Model Answer Paper
Subject: Applied Mechanics (312312)
Scheme: I-Scheme (Civil, Mechanical, Automobile)
Examination: Summer 2025
Q.1. Attempt any FIVE of the following (10 Marks)
Note: 2 Marks each. Answers must be concise (3-5 lines).
(a) Define Statics and Dynamics.
Answer:
- Statics: It is that branch of Engineering Mechanics which deals with the study of forces and their effects on a body when the body is at rest.
- Dynamics: It is that branch of Engineering Mechanics which deals with the study of forces and their effects on a body when the body is in motion.
(b) State Lami’s theorem.
Statement:
"If three coplanar concurrent forces acting at a point are in equilibrium, then each force is directly proportional to the sine of the angle between the other two forces."
Mathematical Formula:
P / sin(α) = Q / sin(β) = R / sin(γ)
Where P, Q, R are the forces and α, β, γ are the angles opposite to them respectively.
(c) Define Centroid and Centre of gravity.
Answer:
- Centroid (G): It is the geometrical centre of a plane figure (2D area) like a rectangle, triangle, or circle, where the entire area is assumed to be concentrated.
- Centre of Gravity (C.G.): It is the specific point through which the entire weight of a solid body (3D object) acts, irrespective of the position of the body.
(d) Define Force and state its S.I. unit.
Definition: Force is an external agency which changes or tends to change the state of rest or of uniform motion of a body. It is a vector quantity.
S.I. Unit: Newton (N).
(e) Define angle of Repose.
Definition: The Angle of Repose is the maximum angle made by an inclined plane with the horizontal such that a body placed on it is just at the point of sliding down (impending motion) under the action of its own weight only. It is mathematically equal to the angle of friction.
(f) State ideal machine and write its any two characteristics.
Ideal Machine: A machine whose efficiency is 100% is called an ideal machine. In such a machine, there is no loss of energy due to friction.
Characteristics (Any Two):
- Efficiency (η) = 100%
- Mechanical Advantage (M.A.) = Velocity Ratio (V.R.)
- Output Work = Input Work
(g) Define effort and load.
Definitions:
- Effort (P): The force applied to a machine to perform work (usually to lift a weight) is called Effort.
- Load (W): The resistance to be overcome or the weight to be lifted by the machine is called Load.
Q.2. Attempt any THREE of the following (12 Marks)
Note: 4 Marks each. Detailed steps required.
(a) For a certain machine, V.R. is 140. To lift a load of 10 kN, an effort of 100 N is required. Calculate the effort required to lift a load of 45 kN.
Given:
1. Velocity Ratio (V.R.) = 140
2. Case 1: Load (W1) = 10 kN = 10,000 N; Effort (P1) = 100 N
3. Case 2: Load (W2) = 45 kN = 45,000 N
1. Velocity Ratio (V.R.) = 140
2. Case 1: Load (W1) = 10 kN = 10,000 N; Effort (P1) = 100 N
3. Case 2: Load (W2) = 45 kN = 45,000 N
To Find:
Effort required for Case 2 (P2)
Effort required for Case 2 (P2)
Formulae:
M.A. = W / P
Efficiency (η) = (M.A. / V.R.) × 100
Calculations:
Step 1: Calculate Efficiency from Case 1
M.A.1 = W1 / P1 = 10000 / 100 = 100
η = (M.A.1 / V.R.) × 100
η = (100 / 140) × 100
η = 71.43%
Step 2: Calculate Effort for Case 2
(Assumption: For the same machine, efficiency is considered constant or we use the Law of Machine. Since P2 is asked directly without Law of Machine data, we assume constant efficiency.)
M.A.2 = η × V.R. / 100
M.A.2 = 0.7143 × 140 = 100
Since M.A.2 = W2 / P2
100 = 45000 / P2
P2 = 45000 / 100
M.A.1 = W1 / P1 = 10000 / 100 = 100
η = (M.A.1 / V.R.) × 100
η = (100 / 140) × 100
η = 71.43%
Step 2: Calculate Effort for Case 2
(Assumption: For the same machine, efficiency is considered constant or we use the Law of Machine. Since P2 is asked directly without Law of Machine data, we assume constant efficiency.)
