Msbte Study Resources
Model Answer Paper
Subject: Applied Mechanics (312312)
Scheme: I-Scheme
Examination: Summer 2025
Q.1. Attempt any FIVE of the following (10 Marks)
Note: 2 Marks each. Answers must be concise (3-5 lines).
Reference: Question Paper Image Q1
(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:
\[ \frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma} \]
Where \( P, Q, R \) are the forces and \( \alpha, \beta, \gamma \) 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 (\( \phi \)) 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.
(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 \( \eta = 100\% \)
- \( M.A. = V.R. \)
- \( \text{Input Work} = \text{Output Work} \)
(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.
Reference: Question Paper Image Q2
(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. \( V.R. = 140 \)
2. Case 1: \( W_1 = 10 \text{ kN} = 10,000 \text{ N} \); \( P_1 = 100 \text{ N} \)
3. Case 2: \( W_2 = 45 \text{ kN} = 45,000 \text{ N} \)
1. \( V.R. = 140 \)
2. Case 1: \( W_1 = 10 \text{ kN} = 10,000 \text{ N} \); \( P_1 = 100 \text{ N} \)
3. Case 2: \( W_2 = 45 \text{ kN} = 45,000 \text{ N} \)
To Find:
Effort required for Case 2 (\( P_2 \))
Effort required for Case 2 (\( P_2 \))
Calculations:
Step 1: Calculate Efficiency from Case 1
\[ M.A._1 = \frac{W_1}{P_1} = \frac{10000}{100} = 100 \] \[ \eta = \frac{M.A._1}{V.R.} \times 100 = \frac{100}{140} \times 100 = 71.43\% \]
Step 2: Calculate Effort for Case 2
(Assuming constant efficiency)
\[ M.A._2 = \eta \times V.R. = 0.7143 \times 140 = 100 \] \[ P_2 = \frac{W_2}{M.A._2} = \frac{45000}{100} \]
\[ M.A._1 = \frac{W_1}{P_1} = \frac{10000}{100} = 100 \] \[ \eta = \frac{M.A._1}{V.R.} \times 100 = \frac{100}{140} \times 100 = 71.43\% \]
Step 2: Calculate Effort for Case 2
(Assuming constant efficiency)
\[ M.A._2 = \eta \times V.R. = 0.7143 \times 140 = 100 \] \[ P_2 = \frac{W_2}{M.A._2} = \frac{45000}{100} \]
Final Answer: Effort Required \( P_2 = 450 \text{ N} \)
(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 \( W = 5 \text{ kN} = 5000 \text{ N} \)
4. \( V.R. = 20 \)
1. Law of Machine: \( P = 0.08W + 5 \)
2. Slope \( m = 0.08 \), Constant \( C = 5 \)
3. Load \( W = 5 \text{ kN} = 5000 \text{ N} \)
4. \( V.R. = 20 \)
Calculations:
1. Effort (P):
Substitute W in the law:
\[ P = 0.08(5000) + 5 \] \[ P = 400 + 5 = 405 \text{ N} \]
2. Maximum M.A.:
\[ \text{Max M.A.} = \frac{1}{m} = \frac{1}{0.08} = 12.5 \]
3. Type of Machine:
\[ \text{Max Efficiency} = \frac{\text{Max M.A.}}{V.R.} \times 100 \] \[ \eta_{max} = \frac{12.5}{20} \times 100 = 62.5\% \] Since \( \eta > 50\% \), the machine is Reversible.
Substitute W in the law:
\[ P = 0.08(5000) + 5 \] \[ P = 400 + 5 = 405 \text{ N} \]
2. Maximum M.A.:
\[ \text{Max M.A.} = \frac{1}{m} = \frac{1}{0.08} = 12.5 \]
3. Type of Machine:
\[ \text{Max Efficiency} = \frac{\text{Max M.A.}}{V.R.} \times 100 \] \[ \eta_{max} = \frac{12.5}{20} \times 100 = 62.5\% \] Since \( \eta > 50\% \), the machine is Reversible.
