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.