M.A.2 = η × V.R. / 100
M.A.2 = 0.7143 × 140 = 100
Since M.A.2 = W2 / P2
100 = 45000 / P2
P2 = 45000 / 100
Final Answer: Effort Required (P2) = 450 N
(b) For a simple lifting machine the law of machine is P = (0.08 W + 5) N. Calculate the effort required to lift a load of 5 kN. Also calculate the max. M.A. and identify the machine type when V.R. = 20.
Given:
1. Law of Machine: P = 0.08W + 5
2. Slope (m) = 0.08; Constant (C) = 5
3. Load (W) = 5 kN = 5000 N
4. V.R. = 20
1. Law of Machine: P = 0.08W + 5
2. Slope (m) = 0.08; Constant (C) = 5
3. Load (W) = 5 kN = 5000 N
4. V.R. = 20
To Find:
1. Effort (P)
2. Maximum M.A.
3. Type of Machine
1. Effort (P)
2. Maximum M.A.
3. Type of Machine
Calculations:
1. Effort (P):
Substitute W = 5000 N in the law:
P = 0.08(5000) + 5
P = 400 + 5
P = 405 N
2. Maximum M.A.:
Max M.A. = 1 / m
Max M.A. = 1 / 0.08
Max M.A. = 12.5
3. Type of Machine:
Max Efficiency = (Max M.A. / V.R.) × 100
Max Efficiency = (12.5 / 20) × 100
Max Efficiency = 62.5%
Since Efficiency > 50%, the machine is Reversible.
Substitute W = 5000 N in the law:
P = 0.08(5000) + 5
P = 400 + 5
P = 405 N
2. Maximum M.A.:
Max M.A. = 1 / m
Max M.A. = 1 / 0.08
Max M.A. = 12.5
3. Type of Machine:
Max Efficiency = (Max M.A. / V.R.) × 100
Max Efficiency = (12.5 / 20) × 100
Max Efficiency = 62.5%
Since Efficiency > 50%, the machine is Reversible.
Final Answer: P = 405 N; Max M.A. = 12.5; Type = Reversible Machine
(c) In a single purchase crab length of effort handle is 40 cm; diameter of load drum = 20 cm, number of teeth in pinion = 16 nos, number of spur = 80 nos. Find (i) Velocity Ratio (ii) Effort required to raise (lift) load of 2 kN with efficiency of 75%.
Given:
1. Length of Handle (L) = 40 cm
2. Diameter of Drum (D) = 20 cm ⇒ Radius (r) = 10 cm
3. Teeth on Pinion (T1) = 16
4. Teeth on Spur (T2) = 80
5. Load (W) = 2 kN = 2000 N
6. Efficiency (η) = 75% = 0.75
1. Length of Handle (L) = 40 cm
2. Diameter of Drum (D) = 20 cm ⇒ Radius (r) = 10 cm
3. Teeth on Pinion (T1) = 16
4. Teeth on Spur (T2) = 80
5. Load (W) = 2 kN = 2000 N
6. Efficiency (η) = 75% = 0.75
To Find:
1. Velocity Ratio (V.R.)
2. Effort (P)
1. Velocity Ratio (V.R.)
2. Effort (P)
Formula:
V.R. = (2L / D) × (T2 / T1)
Calculations:
1. Velocity Ratio:
V.R. = (2 × 40 / 20) × (80 / 16)
V.R. = (80 / 20) × 5
V.R. = 4 × 5
V.R. = 20
2. Effort (P):
η = (M.A. / V.R.)
0.75 = M.A. / 20
M.A. = 0.75 × 20 = 15
Now, M.A. = W / P
15 = 2000 / P
P = 2000 / 15
P = 133.33 N
V.R. = (2 × 40 / 20) × (80 / 16)
V.R. = (80 / 20) × 5
V.R. = 4 × 5
V.R. = 20
2. Effort (P):
η = (M.A. / V.R.)
0.75 = M.A. / 20
M.A. = 0.75 × 20 = 15
Now, M.A. = W / P
15 = 2000 / P
P = 2000 / 15
P = 133.33 N
Final Answer: V.R. = 20; Effort (P) = 133.33 N
(d) Explain law of machine. State its use.
Explanation:
The Law of Machine is the mathematical relationship between the Effort applied (P) and the Load lifted (W) by a machine. For most simple machines, this relationship is linear and forms a straight line when plotted on a graph.
It is expressed by the equation:
P = mW + C
Where:
- P = Effort applied to lift the load.
- W = Load to be lifted.
- m = Slope of the graph (related to the friction in the machine).