Final Answer: \( P = 405 \text{ N} \); Max M.A. = 12.5; Type = Reversible
(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
3. Teeth on Pinion (\( T_1 \)) = 16
4. Teeth on Spur (\( T_2 \)) = 80
5. Load (\( W \)) = 2 kN = 2000 N
6. Efficiency (\( \eta \)) = 75% = 0.75
1. Length of Handle (\( L \)) = 40 cm
2. Diameter of Drum (\( D \)) = 20 cm
3. Teeth on Pinion (\( T_1 \)) = 16
4. Teeth on Spur (\( T_2 \)) = 80
5. Load (\( W \)) = 2 kN = 2000 N
6. Efficiency (\( \eta \)) = 75% = 0.75
Calculations:
1. Velocity Ratio (V.R.):
Formula: \( V.R. = \frac{2L}{D} \times \frac{T_2}{T_1} \)
\[ V.R. = \frac{2 \times 40}{20} \times \frac{80}{16} \] \[ V.R. = 4 \times 5 = 20 \]
2. Effort (P):
Using efficiency formula: \( \eta = \frac{M.A.}{V.R.} \)
\[ 0.75 = \frac{M.A.}{20} \Rightarrow M.A. = 15 \] Since \( M.A. = W/P \):
\[ 15 = \frac{2000}{P} \Rightarrow P = \frac{2000}{15} = 133.33 \text{ N} \]
Formula: \( V.R. = \frac{2L}{D} \times \frac{T_2}{T_1} \)
\[ V.R. = \frac{2 \times 40}{20} \times \frac{80}{16} \] \[ V.R. = 4 \times 5 = 20 \]
2. Effort (P):
Using efficiency formula: \( \eta = \frac{M.A.}{V.R.} \)
\[ 0.75 = \frac{M.A.}{20} \Rightarrow M.A. = 15 \] Since \( M.A. = W/P \):
\[ 15 = \frac{2000}{P} \Rightarrow P = \frac{2000}{15} = 133.33 \text{ N} \]
Final Answer: V.R. = 20; Effort \( P = 133.33 \text{ N} \)
(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. It is expressed by the equation:
\[ P = mW + C \]
Where:
- \( P \) = Effort applied
- \( W \) = Load lifted
- \( m \) = Slope of the graph (related to friction)
- \( C \) = Y-intercept (effort required to start the machine/friction)
Uses:
- To find the effort required to lift a specific load.
- To determine Max M.A. (\( 1/m \)) and Max Efficiency.
- To understand the frictional characteristics of the machine.
Q.3. Attempt any THREE of the following (12 Marks)
Reference: Question Paper Image Q3 (Includes Fig-1, Fig-2, Fig-3)
(a) A weight of 100 N is attached by two strings (Fig.-1). Calculate the tension in the strings.
Diagram Analysis:
- Lami's Angles:
Angle opposite \( W \) = \( 90^{\circ} \)
Angle opposite \( T_A \) = \( 90^{\circ} + 60^{\circ} = 150^{\circ} \)
Angle opposite \( T_B \) = \( 90^{\circ} + 30^{\circ} = 120^{\circ} \)
- Lami's Angles:
Angle opposite \( W \) = \( 90^{\circ} \)
Angle opposite \( T_A \) = \( 90^{\circ} + 60^{\circ} = 150^{\circ} \)
Angle opposite \( T_B \) = \( 90^{\circ} + 30^{\circ} = 120^{\circ} \)
Calculations (Using Lami's Theorem):
\[ \frac{100}{\sin 90^{\circ}} = \frac{T_A}{\sin 150^{\circ}} = \frac{T_B}{\sin 120^{\circ}} \]
Find \( T_A \):
\[ T_A = \frac{100 \times \sin 150^{\circ}}{\sin 90^{\circ}} = 100 \times 0.5 = 50 \text{ N} \]
Find \( T_B \):
\[ T_B = \frac{100 \times \sin 120^{\circ}}{\sin 90^{\circ}} = 100 \times 0.866 = 86.6 \text{ N} \]
\[ T_A = \frac{100 \times \sin 150^{\circ}}{\sin 90^{\circ}} = 100 \times 0.5 = 50 \text{ N} \]
Find \( T_B \):
\[ T_B = \frac{100 \times \sin 120^{\circ}}{\sin 90^{\circ}} = 100 \times 0.866 = 86.6 \text{ N} \]
Final Answer: \( T_A = 50 \text{ N}, T_B = 86.6 \text{ N} \)
(b) Find the support reaction for the given simply supported beam. (Ref. Fig.-2)
Given:
1. 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 for 2m length (from 2m to 4m)
1. 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 for 2m length (from 2m to 4m)
Calculations:
1. Convert UDL to Point Load:
Total UDL = \( 10 \times 2 = 20 \text{ kN} \)
Acts at center of UDL = 2m + (2/2)m = 3m from A.
