Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Stactics
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Stactics
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Subject: Further Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 4
Theme: Mechanics
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A force is a push or pull that tends to change the state of rest or uniform motion of an object. It is a vector quantity, possessing both magnitude and direction. The standard unit of force is the Newton (N). Examples of forces in everyday Nigerian life include: The effort applied to push a wheelbarrow carrying agricultural produce in a market. The tension in the ropes supporting a water tank on a stand. The weight of a load of cement on a construction site. The resultant force is the single force that produces the same effect as all the individual forces acting on an object combined. It is the vector sum of all the forces.
Methods for finding the resultant force: Graphical Methods: Parallelogram Law: For two concurrent forces, the diagonal of the parallelogram formed by the two force vectors (originating from the same point) represents the resultant.
Triangle Law: If two forces are represented by two sides of a triangle taken in order, their resultant is represented by the third side taken in the opposite order.
Polygon Law: For multiple forces, if they are represented by the sides of a polygon taken in order, and the polygon closes, the resultant is zero (equilibrium). If the polygon does not close, the resultant is represented by the vector that closes the polygon, taken in the opposite order.
Analytical Method (Resolution of Forces): This involves resolving each force into its perpendicular components (usually horizontal, $F_x$, and vertical, $F_y$). The sum of all horizontal components gives the net horizontal force ($\Sigma F_x$). The sum of all vertical components gives the net vertical force ($\Sigma F_y$). The magnitude of the resultant force, $R$, is given by $R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2}$. The direction of the resultant force, $\theta$, is given by $\tan \theta = \frac{\Sigma F_y}{\Sigma F_x}$. Worked Example 2.2.1 (Resultant Force - Analytical Method): Two children are pulling a toy truck. Child A pulls with a force of 50 N at an angle of 30° to the horizontal. Child B pulls with a force of 70 N at an angle of 60° to the horizontal, on the same side. Find the magnitude and direction of the resultant force.
Solution: Resolve Child A's force ($F_A = 50$ N) into components: Horizontal component ($F_{Ax}$) = $F_A \cos 30^\circ = 50 \times 0.866 = 43.3$ N Vertical component ($F_{Ay}$) = $F_A \sin 30^\circ = 50 \times 0.5 = 25$ N Resolve Child B's force ($F_B = 70$ N) into components: Horizontal component ($F_{Bx}$) = $F_B \cos 60^\circ = 70 \times 0.5 = 35$ N Vertical component ($F_{By}$) = $F_B \sin 60^\circ = 70 \times 0.866 = 60.62$ N Sum the horizontal components ($\Sigma F_x$): $\Sigma F_x = F_{Ax} + F_{Bx} = 43.3 + 35 = 78.3$ N Sum the vertical components ($\Sigma F_y$): $\Sigma F_y = F_{Ay} + F_{By} = 25 + 60.62 = 85.62$ N Calculate the magnitude of the resultant force ($R$): $R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{(78.3)^2 + (85.62)^2}$ $R = \sqrt{6130.89 + 7330.7844} = \sqrt{13461.6744} \approx 116.03$ N Calculate the direction of the resultant force ($\theta$) with respect to the horizontal: $\tan \theta = \frac{\Sigma F_y}{\Sigma F_x} = \frac{85.62}{78.3} \approx 1.0935$ $\theta = \arctan(1.0935) \approx 47.57^\circ$ Therefore, the resultant force is approximately 116.03 N at an angle of 47.57° to the horizontal. A body is in equilibrium when it is either at rest or moving with a constant velocity. For a body to be in equilibrium under the action of concurrent forces (forces acting at a single point), two conditions must be met: The vector sum of all forces acting on the body must be zero (translational equilibrium). This means $\Sigma F_x = 0$ and $\Sigma F_y = 0$. The vector sum of all moments (torques) about any point must be zero (rotational equilibrium). This means $\Sigma M = 0$. The moment of a force (or torque) is the turning effect produced by a force about a pivot point (or axis of rotation).
Formula: Moment ($M$) = Force ($F$) $\times$ Perpendicular distance ($d$) from the pivot to the line of action of the force. $M = F \times d$.
Units: Newton-metre (Nm).