- C = Y-intercept (represents the effort required to overcome friction at zero load, i.e., machine friction).
Uses of Law of Machine:
- To find the effort required to lift a specific known load.
- To determine the maximum Mechanical Advantage (Max M.A. = 1/m).
- To determine the maximum Efficiency of the machine.
- To understand the frictional characteristics of the machine.
Q.3. Attempt any THREE of the following (12 Marks)
(a) A weight of 100 N is attached by two strings (Fig.-1). Calculate the tension in the strings.
Given:
1. Load (W) = 100 N
2. Geometry from Fig-1: Two strings attached to a ceiling.
3. String A angle with ceiling = 30°
4. String B angle with ceiling = 60°
1. Load (W) = 100 N
2. Geometry from Fig-1: Two strings attached to a ceiling.
3. String A angle with ceiling = 30°
4. String B angle with ceiling = 60°
To Find:
Tensions TA and TB.
Tensions TA and TB.
Diagram Analysis (FBD at point C):
- Angle of String A with horizontal = 30°
- Angle of String B with horizontal = 60°
- Calculate Lami's Angles:
1. Angle opposite W (between TA & TB): 180° - (30° + 60°) = 90°
2. Angle opposite TA (between TB & W): 90° + 60° = 150°
3. Angle opposite TB (between TA & W): 90° + 30° = 120°
- Angle of String A with horizontal = 30°
- Angle of String B with horizontal = 60°
- Calculate Lami's Angles:
1. Angle opposite W (between TA & TB): 180° - (30° + 60°) = 90°
2. Angle opposite TA (between TB & W): 90° + 60° = 150°
3. Angle opposite TB (between TA & W): 90° + 30° = 120°
Calculations (Using Lami's Theorem):
W / sin(90°) = TA / sin(150°) = TB / sin(120°)
Find TA:
TA / sin(150°) = 100 / sin(90°)
TA = (100 × 0.5) / 1
TA = 50 N
Find TB:
TB / sin(120°) = 100 / sin(90°)
TB = (100 × 0.866) / 1
TB = 86.6 N
TA / sin(150°) = 100 / sin(90°)
TA = (100 × 0.5) / 1
TA = 50 N
Find TB:
TB / sin(120°) = 100 / sin(90°)
TB = (100 × 0.866) / 1
TB = 86.6 N
Final Answer: TA = 50 N; TB = 86.6 N
(b) Find the support reaction for the given simply supported beam. (Ref. Fig.-2)
Given:
1. Beam AB Span = 4m
2. Point Load 40 kN at 1m from A
3. Point Load 20 kN at 2m from A
4. UDL 10 kN/m over length 2m (from 2m to 4m)
1. Beam AB Span = 4m
2. Point Load 40 kN at 1m from A
3. Point Load 20 kN at 2m from A
4. UDL 10 kN/m over length 2m (from 2m to 4m)
To Find: Reactions RA and RB.
Calculations:
1. Convert UDL to Point Load:
Total UDL = 10 kN/m × 2 m = 20 kN.
It acts at the center of the UDL span (i.e., at 1m from start of UDL).
Distance from A = 2m (start of UDL) + 1m (half width) = 3m.
2. Take Moments about A (ΣMA = 0):
(40 × 1) + (20 × 2) + (UDL_Load × Distance) - (RB × Total Span) = 0
(40 × 1) + (20 × 2) + (20 × 3) - (RB × 4) = 0
40 + 40 + 60 - 4RB = 0
140 = 4RB
RB = 140 / 4
RB = 35 kN
3. Vertical Equilibrium (ΣFy = 0):
Upward Forces = Downward Forces
RA + RB = 40 + 20 + 20 (UDL)
RA + 35 = 80
RA = 80 - 35
RA = 45 kN
Total UDL = 10 kN/m × 2 m = 20 kN.
It acts at the center of the UDL span (i.e., at 1m from start of UDL).
Distance from A = 2m (start of UDL) + 1m (half width) = 3m.
2. Take Moments about A (ΣMA = 0):
(40 × 1) + (20 × 2) + (UDL_Load × Distance) - (RB × Total Span) = 0
(40 × 1) + (20 × 2) + (20 × 3) - (RB × 4) = 0
40 + 40 + 60 - 4RB = 0
140 = 4RB
RB = 140 / 4
RB = 35 kN
3. Vertical Equilibrium (ΣFy = 0):
Upward Forces = Downward Forces
RA + RB = 40 + 20 + 20 (UDL)
RA + 35 = 80
RA = 80 - 35
RA = 45 kN
Final Answer: RA = 45 kN; RB = 35 kN
(c) Find the reaction at roller and hinge support of a beam loaded as shown in Fig.-3.