2. Moments about A (\( \Sigma M_A = 0 \)):
Sum of clockwise moments = Sum of anti-clockwise moments
\[ (40 \times 1) + (20 \times 2) + (20 \times 3) = R_B \times 4 \] \[ 40 + 40 + 60 = 4R_B \] \[ 140 = 4R_B \Rightarrow R_B = 35 \text{ kN} \]
3. Vertical Equilibrium (\( \Sigma F_y = 0 \)):
\[ R_A + R_B = \text{Total Load} \] \[ R_A + 35 = 40 + 20 + 20 \] \[ R_A = 80 - 35 = 45 \text{ kN} \]
Total UDL = \( 10 \times 2 = 20 \text{ kN} \)
Acts at center of UDL = 2m + (2/2)m = 3m from A.
2. Moments about A (\( \Sigma M_A = 0 \)):
Sum of clockwise moments = Sum of anti-clockwise moments
\[ (40 \times 1) + (20 \times 2) + (20 \times 3) = R_B \times 4 \] \[ 40 + 40 + 60 = 4R_B \] \[ 140 = 4R_B \Rightarrow R_B = 35 \text{ kN} \]
3. Vertical Equilibrium (\( \Sigma F_y = 0 \)):
\[ R_A + R_B = \text{Total Load} \] \[ R_A + 35 = 40 + 20 + 20 \] \[ R_A = 80 - 35 = 45 \text{ kN} \]
Final Answer: \( R_A = 45 \text{ kN}, R_B = 35 \text{ kN} \)
(c) Find the reaction at roller and hinge support of a beam loaded as shown in Fig.-3.
Calculations:
1. Components of 200 kN:
\( H = 200 \cos 45^{\circ} = 141.42 \text{ kN} (\rightarrow) \)
\( V = 200 \sin 45^{\circ} = 141.42 \text{ kN} (\downarrow) \)
2. Moments about A (\( \Sigma M_A = 0 \)):
\[ (141.42 \times 3) + (100 \times 6) - (R_B \times 8) = 0 \] \[ 424.26 + 600 = 8R_B \] \[ 1024.26 = 8R_B \Rightarrow R_B = 128.03 \text{ kN} \]
3. Reactions at Hinge A:
Vertical: \( V_A + R_B = 141.42 + 100 \)
\( V_A = 241.42 - 128.03 = 113.39 \text{ kN} (\uparrow) \)
Horizontal: \( H_A = 141.42 \text{ kN} (\leftarrow) \)
4. Resultant Reaction at A:
\[ R_A = \sqrt{V_A^2 + H_A^2} = \sqrt{113.39^2 + 141.42^2} = 181.26 \text{ kN} \]
\( H = 200 \cos 45^{\circ} = 141.42 \text{ kN} (\rightarrow) \)
\( V = 200 \sin 45^{\circ} = 141.42 \text{ kN} (\downarrow) \)
2. Moments about A (\( \Sigma M_A = 0 \)):
\[ (141.42 \times 3) + (100 \times 6) - (R_B \times 8) = 0 \] \[ 424.26 + 600 = 8R_B \] \[ 1024.26 = 8R_B \Rightarrow R_B = 128.03 \text{ kN} \]
3. Reactions at Hinge A:
Vertical: \( V_A + R_B = 141.42 + 100 \)
\( V_A = 241.42 - 128.03 = 113.39 \text{ kN} (\uparrow) \)
Horizontal: \( H_A = 141.42 \text{ kN} (\leftarrow) \)
4. Resultant Reaction at A:
\[ R_A = \sqrt{V_A^2 + H_A^2} = \sqrt{113.39^2 + 141.42^2} = 181.26 \text{ kN} \]
Final Answer: \( R_B = 128.03 \text{ kN}; R_A = 181.26 \text{ kN} \)
Q.4. Attempt any THREE of the following (12 Marks)
Reference: Question Paper Image Q4 (Includes Fig-4 to Fig-7)
(a) Determine the resultant of coplanar non-concurrent forces as shown in Fig.-4.