Direction: Moments are typically classified as clockwise (CW) or anticlockwise (ACW). By convention, one direction is taken as positive and the other as negative (e.g., ACW positive, CW negative).
Principle of Moments: For a body to be in rotational equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point. Worked Example 2.4.1 (Moment of a Force): A mechanic uses a spanner of length 30 cm (0.3 m) to loosen a stubborn bolt on a Keke Napep engine. He applies a force of 80 N perpendicular to the spanner at its end. Calculate the moment of the force about the centre of the bolt.
Solution: Identify the force ($F$) and perpendicular distance ($d$): Force, $F = 80$ N Perpendicular distance (length of spanner), $d = 30$ cm = 0.3 m Apply the moment formula: Moment ($M$) = $F \times d = 80 \text{ N} \times 0.3 \text{ m} = 24$ Nm. The moment of the force about the centre of the bolt is 24 Nm. This is the turning effect that helps loosen the bolt. Worked Example 2.4.2 (Moments of Multiple Forces): A uniform plank of length 4 m and weight 200 N is supported horizontally by two supports at its ends. A load of 500 N is placed at 1.5 m from one end (say, end A). Calculate the reactions at the two supports. (Assume the plank's weight acts at its centre).
Solution: Draw a free-body diagram: Plank length = 4 m. Weight of plank = 200 N, acting at 2 m from A (centre). Load = 500 N, acting at 1.5 m from A. Let the reactions at supports A and B be $R_A$ and $R_B$ respectively. Apply the condition for translational equilibrium (sum of vertical forces = 0): Upward forces = Downward forces $R_A + R_B = 200 \text{ N} + 500 \text{ N}$ $R_A + R_B = 700 \text{ N}$ (Equation 1) Apply the principle of moments (sum of clockwise moments = sum of anticlockwise moments) about a convenient pivot point. Let's take moments about A to eliminate $R_A$.
Clockwise moments (CW): Moment due to plank's weight: $200 \text{ N} \times 2 \text{ m} = 400 \text{ Nm}$ Moment due to load: $500 \text{ N} \times 1.5 \text{ m} = 750 \text{ Nm}$ Total CW moment = $400 + 750 = 1150 \text{ Nm}$ Anticlockwise moments (ACW): Moment due to reaction $R_B$: $R_B \times 4 \text{ m} = 4R_B$ Equating CW and ACW moments: $4R_B = 1150$ $R_B = \frac{1150}{4} = 287.5$ N Substitute $R_B$ into Equation 1 to find $R_A$: $R_A + 287.5 = 700$ $R_A = 700 - 287.5 = 412.5$ N Therefore, the reactions at the supports are $R_A = 412.5$ N and $R_B = 287.5$ N.
Understanding Statics is vital for numerous applications relevant to Nigeria's development and daily life: Civil Engineering and Construction: Building and Bridge Design: Civil engineers use principles of Statics to ensure the stability and structural integrity of buildings, bridges (e.g., those spanning rivers like the Niger or Benue), and other infrastructure. They calculate the forces acting on beams, columns, and foundations to prevent collapse, especially important in regions prone to heavy loads or environmental stresses.
Scaffolding and Temporary Structures: Designing safe scaffolding for construction workers or stable temporary market stalls requires careful consideration of forces and moments to prevent accidents.
Automotive and Mechanical Engineering: Vehicle Stability: The stability of vehicles like trailers, buses, and Keke Napeps on uneven roads or during turning relies on static analysis to understand the distribution of weight and forces. Mechanics use principles of moments when using spanners to tighten bolts on engines or suspension systems.
Machine Design: Designing stable and efficient machinery, from agricultural equipment like tractors and ploughs to industrial processing units, involves ensuring all components are in equilibrium under operating loads. Everyday Applications and Traditional Practices: Load Carrying and Balancing: Individuals carrying heavy loads on their heads (common in Nigerian markets) intuitively apply principles of balancing and understanding the center of gravity to maintain equilibrium.
Leverage in Tools: Simple tools like crowbars for removing nails or wheelbarrows for transporting goods utilize the principle of moments to multiply force and make work easier. Understanding this helps in designing and using such tools effectively.
Designing Furniture: Architects and carpenters use static principles to design stable tables, chairs, and beds that can withstand typical loads without wobbling or breaking.