Grade 13 — Term 2
10 study notes · 70 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 10 lessons (NIE Units 2, 3, 4, 5). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Systems of coplanar forces
What you must know
- Couple definition and properties: A couple consists of two equal, opposite, and parallel forces whose lines of action do not coincide; moment of a couple is G = F * d (where d is perpendicular arm between forces), INDEPENDENT of the choice of reference point.
- Equilibrium conditions for Coplanar Forces: A system of coplanar forces is in equilibrium if and only if: 1) Rx = ∑ X = 0, 2) Ry = ∑ Y = 0, and 3) ∑ Mo = 0 about ANY point.
- Three non-parallel coplanar forces in equilibrium: MUST be Concurrent (their lines of action must meet at a single point) and coplanar.
Try one question
Three coplanar forces of magnitudes 6 N, 8 N, and 10 N act at a point in equilibrium. What is the angle between the 6 N and 8 N forces?
- A. 60°
- B. 180°
- C. 120°
- D. 90°
Show the answer
Answer: D. 90°. For three forces in equilibrium, any one force must be equal in magnitude and opposite in direction to the resultant of the other two: R² = F₁² + F₂² + 2F₁F₂ cos θ. Here, 10² = 6² + 8² + 2(6)(8) cos θ => 100 = 36 + 64 + 96 cos θ => 100 = 100 + 96 cos θ => 96 cos θ = 0 => cos θ = 0 => θ = 90°.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Systems of coplanar forces (statics)
What you must know
- Equation of line of action of resultant: The line of action of the single resultant force R has Cartesian equation: x*Ry - y*Rx = G (taking origin as reference point where G is sum of moments about origin).
- Triangle of Forces: If three coplanar forces acting on a particle are in equilibrium, they can be represented in magnitude and direction by the sides of a closed triangle taken in order.
- Polygon of Forces: If n coplanar forces acting on a particle are in equilibrium, the vector polygon formed by drawing the forces head-to-tail must be completely closed.
Try one question
According to Lami's Theorem for three coplanar concurrent forces P, Q, and R in static equilibrium, what is the relation connecting the forces to the angles α, β, and γ between the other two forces?
- A. P / cos α = Q / cos β = R / cos γ
- B. P / sin α = Q / sin β = R / sin γ
- C. P sin α + Q sin β + R sin γ = 0
- D. P · α = Q · β = R · γ
Show the answer
Answer: B. P / sin α = Q / sin β = R / sin γ. Lami's theorem states that if three coplanar forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces: P / sin α = Q / sin β = R / sin γ.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Stresses in the rods of a framework
What you must know
- Zero-force member recognition rules: 1) If only two non-collinear rods meet at an unloaded joint, stress in BOTH rods is ZERO; 2) If three rods meet at an unloaded joint where two are collinear, stress in the third rod is ZERO.
- Method of Sections (Ritter's Method): Cut the framework through at most three rods whose stresses are unknown; treat either isolated sub-structure as a rigid body in equilibrium, resolving and taking moments about rod intersection points.
- Support reactions determination: Always calculate external support reactions first by treating the ENTIRE truss as a single rigid body before analyzing internal joints.
Try one question
In the analysis of a pin-jointed light framework in static equilibrium, how is a member designated when it exerts an outward pushing force against the joint pins at its ends?
- A. A neutral cantilever
- B. A tie under tension
- C. A shear pin
- D. A strut under compression (thrust)
Show the answer
Answer: D. A strut under compression (thrust). A framework member under compression is squeezed by external joint forces and consequently pushes outward on the joint pins to maintain equilibrium; such a member is called a strut (bearing thrust). A member pulled by joint pins is a tie (bearing tension).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Centre of mass of uniform bodies
What you must know
- Mass vs Area/Length proportionality: For uniform wires of density λ, mass m = λ * length; for uniform laminas of surface density σ, mass m = σ * area; densities cancel in xG and yG ratios.
- Triangular Lamina vs Three-Rod Framework: Triangular LAMINA has C.M. at centroid (h/3 from base); Triangular perimeter FRAMEWORK made of 3 rods has C.M. at INCENTER of medial triangle formed by rod midpoints.
- Mass Moment principle: The algebraic sum of mass moments about any plane passing through the Center of Mass is identically ZERO.
Try one question
Where is the center of mass of a uniform thin semicircular wire of radius r located along its axis of symmetry from the center of the bounding diameter?
