Question

01) A esfera da figura está sob ação de um sistema de forças. Sabe-se que a massa da
esfera é de 2 Kg. Determinar: a) Sentido da força resultante, b) Valor da força resultante
c) Aceleração da esfera. (Dados: cos 60° =1
2
, √19 = 4,3)

Answer 1

$\mathbf{Dados}\Longrightarrow m=2\ \text{kg};\ \vec{F}_1=(8\hat{i})\ \text{N};\ \vec{F}_2=(-2\hat{i})\ \text{N};\ \theta=60^{\circ}.$

$\mathrm{Seja}\ |\vec{F}_3|=4\ \text{N}\ \mathrm{com}\ \vec{F}_3\ \mathrm{formando\ 60^{\circ}\ com\ a\ horizontal.}$

$\mathrm{As\ componentes\ horizontal\ e\ vertical\ de}\ \vec{F}_3\ \mathrm{ser\tilde{a}o,\ respectivamente}\text{:}$

$|\vec{F}_{3x}|=|\vec{F}_3|\cos{60^{\circ}}=4\bigg(\dfrac{1}{2}\bigg)\Longrightarrow |\vec{F}_{3x}|=2\ \text{N}$

$|\vec{F}_{3y}|=|\vec{F}_3|\sin{60^{\circ}}=4\bigg(\dfrac{\sqrt{3}}{2}\bigg)\Longrightarrow |\vec{F}_{3y}|=2\sqrt{3}\ \text{N}$

$\mathrm{Desse\ modo}\text{:}$

$\vec{F}_3=(2\hat{i}+2\sqrt{3}\hat{j})\ \text{N}$

$\mathrm{A\ for\c{c}a\ resultante\ ser\acute{a}}\text{:}$

$\vec{F}_R=\vec{F}_1+\vec{F}_2+\vec{F}_3=(8\hat{i})+(-2\hat{i})+(2\hat{i}+2\sqrt{3}\hat{j})$

$\Longrightarrow \vec{F}_R=\big((8-2+2)\hat{i}+2\sqrt{3}\hat{j})\ \therefore\ \boxed{\vec{F}_R=(8\hat{i}+2\sqrt{3}\hat{j})\ \text{N}}$

$\mathbf{a)\ Sentido\ da\ for\c{c}a\ resultante}$

$\vec{F}_R\ \mathrm{forma\ um\ \hat{a}ngulo\ \phi\ com\ a\ horizontal\ tal\ que}\text{:}$

$\tan{\phi}=\dfrac{|\vec{F}_{Ry}|}{|\vec{F}_{Rx}|}=\dfrac{2\sqrt{3}}{8}=\dfrac{\sqrt{3}}{4}\Longrightarrow \boxed{\phi=\arctan{\bigg(\dfrac{\sqrt{3}}{4}\bigg)\approx23.5^{\circ}}}$

$\mathrm{O\ sentido\ de}\ \vec{F}_R\ \mathrm{est\acute{a}\ representado\ na\ imagem\ em\ anexo.}$

$\mathbf{b)\ M\acute{o}dulo\ da\ for\c{c}a\ resultante}$

$|\vec{F}_R|=\sqrt{|\vec{F}_{Rx}|^2+|\vec{F}_{Ry}|^2}=\sqrt{8^2+(2\sqrt{3})^2}$

$\Longrightarrow |\vec{F}_R|=\sqrt{76}\ \therefore\ \boxed{|\vec{F}_R|=2\sqrt{19}\approx8.72\ \mathrm{N}}$

$\mathbf{c)\ Acelera\c{c}\tilde{a}o\ da\ esfera}$

$\vec{F}_R=m\vec{a}\Longrightarrow \vec{a}=\dfrac{1}{m}\vec{F}_R$

$\Longrightarrow \vec{a}=\dfrac{1}{2}(8\hat{i}+2\sqrt{3}\hat{j})\ \therefore\ \boxed{\vec{a}=(4\hat{i}+\sqrt{3}\hat{j})\ \mathrm{m/s^2}}$

$|\vec{a}|=\sqrt{|\vec{a}_x|^2+|\vec{a}_y|^2}=\sqrt{4^2+(\sqrt{3})^2}\ \therefore\ \boxed{|\vec{a}|=\sqrt{19}\approx4.36\ \mathrm{m/s^2}}$