Given:
1. Hinge Support at A, Roller Support at B.
2. Span = 3 + 3 + 2 = 8 m.
3. Inclined Load: 200 kN at 45° passing through a point 3m from A.
4. Vertical Load: 100 kN at 6m from A.
1. Hinge Support at A, Roller Support at B.
2. Span = 3 + 3 + 2 = 8 m.
3. Inclined Load: 200 kN at 45° passing through a point 3m from A.
4. Vertical Load: 100 kN at 6m from A.
Calculations:
1. Resolve Inclined Load (200 kN):
Horizontal Component (H) = 200 cos(45°) = 141.42 kN (→)
Vertical Component (V) = 200 sin(45°) = 141.42 kN (↓)
2. Take Moments about A (ΣMA = 0):
(Assuming clockwise moments positive)
(Vertical Comp × 3) + (100 × 6) - (RB × 8) = 0
(141.42 × 3) + 600 - 8RB = 0
424.26 + 600 = 8RB
RB = 1024.26 / 8
RB = 128.03 kN (Vertical reaction at Roller)
3. Reactions at Hinge A:
Vertical Equilibrium (ΣFy = 0):
VA + RB = 141.42 + 100
VA = 241.42 - 128.03 = 113.39 kN (↑)
Horizontal Equilibrium (ΣFx = 0):
HA - 141.42 = 0
HA = 141.42 kN (←)
4. Resultant Reaction at A (RA):
RA = √(VA² + HA²)
RA = √(113.39² + 141.42²)
RA = √(12857 + 19999) = √32856
RA = 181.26 kN
Angle θ = tan-1(VA/HA) = 38.7°
Horizontal Component (H) = 200 cos(45°) = 141.42 kN (→)
Vertical Component (V) = 200 sin(45°) = 141.42 kN (↓)
2. Take Moments about A (ΣMA = 0):
(Assuming clockwise moments positive)
(Vertical Comp × 3) + (100 × 6) - (RB × 8) = 0
(141.42 × 3) + 600 - 8RB = 0
424.26 + 600 = 8RB
RB = 1024.26 / 8
RB = 128.03 kN (Vertical reaction at Roller)
3. Reactions at Hinge A:
Vertical Equilibrium (ΣFy = 0):
VA + RB = 141.42 + 100
VA = 241.42 - 128.03 = 113.39 kN (↑)
Horizontal Equilibrium (ΣFx = 0):
HA - 141.42 = 0
HA = 141.42 kN (←)
4. Resultant Reaction at A (RA):
RA = √(VA² + HA²)
RA = √(113.39² + 141.42²)
RA = √(12857 + 19999) = √32856
RA = 181.26 kN
Angle θ = tan-1(VA/HA) = 38.7°
Final Answer: Roller Reaction (RB) = 128.03 kN; Hinge Reaction (RA) = 181.26 kN
Q.4. Attempt any THREE of the following (12 Marks)
(a) Determine the resultant of coplanar non-concurrent forces as shown in Fig.-4.
Given:
1. Force 1: 7 N acting vertically down.
2. Force 2: 15 N acting at 30° to horizontal (up-right).
3. Force 3: 9 N acting vertically down.
1. Force 1: 7 N acting vertically down.
2. Force 2: 15 N acting at 30° to horizontal (up-right).
3. Force 3: 9 N acting vertically down.
To Find: Resultant Magnitude (R) and Direction (θ).
Calculations:
1. Sum of Horizontal Forces (ΣFx):
ΣFx = 15 cos(30°)
ΣFx = 12.99 N (→)
2. Sum of Vertical Forces (ΣFy):
(Sign Convention: Up +, Down -)
ΣFy = -7 - 9 + 15 sin(30°)
ΣFy = -16 + 7.5
ΣFy = -8.5 N (↓)
3. Magnitude of Resultant (R):
R = √((ΣFx)² + (ΣFy)²)
R = √(12.99² + (-8.5)²)
R = √(168.74 + 72.25) = √240.99
R = 15.52 N
4. Direction (θ):
θ = tan-1|ΣFy / ΣFx|
θ = tan-1(8.5 / 12.99)
θ = 33.20°
Since Fx is +ve and Fy is -ve, Resultant lies in the 4th Quadrant.
