Olá,
Talyson
Observe as forças agindo no bloco
[tex3]\text{B}:[/tex3]
- Decomposição de Forças (2).png (14.51 KiB) Exibido 1757 vezes
Escrevendo as equações que representam os equilíbrios, temos:
[tex3]\begin{cases}\,\text{Equilíbrio do bloco A: } \quad \text{T}_1 = \text{m}_{\text{A}} \cdot \text{g} \quad {\color{red} \text{(I)}}\\\\\
\text{Equilíbrio do bloco C: } \quad \text{T}_2 = \text{m}_{\text{C}} \cdot \text{g} \quad {\color{red} \text{(II)}} \\\\\
\text{Equilíbrio do bloco B na horizontal: } \quad \text{T}_2 \cdot \cos \theta = \text{T}_1 \quad {\color{red} \text{(III)}} \\\\\
\text{Equilíbrio do bloco B na vertical: } \quad \text{F}_{\text{N}} + \text{T}_2 \cdot \sen \theta = \text{P}_{\text{B}} \quad {\color{red} \text{(IV)}}
\end{cases}[/tex3]
Desenvolvendo
[tex3]{\color{red} \text{(III)}}, \,[/tex3] vem:
[tex3]\text{T}_2 \cdot \cos \theta = \text{T}_1 \,\,\,\, \Rightarrow \,\,\,\, \text{m}_{\text{C}} \cdot \text{g} \cdot \cos \theta = \text{m}_{\text{A}} \cdot \text{g} \,\,\,\, \Rightarrow \,\,\,\, 50 \cdot 10 \cdot \cos \theta = 40 \cdot 10 \,\,\,\, \therefore \,\,\,\, \boxed{\cos \theta = 0,8}[/tex3]
Da relação fundamental da trigonometria, temos:
[tex3]\sen^2 \theta + \cos^2 \theta = 1 \,\,\,\, \Rightarrow \,\,\,\, \sen^2 \theta + (0,8)^2 = 1 \,\,\,\, \therefore \,\,\, \boxed{\sen \theta = 0,6}[/tex3]
Substituindo esse valor em
[tex3]{\color{red} \text{(IV)}}, \,[/tex3] obtemos:
[tex3]\text{F}_{\text{N}} + \text{T}_2 \cdot \sen \theta = \text{P} \,\,\,\, \Rightarrow \,\,\,\, \text{F}_{\text{N}} + 50 \cdot 10 \cdot 0,6 = 45 \cdot 10 \,\,\,\, \therefore \,\,\,\, \boxed{\text{F}_{\text{N}} = 150 \, \text{N}} [/tex3]