- A. r / 2
- B. 4r / (3π)
- C. 2r / π
- D. 3r / 8
Show the answer
Answer: C. 2r / π. By integration over the wire element dm = λ r dθ from θ = 0 to π: y_cm = (∫ r sin θ · λ r dθ) / (λ π r) = r² [-cos θ]₀^π / (π r) = r (1 - (-1)) / π = 2r / π. (Note: 4r/(3π) is for a uniform semicircular lamina).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Centre of mass by integration
What you must know
- Solid Hemisphere derivation setup: Slice solid hemisphere into thin circular discs of thickness dx at distance x from center; radius of disc is y = √(a^2 - x^2); volume element dV = π y^2 dx = π(a^2 - x^2)dx; xG = ∫_0^a x π(a^2 - x^2)dx / (2/3 π a^3) = (π a^4 / 4) / (2/3 π a^3) = 3a / 8.
- Hollow vs Solid C.M. rule: For corresponding geometrical shapes, the C.M. of a hollow shell is always further from the apex/center (closer to mass concentration) than the solid counterpart (e.g., solid cone h/4 vs hollow cone h/3).
- Angle in radians requirement: In formulas a(sin α)/α and (2/3)a(sin α)/α, the semi-vertical angle α MUST be expressed in RADIANS.
Try one question
Using integration, at what distance from the flat base center is the center of mass of a uniform solid hemisphere of radius r located along its axis of symmetry?
- A. 3r / 8
- B. 3r / 4
- C. 4r / (3π)
- D. r / 2
Show the answer
Answer: A. 3r / 8. Slicing the hemisphere into thin disks of thickness dy at distance y from the base: disk volume dV = π (r² - y²) dy. Total volume V = (2/3) π r³. The center of mass is y_cm = [∫₀^r y π (r² - y²) dy] / [(2/3) π r³] = [r⁴/2 - r⁴/4] / [(2/3) r³] = (r⁴/4) / (2r³/3) = 3r / 8.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Centre of mass of composite bodies
What you must know
- Negative mass technique for cutouts: Treat removed holes or cavity sections as objects with NEGATIVE mass m = -m2; sum normally using xG = (∑ m_i * x_i) / (∑ m_i).
- Proportional mass substitution: If thickness and density are uniform, substitute Area directly for 2D laminas and Volume for 3D solids; relative ratios can be simplified by canceling common factors.
- Hemi-sphere with cylinder or cone: For a toy consisting of a hemisphere of radius r joined to a cone of base radius r and height h, C.M. lies on common axis of symmetry; distance from joint interface is determined by balancing moments.
Try one question
From a uniform circular lamina of radius 2a and center O, a circular hole of radius a touching the outer boundary is removed. How far from O is the center of mass of the remaining lamina?
- A. a / 2 in the direction opposite to the hole
- B. a / 3 in the direction opposite to the hole
- C. a / 4 in the direction of the hole
- D. 2a / 3 in the direction opposite to the hole
Show the answer
Answer: B. a / 3 in the direction opposite to the hole. Let O be the origin (0, 0). Mass is proportional to area: Original disk M₁ ∝ π(2a)² = 4πa² at (0, 0). Removed hole M₂ ∝ πa² with center at (a, 0). Remaining mass M_rem = 4πa² - πa² = 3πa². Center of mass x_cm: M_rem · x_cm = M₁ · 0 - M₂ · a => 3πa² · x_cm = -πa³ => x_cm = -a / 3.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Centre of gravity
What you must know
- Rigid body weight representation: In equilibrium problems, the distributed gravitational forces on all particles are replaced by a SINGLE downward force W = Mg acting vertically through G.
- Support point alignment: A body supported at a single point or pivot can only remain in equilibrium if the vertical line through the support point passes directly through G.
- Work done against gravity: Work done in tilting, raising, or turning a body of mass M equals M * g * ΔhG, where ΔhG is the vertical change in height of its Center of Gravity.
Try one question
Under what physical condition does the center of gravity (CG) of an extended rigid body coincide exactly with its center of mass (CM)?