Calculations:
1. Sum of Horizontal Forces (\( \Sigma F_x \)):
\[ \Sigma F_x = 15 \cos 30^{\circ} = 12.99 \text{ N} \]
2. Sum of Vertical Forces (\( \Sigma F_y \)):
\[ \Sigma F_y = -7 - 9 + 15 \sin 30^{\circ} = -16 + 7.5 = -8.5 \text{ N} \]
3. Magnitude of Resultant (R):
\[ R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{12.99^2 + (-8.5)^2} \] \[ R = \sqrt{168.74 + 72.25} = 15.52 \text{ N} \]
4. Direction (\( \theta \)):
\[ \theta = \tan^{-1} \left| \frac{8.5}{12.99} \right| = 33.20^{\circ} \quad \text{(4th Quadrant)} \]
\[ \Sigma F_x = 15 \cos 30^{\circ} = 12.99 \text{ N} \]
2. Sum of Vertical Forces (\( \Sigma F_y \)):
\[ \Sigma F_y = -7 - 9 + 15 \sin 30^{\circ} = -16 + 7.5 = -8.5 \text{ N} \]
3. Magnitude of Resultant (R):
\[ R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{12.99^2 + (-8.5)^2} \] \[ R = \sqrt{168.74 + 72.25} = 15.52 \text{ N} \]
4. Direction (\( \theta \)):
\[ \theta = \tan^{-1} \left| \frac{8.5}{12.99} \right| = 33.20^{\circ} \quad \text{(4th Quadrant)} \]
(b) A block is weighing 1000 N... pull of 400 N at 30°... Find coeff of friction, normal reaction... (Ref Fig.-5)
Calculations:
1. Resolve Force P:
\( P_x = 400 \cos 30^{\circ} = 346.41 \text{ N} (\rightarrow) \)
\( P_y = 400 \sin 30^{\circ} = 200 \text{ N} (\uparrow) \)
2. Normal Reaction (\( R_N \)):
Using \( \Sigma F_y = 0 \):
\( R_N + P_y - W = 0 \Rightarrow R_N = 1000 - 200 = 800 \text{ N} \)
3. Frictional Force (\( F_r \)):
Using \( \Sigma F_x = 0 \):
\( P_x - F_r = 0 \Rightarrow F_r = 346.41 \text{ N} \)
4. Coefficient of Friction (\( \mu \)):
\[ \mu = \frac{F_r}{R_N} = \frac{346.41}{800} = 0.433 \]
5. Resultant (\( S \)):
\[ S = \sqrt{R_N^2 + F_r^2} = \sqrt{800^2 + 346.41^2} = 871.78 \text{ N} \]
\( P_x = 400 \cos 30^{\circ} = 346.41 \text{ N} (\rightarrow) \)
\( P_y = 400 \sin 30^{\circ} = 200 \text{ N} (\uparrow) \)
2. Normal Reaction (\( R_N \)):
Using \( \Sigma F_y = 0 \):
\( R_N + P_y - W = 0 \Rightarrow R_N = 1000 - 200 = 800 \text{ N} \)
3. Frictional Force (\( F_r \)):
Using \( \Sigma F_x = 0 \):
\( P_x - F_r = 0 \Rightarrow F_r = 346.41 \text{ N} \)
4. Coefficient of Friction (\( \mu \)):
\[ \mu = \frac{F_r}{R_N} = \frac{346.41}{800} = 0.433 \]
5. Resultant (\( S \)):
\[ S = \sqrt{R_N^2 + F_r^2} = \sqrt{800^2 + 346.41^2} = 871.78 \text{ N} \]
(c) Locate the position of centroid for the following L-section (Ref. Fig.-6).
