Flashcards · 10 cards

Statics: forces, equilibrium & moments

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Rule
The resultant of forces $P$ and $Q$ with an angle $\theta$ between them?
Answer
$R^2 = P^2 + Q^2 + 2PQ\cos\theta$ : the parallelogram of forces.
The parallelogram of forcesR² = P² + Q² + 2PQ cos θ
From the lesson: Resultants of forces
Rule
How do you find the resultant of several forces?
Answer
Split each into east and north parts ( $F\sin\theta$ and $F\cos\theta$ on a bearing $\theta$ ), add each set, then combine them with Pythagoras.
Resolving a force on a bearing(F, θ) = F sin θ east + F cos θ north
From the lesson: Resultants of forces
Rule
A weight $W$ hangs from two strings. The equations for equilibrium?
Answer
Resolve both ways. Across: $T_1\cos\alpha = T_2\cos\beta$ . Up: $T_1\sin\alpha + T_2\sin\beta = W$ .
A weight on two stringsAcross: T₁ cos α = T₂ cos β. Up: T₁ sin α + T₂ sin β = W
From the lesson: Equilibrium of a particle
Rule
Lami's theorem?
Answer
For three forces in equilibrium, each force divided by the sine of the angle between the other two is the same: $\dfrac{P}{\sin\alpha} = \dfrac{Q}{\sin\beta} = \dfrac{W}{\sin\gamma}$ .
Lami's theoremP ÷ sin α = Q ÷ sin β = W ÷ sin γ
From the lesson: Equilibrium of a particle
Rule
Three forces in equilibrium drawn as a triangle: how are the angles related?
Answer
They form a closed triangle. Each angle between two forces and the matching angle inside the triangle add up to $180^\circ$ .
Three forces in equilibriumInside the triangle 60°; between the forces 180° − 60° = 120°
From the lesson: Equilibrium of a particle
Rule
A block rests on a rough slope at angle $\theta$ . The reaction and the friction?
Answer
Normal reaction $R = mg\cos\theta$ ; limiting friction $F = \mu R$ , acting against the motion. Down the slope, the weight's part is $mg\sin\theta$ .
A block on a rough slopeR = mg cos θ; limiting friction F = μR, against the motion
From the lesson: Equilibrium of a particle
Rule
The moment of a force about a point, and the condition for balance?
Answer
Force × perpendicular distance from the point. For balance, clockwise moments = anticlockwise moments (and the forces balance too).
Balancing on a pivotF₁ × d₁ = F₂ × d₂
From the lesson: Moments and beams
Know it
Where does the weight of a uniform beam act?
Answer
At its middle.
From the lesson: Moments and beams
Which method?
WAEC 2019 · Paper 2 · Q8
In the diagram, a mass of $12\text{ kg}$ hanging from a light inextensible string is pulled aside by a horizontal force $R$ , such that the string is inclined at $45^\circ$ to the vertical. If the system is in equilibrium, calculate the: $[\text{Take } g = 10\text{ m s}^{-2}]$
tension in the string;
value of $R$ .
How do you find the tension?
Answer
Resolve vertically: only the tension has an upward part, so $T\cos 45^\circ = 120$ . Then resolve horizontally to find $R$ .
From the lesson: Equilibrium of a particle Try the question
Which method?
WAEC 2019 · Paper 2 · Q6
A uniform beam $WX$ , of length $90\text{ cm}$ and weight $50\text{ N}$ , is suspended on a pivot $35\text{ cm}$ from $W$ . It is kept in equilibrium by means of forces $T$ and $20\text{ N}$ applied at $Y$ and $Z$ respectively. $|WY| = 10\text{ cm}$ and $|XZ| = 10\text{ cm}$ . Find the value of $T$ .
How do you start?
Answer
Take moments about the pivot. Measure each force's distance from the pivot (35 cm from $W$ ); the weight acts at the middle, 45 cm from $W$ .
From the lesson: Moments and beams Try the question