
Squaring both sides gives;
R2 = 2R2 (1+ cosθ)
1/2 = 1 + cosθ [by crossing and shifting]
cosθ = -1/2
θ = cos-1(-1/2)
therefore; θ=120°
Question 4.
In order to determine the effects of a force acting on body, we must know:
Previously Asked in : HPSSSC JE 11.04.2021 ||
Its magnitude and direction of the line along which it acts.
Its nature (whether push or pull)
Point through which it acts on the body.
All of these
Answer (Detailed Solution Below)
Option All of these is correct
Detailed Explanation :
Concept:
For finding/determining the effect of a force acting on a point, we must know the following properties of force:
- the magnitude of the force.
- line of action of the force.
- nature of the force, i. e., whether the force is push or pull.
Key Points
Magnitude:
This refers to the strength or intensity of the force, often measured in Newtons (N).
Direction:
This specifies the orientation of the force, indicating which way it’s acting (e.g., North, West, at a specific angle).
Point of Application:
This identifies the specific location on the body where the force is applied.
Nature of the Force:
This clarifies whether the force is a push or a pull, which can affect the type of stress or strain exerted on the body.
Question 5.
Which of following is not scalar quantity?
Previously Asked in : GSSSB AAE 07. 02 .2021 ||
Time
Mass
Area
Velocity
Answer (Detailed Solution Below)
Option Velocity is correct
Detailed Explanation :
Concept:
Velocity is not scalar quantity.
A scalar quantity is defined as the physical quantity
that has only magnitude, for example, mass, energy,
time, area and speed etc.
A vector quantity is defined as the physical quantity
that has both magnitude as well as direction like force
momentum torque, impulse weight etc.
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Question 10.
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