Terracotta Jali Brick

 

6

438753-001

Terracotta Jali Brick

1. Ajit Vilas Karande 2. Ganesh Dhareshwar Lakade 3. Chetan Pitambar Pise 4. Amol Ashok Kamble

28/11/2024

221912

 

Hexagonal Brick

 

1

438588-001

Hexagonal Brick

1. Yashwant Prabhakar Pawar 2. Satyawan Dagadu Jagdale 3. Chetan Pitambar Pise 4. Chandrakant Mahadeo Deshmukh 5. Sachin Arun Ghadge 6. Ganesh Dhareshwar Lakade 7. Shriganesh Shantikumar Kadam 8. Jaywant Prabhakar Pawar 9. Amol Ashok Kamble

27/11/2024

221790

Problems on Engineering Mechanics

Problem: Newton’s Second Law

Problem Statement:
A 5 kg block is pulled by a 20 N force on a frictionless surface. Find its acceleration.

Solution:
Using Newton’s Second Law:

$$F = ma$$ $$20 = 5a$$ $$a = 4 \text{ m/s}^2$$

Answer: Acceleration = 4 m/s²

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Problem: Equilibrium of Forces

Problem Statement:
A 10 kg block is suspended by two ropes making 30° and 45° angles with the ceiling. Find the tension in each rope.

Solution:
Let $T_1$ and $T_2$ be the tensions in the two ropes.
Using equilibrium equations:

$$T_1 \cos 30^\circ + T_2 \cos 45^\circ = 98 \text{ N} (Weight of the block)$$ $$T_1 \sin 30^\circ = T_2 \sin 45^\circ$$

Solving these two equations gives:

$$T_1 = 113.14 \text{ N}, \quad T_2 = 80.21 \text{ N}$$

Answer: $T_1 = 113.14$ N, $T_2 = 80.21$ N

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Problem: Friction

Problem Statement:
A 15 kg box is placed on a rough surface with a coefficient of friction 0.4. Find the force required to just move the box.

Solution:
Friction force is given by:

$$F_f = \mu N$$ $$F_f = (0.4)(15 \times 9.81)$$ $$F_f = 58.86 \text{ N}$$

Answer: 58.86 N

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Problem: Centroid Calculation

Problem Statement:
Find the centroid of a right triangle with base 6 m and height 4 m, measured from the base.

Solution:
For a right triangle, the centroid is given by:

$$x_c = \frac{b}{3}, \quad y_c = \frac{h}{3}$$ $$x_c = \frac{6}{3} = 2 \text{ m}, \quad y_c = \frac{4}{3} = 1.33 \text{ m}$$

Answer: Centroid = (2 m, 1.33 m)

Problem: Pushing a Crate on a Rough Surface

Problem Statement:
A 20 kg wooden crate is pushed along a rough horizontal floor with a force of 100 N at an angle of 30° from the horizontal. If the coefficient of friction between the crate and the floor is 0.3, determine:

1. The normal force acting on the crate.
2. The acceleration of the crate.

Solution:

Step 1: Compute Normal Force

The weight of the crate:

$$W = mg = (20)(9.81) = 196.2 \text{ N}$$

The vertical component of the applied force reduces the normal force:

$$N = W - 100 \sin 30^\circ$$ $$N = 196.2 - 50 = 146.2 \text{ N}$$

Step 2: Compute Friction Force

$$F_f = \mu N = (0.3)(146.2) = 43.86 \text{ N}$$

Step 3: Compute Acceleration

The net horizontal force:

$$F_{\text{net}} = 100 \cos 30^\circ - 43.86$$ $$= 86.6 - 43.86 = 42.74 \text{ N}$$

Applying Newton’s Second Law:

$$F = ma$$ $$42.74 = (20)a$$ $$a = 2.14 \text{ m/s}^2$$

Answer:

1. Normal force = 146.2 N
2. Acceleration = 2.14 m/s²

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Problem: Beam with an Uneven Load

Problem Statement:
A 4-meter-long beam is simply supported at both ends. A 500 N force is applied 1 m from the left support, and a 200 N force is applied 3 m from the left support. Find the reaction forces at both supports.

Solution:
Using the sum of moments about the left support:

$$\sum M_A = 0$$

Let $R_B$ be the reaction at the right support:

$$(500)(1) + (200)(3) = R_B (4)$$ $$500 + 600 = 4 R_B$$ $$R_B = 275 \text{ N}$$

Using the sum of vertical forces:

$$R_A + R_B = 500 + 200$$ $$R_A + 275 = 700$$ $$R_A = 425 \text{ N}$$

Answer:
Reaction at left support $R_A = 425 N$
Reaction at right support $R_B = 275 N$

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Problem: Inclined Plane with a Pulley System

Problem Statement:
A 10 kg block is placed on a 30° incline and connected to a 5 kg hanging mass via a frictionless pulley. If the coefficient of friction between the block and the incline is 0.2, determine whether the system moves and find its acceleration.

Solution:

Step 1: Forces Acting on the 10 kg Block

• Gravity component along incline: $$F_{\text{gravity}} = mg \sin 30^\circ = (10)(9.81) \sin 30^\circ = 49.05 \text{ N}$$
• Normal force: $$N = mg \cos 30^\circ = (10)(9.81) \cos 30^\circ = 84.95 \text{ N}$$
• Friction force: $$F_f = \mu N = (0.2)(84.95) = 16.99 \text{ N}$$

Step 2: Forces Acting on the 5 kg Hanging Mass

$$F = mg = (5)(9.81) = 49.05 \text{ N}$$

Step 3: Determine Motion

The net force pulling the system:

$$F_{\text{net}} = 49.05 - (49.05 + 16.99)$$ $$F_{\text{net}} = -16.99 \text{ N}$$

Since the net force is negative, the block does not move; friction is preventing motion.

Answer: The system does not move.