Question

three forces pull on object: f₁ = 50-N towards the +x; F₂ = 50-N toward the +y and F₃ = 70-N, 30° below -x. what force will balance the three? [Analythical method]

Answer 1

Solution.

$F_1=50N;$

$F_2=50N;$

$F_3=70N;$

$\phi=30^o;$

$F-?;$

$\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3};$

$\overrightarrow{F_{12}}=\overrightarrow{F_1}+\overrightarrow{F_2};$

Since the forces F₁ and F₂ are perpendicular, the equivalent of these forces is found by Pythagoras' theorem.

$F_{12}=\sqrt{F_1^2+F_2^2};$

$F_{12}=\sqrt{(50N)^2+(50N)^2}=50\sqrt{2}N;$

By the cosine theorem, we find the equivalent module:

$F^2=F_{12}^2+F_3^2-2F_{12}F_3cos\theta$ ;

$F^2=(50\sqrt{2}N)^2+(70N)^2-2\sdot50\sqrt{2}N\sdot70N\sdot cos15^o=$

$=339N^2;$

$F=18.4N;$

$\dfrac{F_3}{sin\alpha}=\dfrac{F}{sin\theta};\implies sin\alpha=\dfrac{F_3sin\theta}{F};$

$sin\alpha=\dfrac{70N\sdot 0.2588}{18.4N}=0.9845;$

$\alpha=80^o$

Equivalent force is directed at an angle of 55^o above -x.

Answer: $F=18.4N$ at an angle of 55o above -x.