Calculations:
Split into 2 Rectangles:
1. Vertical Leg (\( A_1 \)): \( 120 \times 20 \).
2. Horizontal Leg (\( A_2 \)): \( 80 \times 20 \) (Subtracting overlap width: 100 - 20).
Area & Centroid of 1:
\( A_1 = 2400 \text{ mm}^2 \)
\( x_1 = 10 \text{ mm} \), \( y_1 = 60 \text{ mm} \)
Area & Centroid of 2:
\( A_2 = 1600 \text{ mm}^2 \)
\( x_2 = 20 + 40 = 60 \text{ mm} \), \( y_2 = 10 \text{ mm} \)
Coordinates (\( \bar{x}, \bar{y} \)):
\[ \bar{x} = \frac{A_1x_1 + A_2x_2}{A_1 + A_2} = \frac{2400(10) + 1600(60)}{4000} = 30 \text{ mm} \] \[ \bar{y} = \frac{A_1y_1 + A_2y_2}{A_1 + A_2} = \frac{2400(60) + 1600(10)}{4000} = 40 \text{ mm} \]
1. Vertical Leg (\( A_1 \)): \( 120 \times 20 \).
2. Horizontal Leg (\( A_2 \)): \( 80 \times 20 \) (Subtracting overlap width: 100 - 20).
Area & Centroid of 1:
\( A_1 = 2400 \text{ mm}^2 \)
\( x_1 = 10 \text{ mm} \), \( y_1 = 60 \text{ mm} \)
Area & Centroid of 2:
\( A_2 = 1600 \text{ mm}^2 \)
\( x_2 = 20 + 40 = 60 \text{ mm} \), \( y_2 = 10 \text{ mm} \)
Coordinates (\( \bar{x}, \bar{y} \)):
\[ \bar{x} = \frac{A_1x_1 + A_2x_2}{A_1 + A_2} = \frac{2400(10) + 1600(60)}{4000} = 30 \text{ mm} \] \[ \bar{y} = \frac{A_1y_1 + A_2y_2}{A_1 + A_2} = \frac{2400(60) + 1600(10)}{4000} = 40 \text{ mm} \]
Final Answer: Centroid G(30 mm, 40 mm)
Q.5. Attempt any TWO of the following (12 Marks)
Note: 6 Marks each. Comprehensive solution required.
Reference: Question Paper Image Q5 (Includes Fig-9 and Fig-10)
(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.
Diagram Analysis:
- Weight W acts down. Components: \( W \cos \alpha \) (perpendicular), \( W \sin \alpha \) (down plane).
- Normal Reaction R acts perpendicular UP.
- Friction F acts DOWN plane (opposing upward motion).
- Force P acts UP plane.
- Weight W acts down. Components: \( W \cos \alpha \) (perpendicular), \( W \sin \alpha \) (down plane).
- Normal Reaction R acts perpendicular UP.
- Friction F acts DOWN plane (opposing upward motion).
- Force P acts UP plane.
Calculations:
1. Normal Reaction (\( R \)):
\[ R = W \cos 20^{\circ} = 450 \times 0.9397 = 422.86 \text{ N} \]
2. Limiting Friction (\( F \)):
\[ F = \mu R = 0.24 \times 422.86 = 101.49 \text{ N} \]
3. Applied Force (\( P \)):
\[ P = W \sin 20^{\circ} + F \] \[ P = (450 \times 0.342) + 101.49 \] \[ P = 153.9 + 101.49 = 255.39 \text{ N} \]
\[ R = W \cos 20^{\circ} = 450 \times 0.9397 = 422.86 \text{ N} \]
2. Limiting Friction (\( F \)):
\[ F = \mu R = 0.24 \times 422.86 = 101.49 \text{ N} \]
3. Applied Force (\( P \)):
\[ P = W \sin 20^{\circ} + F \] \[ P = (450 \times 0.342) + 101.49 \] \[ P = 153.9 + 101.49 = 255.39 \text{ N} \]
Final Answer: \( P = 255.39 \text{ N} \)
(b) Find resultant of concurrent force system as shown in Fig.-9. Magnitude and direction by analytical method.
