Weight component perpendicular to a slope of angle \alpha?

mg\cos\alpha, balanced by R if there is no acceleration across the slope.

What is the resultant of a set of forces?

Their vector sum: the single force with the same effect.

Condition for equilibrium of a particle?

Resultant force is zero: \sum Fx=0 and \sum Fy=0.

Three forces \mathbf{P},\mathbf{Q},\mathbf{R} are in equilibrium. Find \mathbf{R}.

\mathbf{R}=-(\mathbf{P}+\mathbf{Q}).

State Newton's second law.

\mathbf{F}=m\mathbf{a}: the resultant force equals mass times acceleration.

Acceleration of a block on a smooth slope at angle \alpha?

g\sin\alpha down the slope, independent of mass.

What does 'smooth' mean in a mechanics model?

There is no friction, so the contact force is only the normal reaction.

What does 'light, inextensible string' mean?

Negligible mass and constant length, so the tension is the same throughout.

Vector equation for position under constant acceleration?

\mathbf{r}=\mathbf{u}t+\frac12\mathbf{a}t^2.

How do you give the direction of a resultant X\mathbf{i}+Y\mathbf{j}?

\tan\theta=\left|\frac{Y}{X}\right|, then state the quadrant, e.g. below the \mathbf{i} direction.

Resolving forces and resultantsAQA A-Level Maths: Flashcards

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Resolve a force $F$ at angle $\theta$ to a direction.

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Resolve a force $F$ at angle $\theta$ to a direction.: $F\cos\theta$ along that direction and $F\sin\theta$ perpendicular to it.
Weight component down a slope of angle $\alpha$ ?: $mg\sin\alpha$ .
Weight component perpendicular to a slope of angle $\alpha$ ?: $mg\cos\alpha$ , balanced by $R$ if there is no acceleration across the slope.
What is the resultant of a set of forces?: Their vector sum: the single force with the same effect.
Magnitude of the vector $X\mathbf{i}+Y\mathbf{j}$ ?: $\sqrt{X^2+Y^2}$ .
Condition for equilibrium of a particle?: Resultant force is zero: $\sum F_x=0$ and $\sum F_y=0$ .
Three forces $\mathbf{P},\mathbf{Q},\mathbf{R}$ are in equilibrium. Find $\mathbf{R}$ .: $\mathbf{R}=-(\mathbf{P}+\mathbf{Q})$ .
State Newton's second law.: $\mathbf{F}=m\mathbf{a}$ : the resultant force equals mass times acceleration.
Acceleration of a block on a smooth slope at angle $\alpha$ ?: $g\sin\alpha$ down the slope, independent of mass.
What does 'smooth' mean in a mechanics model?: There is no friction, so the contact force is only the normal reaction.
What does 'light, inextensible string' mean?: Negligible mass and constant length, so the tension is the same throughout.
Vector equation for position under constant acceleration?: $\mathbf{r}=\mathbf{u}t+\frac12\mathbf{a}t^2$ .
How do you give the direction of a resultant $X\mathbf{i}+Y\mathbf{j}$ ?: $\tan\theta=\left|\frac{Y}{X}\right|$ , then state the quadrant, e.g. below the $\mathbf{i}$ direction.

Exam questions on Resolving forces and resultants

Three coplanar forces $\mathbf{P}=(4\mathbf{i}-3\mathbf{j})$ N, $\mathbf{Q}=(-7\mathbf{i}+5\mathbf{j})$ N and $\mathbf{R}$ act on a particle, which is in equilibrium. The vectors $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors.
Find the angle that $\mathbf{R}$ makes with the direction of $\mathbf{i}$ , and state whether it is above or below that direction.2 marks
A block of mass $6$ kg is released from rest on a smooth plane inclined at $30^\circ$ to the horizontal. Model the block as a particle, with $g=9.8\ \text{m s}^{-2}$ .
Find the acceleration of the block down the plane.2 marks
Two horizontal forces $(6\mathbf{i}+4\mathbf{j})$ N and $(-2\mathbf{i}+8\mathbf{j})$ N act on a particle of mass $2$ kg on a smooth horizontal surface, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular horizontal unit vectors. These are the only horizontal forces. The particle starts from rest at the origin $O$ .
Find the acceleration of the particle as a vector, and its magnitude.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).