Define the following terms: Resultant force Elastic limit Strain Hydrostatic pressure Second moment of area

here are the solutions for Question 1, Question 2, and Question 3.

SECTION A: QUESTION 1: GENERAL

1.1 Resultant force

The resultant force is the single force that produces the same effect on a body as all the individual forces acting on it. It is the vector sum of all forces.

1.2 Elastic limit

The elastic limit is the maximum stress that a material can withstand without undergoing permanent deformation. If the stress exceeds this limit, the material will not return to its original shape once the stress is removed.

1.3 Strain

Strain is a measure of the deformation of a material, defined as the ratio of the change in dimension (e.g., length, volume) to the original dimension. It is a dimensionless quantity.

1.4 Hydrostatic pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth and is calculated as $P = \rho gh$, where $\rho$ is the fluid density, $g$ is the acceleration due to gravity, and $h$ is the depth.

1.5 Second moment of area

The second moment of area, also known as the area moment of inertia, is a geometrical property of a cross-section that describes how its area is distributed with respect to an axis. It is crucial in beam bending calculations to determine a beam's resistance to bending and deflection.

SECTION B: QUESTION 2: KINEMATICS

2.1 A delivery truck travels at a velocity of $90 \text{ km/h}$ on a bearing of $045^\circ$ true. It has to deliver goods to a warehouse that is $120 \text{ km}$ due East of its current position. A cross wind of $20 \text{ km/h}$ from the South acts on the truck.

2.1.1 Determine the true course the driver must steer to reach the warehouse.

Let $V_{TA}$ be the velocity of the truck relative to the air (what the driver steers), with magnitude $90 \text{ km/h}$ and unknown bearing $\theta$. Let $V_W$ be the velocity of the wind. Since it's from the South, it blows North, so $V_W = (0, 20) \text{ km/h}$. Let $V_G$ be the resultant velocity of the truck relative to the ground. Since the destination is due East, $V_G$ must be purely in the East direction, so $V_G = (V_G, 0)$.

The vector equation is $V_G = V_{TA} + V_W$. In components, using standard Cartesian coordinates (East as positive x, North as positive y): $V_{TA,x} = 90 \sin\theta$ (East component, using bearing from North) $V_{TA,y} = 90 \cos\theta$ (North component, using bearing from North)

Step 1: Set up the vector equation in components. $(V_G, 0) = (90 \sin\theta, 90 \cos\theta) + (0, 20)$ $(V_G, 0) = (90 \sin\theta, 90 \cos\theta + 20)$

Step 2: Equate the y-components to find the steering angle $\theta$. $0 = 90 \cos\theta + 20$ $90 \cos\theta = -20$ $\cos\theta = -\frac{20}{90} = -\frac{2}{9}$ $\theta = \arccos\left(-\frac{2}{9}\right)$ \theta \approx \text{102.84^\circ} The driver must steer on a true course (bearing) of $102.84^\circ$.

2.1.2 Calculate how long the journey will take.

Step 1: Calculate the magnitude of the resultant velocity $V_G$ (the effective speed towards East). From the x-components: $V_G = 90 \sin\theta$ Substitute $\theta \approx 102.84^\circ$: $V_G = 90 \sin(102.84^\circ) \text{ km/h}$ $V_G \approx 90 \times 0.975 \text{ km/h}$ $V_G \approx 87.75 \text{ km/h}$

Step 2: Calculate the time taken for the journey. The distance to the warehouse is $D = 120 \text{ km}$ (East). $t = \frac{D}{V_G}$ $t = \frac{120 km}{87.75 km/h}$ $t \approx 1.3676 \text{ hours}$ $t \approx 1 hour and (0.3676 \times 60) \text{ minutes}$ $t \approx \text{1 hour and 22.06 minutes}$

2.2 A steel ball is projected from the top of a building $48 \text{ m}$ high with an initial velocity of $18 \text{ m/s}$ at an angle of $25^\circ$ above the horizontal.