Calculations:
1. Sum of Horizontal Forces (\( \Sigma F_x \)):
\[ \Sigma F_x = 10 \cos 50^{\circ} - 15 \cos 30^{\circ} - 12 - 5 \] (Note: 10kN is at 40° to vertical, so 50° to horizontal)
\[ \Sigma F_x = 6.428 - 12.99 - 17 = -23.56 \text{ kN} \]
2. Sum of Vertical Forces (\( \Sigma F_y \)):
\[ \Sigma F_y = 20 + 10 \sin 50^{\circ} + 15 \sin 30^{\circ} - 20 \] \[ \Sigma F_y = 20 + 7.66 + 7.5 - 20 = 15.16 \text{ kN} \]
3. Resultant (\( R \)):
\[ R = \sqrt{(-23.56)^2 + (15.16)^2} = 28.02 \text{ kN} \]
4. Direction (\( \theta \)):
\[ \theta = \tan^{-1} \left( \frac{15.16}{-23.56} \right) = 32.76^{\circ} \] Since x is negative and y is positive, it lies in the 2nd Quadrant (N of W).
\[ \Sigma F_x = 10 \cos 50^{\circ} - 15 \cos 30^{\circ} - 12 - 5 \] (Note: 10kN is at 40° to vertical, so 50° to horizontal)
\[ \Sigma F_x = 6.428 - 12.99 - 17 = -23.56 \text{ kN} \]
2. Sum of Vertical Forces (\( \Sigma F_y \)):
\[ \Sigma F_y = 20 + 10 \sin 50^{\circ} + 15 \sin 30^{\circ} - 20 \] \[ \Sigma F_y = 20 + 7.66 + 7.5 - 20 = 15.16 \text{ kN} \]
3. Resultant (\( R \)):
\[ R = \sqrt{(-23.56)^2 + (15.16)^2} = 28.02 \text{ kN} \]
4. Direction (\( \theta \)):
\[ \theta = \tan^{-1} \left( \frac{15.16}{-23.56} \right) = 32.76^{\circ} \] Since x is negative and y is positive, it lies in the 2nd Quadrant (N of W).
Q.6. Attempt any TWO of the following (12 Marks)
Reference: Question Paper Image Q6 (Includes Fig-11 and Fig-12)
(a) Calculate the coefficient of friction... force of 150 N applied at an angle of 60°... (Ref. Fig.-11).
Calculations:
1. Resolve P:
\( P_x = 150 \cos 60^{\circ} = 75 \text{ N} \)
\( P_y = 150 \sin 60^{\circ} = 129.9 \text{ N} \) (Upwards)
2. Normal Reaction (\( N \)):
\( \Sigma F_y = 0 \Rightarrow N + P_y = W \)
\( N = 600 - 129.9 = 470.1 \text{ N} \)
3. Coefficient of Friction (\( \mu \)):
Since block just moves, \( F = P_x = 75 \text{ N} \)
\[ \mu = \frac{F}{N} = \frac{75}{470.1} = 0.16 \]
\( P_x = 150 \cos 60^{\circ} = 75 \text{ N} \)
\( P_y = 150 \sin 60^{\circ} = 129.9 \text{ N} \) (Upwards)
2. Normal Reaction (\( N \)):
\( \Sigma F_y = 0 \Rightarrow N + P_y = W \)
\( N = 600 - 129.9 = 470.1 \text{ N} \)
3. Coefficient of Friction (\( \mu \)):
Since block just moves, \( F = P_x = 75 \text{ N} \)
\[ \mu = \frac{F}{N} = \frac{75}{470.1} = 0.16 \]
Final Answer: Coefficient μ = 0.16
(b) Find the centroid of the area as shown in Fig.-12.
Note: The area is composite, consisting of a rectangular area minus a semi-circular area (cut-out section).
