11AGeometry

Geometry and Mechanics

Forces can be added by using vector addition and geometry.

Triangle Rules For a triangle ABC with lengths a, b,c {where side a is opposite angle A etc.} the following rules apply:

math .\qquad c^2=a^2+b^2-2ab \cos C math
 * Cosine Rule **

math . \qquad \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} math
 * Sine Rule **

{A variation of the Sine Rule for 3 forces}
 * Lami's Theorem (Assumes Resultant Force = 0 -- Newton's First Law}**

math . \qquad \dfrac{F_A}{\sin A}=\dfrac{F_B}{\sin B}=\dfrac{F_C}{\sin C} math

Example

Given __**u**__ is a force with |__**u**__| = 4N and a direction of 35º And __**v**__ is a force with |__**v**__| = 7N and a direction of -15º Find the direction and magnitude (1 decimal place) of the resultant vector __**R**__ = __**u**__ + __**v**__ First rearrange vectors to form a triangle using the rules for addition of vectors (add head to tail) math \\ . \qquad R^2=4^2+7^2-2\times4\times7\times\cos130^\circ \\. \\ . \qquad R^2=16+49-56\times\cos130^\circ \\. \\ . \qquad R^2=100.996 \\ .\\ . \qquad R=10.0 math
 * Use Cosine Rule to find R **

math \\ . \qquad \dfrac{10}{\sin130^\circ}=\dfrac{7}{\sin \alpha} \\. \\ . \qquad \sin \alpha = \dfrac{7\sin130^\circ}{10}=0.536 \\. \\ . \qquad \alpha=32.4^\circ math
 * Now use Sine Rule to find a **

math . \qquad \theta=35-32.4=2.6^\circ math
 * Hence calculate ** **q (angle to positive x-axis) **

Hence Resultant Force has magnitude 10 N and a direction of 2.6º above the horizontal.

NOTE: This same question could also have been solved by resolving the two forces into their __**i**__ and __**j**__ components.

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