ΣFx = 15 cos(30°)
ΣFx = 12.99 N (→)
2. Sum of Vertical Forces (ΣFy):
(Sign Convention: Up +, Down -)
ΣFy = -7 - 9 + 15 sin(30°)
ΣFy = -16 + 7.5
ΣFy = -8.5 N (↓)
3. Magnitude of Resultant (R):
R = √((ΣFx)² + (ΣFy)²)
R = √(12.99² + (-8.5)²)
R = √(168.74 + 72.25) = √240.99
R = 15.52 N
4. Direction (θ):
θ = tan-1|ΣFy / ΣFx|
θ = tan-1(8.5 / 12.99)
θ = 33.20°
Since Fx is +ve and Fy is -ve, Resultant lies in the 4th Quadrant.
Final Answer: R = 15.52 N; θ = 33.20° (4th Quadrant)
(b) A block is weighing 1000 N... pull of 400 N at 30°... Find coeff of friction, normal reaction... (Ref Fig.-5)
Given:
1. Weight (W) = 1000 N
2. Applied Force (P) = 400 N at 30° to horizontal.
3. Condition: Just starts to motion (Limiting Equilibrium).
1. Weight (W) = 1000 N
2. Applied Force (P) = 400 N at 30° to horizontal.
3. Condition: Just starts to motion (Limiting Equilibrium).
Calculations:
1. Resolve Applied Force P:
Px = 400 cos(30°) = 346.41 N (→)
Py = 400 sin(30°) = 200 N (↑)
2. Normal Reaction (RN):
Apply ΣFy = 0 (Vertical Equilibrium)
RN + Py - W = 0
RN + 200 - 1000 = 0
RN = 1000 - 200
RN = 800 N
3. Frictional Force (Fr):
Apply ΣFx = 0 (Horizontal Equilibrium)
Px - Fr = 0
Fr = 346.41 N
4. Coefficient of Friction (μ):
μ = Fr / RN
μ = 346.41 / 800
μ = 0.433
5. Total Resultant Reaction (S):
S = √(RN² + Fr²)
S = √(800² + 346.41²)
S = 871.78 N
Px = 400 cos(30°) = 346.41 N (→)
Py = 400 sin(30°) = 200 N (↑)
2. Normal Reaction (RN):
Apply ΣFy = 0 (Vertical Equilibrium)
RN + Py - W = 0
RN + 200 - 1000 = 0
RN = 1000 - 200
RN = 800 N
3. Frictional Force (Fr):
Apply ΣFx = 0 (Horizontal Equilibrium)
Px - Fr = 0
Fr = 346.41 N
4. Coefficient of Friction (μ):
μ = Fr / RN
μ = 346.41 / 800
μ = 0.433
5. Total Resultant Reaction (S):
S = √(RN² + Fr²)
S = √(800² + 346.41²)
S = 871.78 N
(c) Locate the position of centroid for the following L-section (Ref. Fig.-6).
Given:
An L-shaped section. Let's split it into two rectangles.
1. Rectangle 1 (Vertical): Size 120mm x 20mm.
2. Rectangle 2 (Horizontal): Since total width is 100mm and vertical leg is 20mm thick, remaining length = 100 - 20 = 80mm. Size 80mm x 20mm.
An L-shaped section. Let's split it into two rectangles.
1. Rectangle 1 (Vertical): Size 120mm x 20mm.
2. Rectangle 2 (Horizontal): Since total width is 100mm and vertical leg is 20mm thick, remaining length = 100 - 20 = 80mm. Size 80mm x 20mm.