- A. When the body moves with relativistic speed
- B. In a spatially uniform gravitational field
- C. Only when the body is in outer space with zero gravity
- D. Only when the body is spherical and homogeneous
Show the answer
Answer: B. In a spatially uniform gravitational field. The center of mass is the average location of mass distribution (purely geometric/inertial). The center of gravity is the point through which the resultant gravitational force acts. When the gravitational field acceleration g is spatially uniform across the entire body, g factors out of the moment equation, making CG identical to CM.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Stability of bodies in equilibrium
What you must know
- Rolling body stability formula: A body with convex surface of radius of curvature r resting on top of a fixed convex surface of radius R is in STABLE equilibrium if 1/h > 1/r + 1/R (where h is height of C.G. above point of contact); UNSTABLE if 1/h < 1/r + 1/R.
- Toppling vs Sliding on an inclined plane: A body on a plane inclined at angle θ with friction coefficient μ will SLIDE before toppling if tan θ > μ; it will TOPPLE before sliding if the vertical line through C.G. passes outside base before tan θ reaches μ.
- Potential Energy calculus test: If U(θ) is potential energy as function of displacement angle θ: Equilibrium occurs where dU/dθ = 0; Stable if d^2U/dθ^2 > 0 (minimum U); Unstable if d^2U/dθ^2 < 0 (maximum U).
Try one question
A rigid body resting in equilibrium on a horizontal surface is in stable equilibrium if, upon a slight displacement from its equilibrium position:
- A. Its center of gravity rises and the resulting forces generate a restoring moment returning it to equilibrium
- B. Its mass increases instantaneously
- C. Its center of gravity lowers and the resulting moment tips it further away
- D. Its center of gravity remains at the exact same height without any moment
Show the answer
Answer: A. Its center of gravity rises and the resulting forces generate a restoring moment returning it to equilibrium. In stable equilibrium, potential energy is at a local minimum. When slightly disturbed, the center of gravity rises (increasing PE), and the lines of action of the weight and reaction form a couple that creates a restoring moment, bringing the body back to its original equilibrium position.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Angle of inclination of suspended bodies
What you must know
- Cotangent theorem application: In triangle ABC where G divides base BC in ratio m : n and line AG makes angle θ with base: (m + n) cot θ = m cot B - n cot C (or n cot α - m cot β); extremely useful for calculating inclination angles of suspended lamina.
- Loaded suspended lamina: If additional masses are attached to vertices of a suspended lamina, first compute the NEW combined Center of Gravity G', then use the principle that suspension point O and new C.G. G' lie on the vertical line.
- Equilateral triangular lamina suspended from a vertex: Line joining vertex to C.G. is the median (height); hangs symmetrically with base horizontal and altitude vertical.
Try one question
When a rigid body is freely suspended from a fixed pivot point P, what geometric condition must be satisfied by its center of gravity G in static equilibrium?
- A. The line PG must be inclined at exactly 45° to the horizontal
- B. The center of gravity G must lie vertically directly below the suspension point P
- C. The center of gravity G must lie on the horizontal plane passing through P
- D. The point P must coincide with the geometric centroid of the top surface
Show the answer
Answer: B. The center of gravity G must lie vertically directly below the suspension point P. For equilibrium under two forces (the upward reaction at the suspension point P and the downward weight acting through center of gravity G), the two forces must be collinear. Hence, the line segment PG must be vertical, placing G directly below P.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Newtonian model of motion
What you must know
- Connected particles over a pulley: Write individual equations of motion F = ma along the direction of acceleration for each body; tension T is internal to the combined system but external to each isolated body.
- Vehicle on inclined road: Equation of motion along incline: F - R - mg sin α = m * a, where F = P/v is engine driving force, R is resistive force, and α is inclination angle.
- Maximum speed condition: Maximum velocity is attained when acceleration a = 0, so engine driving force equals sum of resistive forces: F_max = R + mg sin α => P = (R + mg sin α) * v_max.
Try one question
A particle moves along a straight line such that its velocity v at position x is given by v² = 16 - 4x. What is the acceleration a of the particle?
- A. -2 m s⁻²
- B. -8 m s⁻²
- C. -4 m s⁻²
- D. +4 m s⁻²
Show the answer
Answer: A. -2 m s⁻². Using the kinematic derivative relation for acceleration as a function of position: a = v (dv/dx) = ½ d(v²)/dx. Differentiating v² = 16 - 4x with respect to x gives d(v²)/dx = -4. Therefore, acceleration a = ½ (-4) = -2 m s⁻².
A model question written by Idasara on the exam pattern. It is not from a past paper.
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