Given: Initial velocity $u = 18 \text{ m/s}$ Angle of projection $\alpha = 25^\circ$ Height of building $H = 48 \text{ m}$ Acceleration due to gravity $g = 9.81 m/s^2$ (downwards)

2.2.1 The maximum height above the ground reached by the ball.

Step 1: Resolve initial velocity into vertical component. $u_y = u \sin\alpha = 18 \sin(25^\circ) \text{ m/s}$ $u_y \approx 18 \times 0.4226 \approx 7.607 \text{ m/s}$

Step 2: Calculate the maximum height reached above the launch point ($h_{max}$). At maximum height, the vertical velocity $v_y = 0$. Using the kinematic equation $v_y^2 = u_y^2 + 2gh_{max}$: $0^2 = (7.607 m/s)^2 + 2(-9.81 m/s^2)h_{max}$ $0 = 57.866 - 19.62 h_{max}$ $19.62 h_{max} = 57.866$ $h_{max} = \frac{57.866}{19.62} \approx 2.949 \text{ m}$

Step 3: Calculate the maximum height above the ground. $\text{Total maximum height} = H + h_{max} = 48 m + 2.949 m = \text{50.95 m}$

2.2.2 The horizontal distance from the foot of the building where the ball hits the ground.

2.2.3 The time taken for the ball to hit the ground.

Step 1: Calculate the time taken for the ball to hit the ground. The total vertical displacement from the launch point to the ground is $\Delta y = -48 \text{ m}$. Using the kinematic equation $\Delta y = u_y t + \frac{1}{2}gt^2$: $-48 = (7.607)t + \frac{1}{2}(-9.81)t^2$ $-48 = 7.607t - 4.905t^2$ Rearrange into a quadratic equation: $4.905t^2 - 7.607t - 48 = 0$ Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-(-7.607) \pm \sqrt{(-7.607)^2 - 4(4.905)(-48)}}{2(4.905)}$ $t = \frac{7.607 \pm \sqrt{57.866 + 941.76}}{9.81}$ $t = \frac{7.607 \pm \sqrt{999.626}}{9.81}$ $t = \frac{7.607 \pm 31.617}{9.81}$ Since time cannot be negative, we take the positive root: $t = \frac{7.607 + 31.617}{9.81} = \frac{39.224}{9.81} \approx \text{3.998 s}$

Step 2: Calculate the horizontal distance. The horizontal velocity $u_x$ is constant. $u_x = u \cos\alpha = 18 \cos(25^\circ) \text{ m/s}$ $u_x \approx 18 \times 0.9063 \approx 16.313 \text{ m/s}$ Horizontal distance $x = u_x t$: $x = 16.313 m/s \times 3.998 \text{ s}$ $x \approx \text{65.22 m}$

2.3 A train accelerates uniformly from rest to a speed of $72 \text{ km/h}$ in $45 \text{ seconds}$. It then travels at this constant speed for $3 \text{ minutes}$ before applying the brakes and coming to rest uniformly in $60 \text{ seconds}$.

2.3.1 Calculate the acceleration of the train.

This refers to the first phase of motion. Initial velocity $u_1 = 0 \text{ m/s}$ (from rest) Final velocity $v_1 = 72 \text{ km/h}$ Convert $v_1$ to m/s: $v_1 = 72 km/h \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 20 \text{ m/s}$ Time $t_1 = 45 \text{ s}$

Step 1: Calculate the acceleration $a_1$. $a_1 = \frac{v_1 - u_1}{t_1}$ $a_1 = \frac{20 m/s - 0 m/s}{45 s}$ a_1 = \frac{20}{45} m/s^2 = \frac{4}{9} m/s^2 \approx \text{0.444 m/s^2}

2.3.2 Determine the total distance travelled by the train from the start until it comes to rest.

This involves three phases of motion:

Phase 1: Acceleration Initial velocity $u_1 = 0 \text{ m/s}$ Final velocity $v_1 = 20 \text{ m/s}$ Time $t_1 = 45 \text{ s}$ Step 1: Calculate distance $s_1$. $s_1 = \left(\frac{u_1 + v_1}{2}\right)t_1$ $s_1 = \left(\frac{0 + 20}{2}\right) \times 45 = 10 \times 45 = 450 \text{ m}$

Phase 2: Constant speed Speed $v_2 = 20 \text{ m/s}$ Time $t_2 = 3 minutes = 3 \times 60 s = 180 \text{ s}$ Step 2: Calculate distance $s_2$. $s_2 = v_2 t_2$ $s_2 = 20 m/s \times 180 s = 3600 \text{ m}$

Phase 3: Deceleration Initial velocity $u_3 = 20 \text{ m/s}$ Final velocity $v_3 = 0 \text{ m/s}$ (comes to rest) Time $t_3 = 60 \text{ s}$ Step 3: Calculate distance $s_3$. $s_3 = \left(\frac{u_3 + v_3}{2}\right)t_3$ $s_3 = \left(\frac{20 + 0}{2}\right) \times 60 = 10 \times 60 = 600 \text{ m}$

Step 4: Calculate the total distance travelled. $S_{total} = s_1 + s_2 + s_3$ $S_{total} = 450 m + 3600 m + 600 m = \text{4650 m}$

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