Calculations:
1. X-Coordinate (\( \bar{x} \)):
Due to symmetry about Y-axis, \( \bar{x} = 100 \text{ mm} \).
2. Y-Coordinate (\( \bar{y} \)):
Component 1 (Rectangle 200x400):
\( A_1 = 200 \times 400 = 80,000 \text{ mm}^2 \)
\( y_1 = 200 \text{ mm} \)
Component 2 (Semicircle Cut-out, r=100):
\( A_2 = \frac{\pi(100)^2}{2} = 15,708 \text{ mm}^2 \) (Negative Area)
Centroid of semi-circle from its base (top edge of rectangle) = \( \frac{4r}{3\pi} = 42.44 \text{ mm} \)
\( y_2 = 400 - 42.44 = 357.56 \text{ mm} \)
3. Combined Centroid:
\[ \bar{y} = \frac{A_1y_1 - A_2y_2}{A_1 - A_2} \] \[ \bar{y} = \frac{(80000 \times 200) - (15708 \times 357.56)}{80000 - 15708} \] \[ \bar{y} = \frac{16,000,000 - 5,616,552}{64,292} \] \[ \bar{y} = \frac{10,383,448}{64,292} = 161.50 \text{ mm} \]
Due to symmetry about Y-axis, \( \bar{x} = 100 \text{ mm} \).
2. Y-Coordinate (\( \bar{y} \)):
Component 1 (Rectangle 200x400):
\( A_1 = 200 \times 400 = 80,000 \text{ mm}^2 \)
\( y_1 = 200 \text{ mm} \)
Component 2 (Semicircle Cut-out, r=100):
\( A_2 = \frac{\pi(100)^2}{2} = 15,708 \text{ mm}^2 \) (Negative Area)
Centroid of semi-circle from its base (top edge of rectangle) = \( \frac{4r}{3\pi} = 42.44 \text{ mm} \)
\( y_2 = 400 - 42.44 = 357.56 \text{ mm} \)
3. Combined Centroid:
\[ \bar{y} = \frac{A_1y_1 - A_2y_2}{A_1 - A_2} \] \[ \bar{y} = \frac{(80000 \times 200) - (15708 \times 357.56)}{80000 - 15708} \] \[ \bar{y} = \frac{16,000,000 - 5,616,552}{64,292} \] \[ \bar{y} = \frac{10,383,448}{64,292} = 161.50 \text{ mm} \]
Final Answer: Centroid \( (\bar{x}, \bar{y}) = (100 \text{ mm}, 161.50 \text{ mm}) \)
(c) Find the centroid of an inverted T-section from the bottom...
Calculations:
1. X-Coordinate (\( \bar{x} \)):
Symmetrical about Y-axis, so \( \bar{x} = 30 \text{ cm} \).
2. Y-Coordinate (\( \bar{y} \)):
Flange (Bottom, 60x10):
\( A_1 = 600 \text{ cm}^2 \), \( y_1 = 5 \text{ cm} \)
Web (Top, 10x60):
\( A_2 = 600 \text{ cm}^2 \), \( y_2 = 10 + 30 = 40 \text{ cm} \)
\[ \bar{y} = \frac{A_1y_1 + A_2y_2}{A_1 + A_2} = \frac{600(5) + 600(40)}{1200} \] \[ \bar{y} = \frac{3000 + 24000}{1200} = 22.5 \text{ cm} \]
Symmetrical about Y-axis, so \( \bar{x} = 30 \text{ cm} \).
2. Y-Coordinate (\( \bar{y} \)):
Flange (Bottom, 60x10):
\( A_1 = 600 \text{ cm}^2 \), \( y_1 = 5 \text{ cm} \)
Web (Top, 10x60):
\( A_2 = 600 \text{ cm}^2 \), \( y_2 = 10 + 30 = 40 \text{ cm} \)
\[ \bar{y} = \frac{A_1y_1 + A_2y_2}{A_1 + A_2} = \frac{600(5) + 600(40)}{1200} \] \[ \bar{y} = \frac{3000 + 24000}{1200} = 22.5 \text{ cm} \]
Final Answer: \( \bar{y} = 22.5 \text{ cm} \) from bottom
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