Calculations:
Rectangle 1 (Vertical Leg):
Area (A1) = 120 × 20 = 2400 mm²
Centroid x1 = 20 / 2 = 10 mm
Centroid y1 = 120 / 2 = 60 mm
Rectangle 2 (Horizontal Leg):
Area (A2) = 80 × 20 = 1600 mm²
Centroid x2 = 20 + (80 / 2) = 20 + 40 = 60 mm
Centroid y2 = 20 / 2 = 10 mm
Total Area (A) = 2400 + 1600 = 4000 mm²
Calculate x̄:
x̄ = (A1x1 + A2x2) / A
x̄ = (2400×10 + 1600×60) / 4000
x̄ = (24000 + 96000) / 4000 = 120000 / 4000
x̄ = 30 mm
Calculate ȳ:
ȳ = (A1y1 + A2y2) / A
ȳ = (2400×60 + 1600×10) / 4000
ȳ = (144000 + 16000) / 4000 = 160000 / 4000
ȳ = 40 mm
Area (A1) = 120 × 20 = 2400 mm²
Centroid x1 = 20 / 2 = 10 mm
Centroid y1 = 120 / 2 = 60 mm
Rectangle 2 (Horizontal Leg):
Area (A2) = 80 × 20 = 1600 mm²
Centroid x2 = 20 + (80 / 2) = 20 + 40 = 60 mm
Centroid y2 = 20 / 2 = 10 mm
Total Area (A) = 2400 + 1600 = 4000 mm²
Calculate x̄:
x̄ = (A1x1 + A2x2) / A
x̄ = (2400×10 + 1600×60) / 4000
x̄ = (24000 + 96000) / 4000 = 120000 / 4000
x̄ = 30 mm
Calculate ȳ:
ȳ = (A1y1 + A2y2) / A
ȳ = (2400×60 + 1600×10) / 4000
ȳ = (144000 + 16000) / 4000 = 160000 / 4000
ȳ = 40 mm
Final Answer: Centroid G(x̄, ȳ) = (30 mm, 40 mm)
Q.5. Attempt any TWO of the following (12 Marks)
Note: 6 Marks each. Comprehensive solution required.
(a) A block of weight 450 N is placed on rough inclined plane making inclination of 20° with horizontal. If μ = 0.24, calculate value of force parallel to plane to move block UP.
Given:
1. Weight (W) = 450 N
2. Angle of Inclination (α) = 20°
3. Coefficient of Friction (μ) = 0.24
4. Condition: Motion is Impending UP the plane.
1. Weight (W) = 450 N
2. Angle of Inclination (α) = 20°
3. Coefficient of Friction (μ) = 0.24
4. Condition: Motion is Impending UP the plane.
To Find: Force (P) applied parallel to the plane.
Diagram Analysis:
- Weight acts vertically down. Components: W cos(α) perpendicular to plane, W sin(α) down the plane.
- Normal Reaction (R) acts perpendicular to plane upwards.
- Friction (F) acts DOWN the plane (opposite to motion).
- Force (P) acts UP the plane.
- Weight acts vertically down. Components: W cos(α) perpendicular to plane, W sin(α) down the plane.
- Normal Reaction (R) acts perpendicular to plane upwards.
- Friction (F) acts DOWN the plane (opposite to motion).
- Force (P) acts UP the plane.
Calculations:
1. Forces Perpendicular to Plane (ΣFy = 0):
R - W cos(20°) = 0
R = 450 × cos(20°)
R = 450 × 0.9397
R = 422.86 N
2. Calculate Limiting Friction (F):
F = μ × R
F = 0.24 × 422.86
F = 101.49 N (Acting Down)
3. Forces Parallel to Plane (ΣFx = 0):
P - W sin(20°) - F = 0
P = W sin(20°) + F
P = (450 × 0.3420) + 101.49
P = 153.90 + 101.49
P = 255.39 N
R - W cos(20°) = 0
R = 450 × cos(20°)
R = 450 × 0.9397
R = 422.86 N
2. Calculate Limiting Friction (F):
F = μ × R
F = 0.24 × 422.86
F = 101.49 N (Acting Down)
3. Forces Parallel to Plane (ΣFx = 0):
P - W sin(20°) - F = 0
P = W sin(20°) + F
P = (450 × 0.3420) + 101.49
P = 153.90 + 101.49
P = 255.39 N
Final Answer: Force Required (P) = 255.39 N
(b) Find resultant of concurrent force system as shown in Fig.-9. Magnitude and direction by analytical method.
Given Forces:
1. 10 kN at 40° from Vertical (1st Quad) → Angle with X-axis = 90-40 = 50°.
2. 20 kN Vertical Up (On Y-axis).
3. 15 kN at 30° from Horizontal (2nd Quad).
4. 12 kN Horizontal Left (Negative X-axis).
5. 20 kN Vertical Down (Negative Y-axis).
6. 5 kN Horizontal Left (Negative X-axis).
1. 10 kN at 40° from Vertical (1st Quad) → Angle with X-axis = 90-40 = 50°.
2. 20 kN Vertical Up (On Y-axis).
3. 15 kN at 30° from Horizontal (2nd Quad).
4. 12 kN Horizontal Left (Negative X-axis).
5. 20 kN Vertical Down (Negative Y-axis).
6. 5 kN Horizontal Left (Negative X-axis).
Resolution Table/Calculations:
1. Sum of Horizontal Forces (ΣFx):
ΣFx = (10 cos 50°) - (15 cos 30°) - 12 - 5
ΣFx = 6.428 - 12.99 - 17
ΣFx = -23.56 kN
2. Sum of Vertical Forces (ΣFy):
ΣFy = +20 + (10 sin 50°) + (15 sin 30°) - 20
ΣFy = 20 + 7.66 + 7.5 - 20
ΣFy = +15.16 kN
3. Resultant Magnitude (R):
R = √((ΣFx)² + (ΣFy)²)
R = √((-23.56)² + (15.16)²)
R = √(555.07 + 229.82)
R = √784.89
R = 28.02 kN
4. Direction (θ):
θ = tan-1|ΣFy / ΣFx|
θ = tan-1(15.16 / 23.56)
θ = 32.76°
Since ΣFx is negative and ΣFy is positive, the resultant lies in the 2nd Quadrant.
ΣFx = (10 cos 50°) - (15 cos 30°) - 12 - 5
ΣFx = 6.428 - 12.99 - 17
ΣFx = -23.56 kN
2. Sum of Vertical Forces (ΣFy):
ΣFy = +20 + (10 sin 50°) + (15 sin 30°) - 20
ΣFy = 20 + 7.66 + 7.5 - 20
ΣFy = +15.16 kN
3. Resultant Magnitude (R):
R = √((ΣFx)² + (ΣFy)²)
R = √((-23.56)² + (15.16)²)
R = √(555.07 + 229.82)
R = √784.89
R = 28.02 kN
4. Direction (θ):
θ = tan-1|ΣFy / ΣFx|
θ = tan-1(15.16 / 23.56)
θ = 32.76°
Since ΣFx is negative and ΣFy is positive, the resultant lies in the 2nd Quadrant.
Final Answer: R = 28.02 kN; θ = 32.76° (N of W)
Q.6. Attempt any TWO of the following (12 Marks)
(a) Calculate the coefficient of friction, if a block weighing 600 N... by a force of 150 N applied at an angle of 60°... (Ref. Fig.-11).
Given:
1. Weight (W) = 600 N
2. Applied Force (P) = 150 N at 60° to horizontal.
3. Condition: Just moves (Limiting Friction).
1. Weight (W) = 600 N
2. Applied Force (P) = 150 N at 60° to horizontal.
3. Condition: Just moves (Limiting Friction).
To Find: Coefficient of friction (μ).
Calculations:
1. Resolve Force P:
Horizontal Component (Px) = 150 cos(60°) = 75 N
Vertical Component (Py) = 150 sin(60°) = 129.90 N (Upwards)
2. Vertical Equilibrium (ΣFy = 0):
Normal Reaction (N) + Py - W = 0
N + 129.90 - 600 = 0
N = 600 - 129.90
N = 470.10 N
3. Horizontal Equilibrium (ΣFx = 0):
Applied Force (Px) = Frictional Force (F)
F = 75 N
4. Coefficient of Friction (μ):
μ = F / N
μ = 75 / 470.10
μ = 0.16
Horizontal Component (Px) = 150 cos(60°) = 75 N
Vertical Component (Py) = 150 sin(60°) = 129.90 N (Upwards)
2. Vertical Equilibrium (ΣFy = 0):
Normal Reaction (N) + Py - W = 0
N + 129.90 - 600 = 0
N = 600 - 129.90
N = 470.10 N
3. Horizontal Equilibrium (ΣFx = 0):
Applied Force (Px) = Frictional Force (F)
F = 75 N
4. Coefficient of Friction (μ):
μ = F / N
μ = 75 / 470.10
μ = 0.16
Final Answer: Coefficient of Friction (μ) = 0.16
(b) Find the centroid of the area as shown in Fig.-12.
Given:
Composite shape consisting of:
1. A Rectangle at the bottom (Size 200mm width x 400mm height).
2. A Semicircle on top (Diameter 200mm ⇒ Radius R = 100mm).
Composite shape consisting of:
1. A Rectangle at the bottom (Size 200mm width x 400mm height).
2. A Semicircle on top (Diameter 200mm ⇒ Radius R = 100mm).
Calculations:
1. Symmetry Analysis:
The figure is symmetrical about the vertical Y-axis passing through the center.
Therefore, x̄ = 200 / 2 = 100 mm
2. Component 1 (Rectangle):
Area (A1) = 200 × 400 = 80,000 mm²
Centroid y1 = 400 / 2 = 200 mm
3. Component 2 (Semicircle):
Area (A2) = (π × R²) / 2 = (π × 100²) / 2 = 15,707.96 mm²
Centroid distance from its base = 4R / 3π = (4 × 100) / (3 × π) = 42.44 mm
Centroid y2 (from bottom) = Height of Rect + 42.44
y2 = 400 + 42.44 = 442.44 mm
4. Calculate Combined Centroid (ȳ):
ȳ = (A1y1 + A2y2) / (A1 + A2)
ȳ = ((80000 × 200) + (15707.96 × 442.44)) / (80000 + 15707.96)
ȳ = (16,000,000 + 6,949,830) / 95,707.96
ȳ = 22,949,830 / 95,707.96
ȳ = 239.79 mm
The figure is symmetrical about the vertical Y-axis passing through the center.
Therefore, x̄ = 200 / 2 = 100 mm
2. Component 1 (Rectangle):
Area (A1) = 200 × 400 = 80,000 mm²
Centroid y1 = 400 / 2 = 200 mm
3. Component 2 (Semicircle):
Area (A2) = (π × R²) / 2 = (π × 100²) / 2 = 15,707.96 mm²
Centroid distance from its base = 4R / 3π = (4 × 100) / (3 × π) = 42.44 mm
Centroid y2 (from bottom) = Height of Rect + 42.44
y2 = 400 + 42.44 = 442.44 mm
4. Calculate Combined Centroid (ȳ):
ȳ = (A1y1 + A2y2) / (A1 + A2)
ȳ = ((80000 × 200) + (15707.96 × 442.44)) / (80000 + 15707.96)
ȳ = (16,000,000 + 6,949,830) / 95,707.96
ȳ = 22,949,830 / 95,707.96
ȳ = 239.79 mm
Final Answer: Centroid G(x̄, ȳ) = (100 mm, 239.79 mm)
(c) Find the centroid of an inverted T-section from the bottom, if flange is (60 × 10) cm and web is (10 × 60) cm.
Given:
Inverted T-section:
1. Bottom Flange: 60 cm wide, 10 cm high.
2. Vertical Web (centered): 10 cm wide, 60 cm high.
Inverted T-section:
1. Bottom Flange: 60 cm wide, 10 cm high.
2. Vertical Web (centered): 10 cm wide, 60 cm high.
Calculations:
1. Symmetry Analysis:
The section is symmetrical about the vertical axis.
x̄ = 60 / 2 = 30 cm
2. Component 1 (Bottom Flange):
Area (A1) = 60 × 10 = 600 cm²
Centroid y1 = 10 / 2 = 5 cm
3. Component 2 (Vertical Web):
Area (A2) = 10 × 60 = 600 cm²
Centroid y2 = Height of Flange + (Height of Web / 2)
y2 = 10 + (60 / 2) = 10 + 30 = 40 cm
4. Calculate Combined Centroid (ȳ):
ȳ = (A1y1 + A2y2) / (A1 + A2)
ȳ = ((600 × 5) + (600 × 40)) / (600 + 600)
ȳ = (3000 + 24000) / 1200
ȳ = 27000 / 1200
ȳ = 22.5 cm
The section is symmetrical about the vertical axis.
x̄ = 60 / 2 = 30 cm
2. Component 1 (Bottom Flange):
Area (A1) = 60 × 10 = 600 cm²
Centroid y1 = 10 / 2 = 5 cm
3. Component 2 (Vertical Web):
Area (A2) = 10 × 60 = 600 cm²
Centroid y2 = Height of Flange + (Height of Web / 2)
y2 = 10 + (60 / 2) = 10 + 30 = 40 cm
4. Calculate Combined Centroid (ȳ):
ȳ = (A1y1 + A2y2) / (A1 + A2)
ȳ = ((600 × 5) + (600 × 40)) / (600 + 600)
ȳ = (3000 + 24000) / 1200
ȳ = 27000 / 1200
ȳ = 22.5 cm
Final Answer: ȳ = 22.5 cm from the bottom.
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