[amp_mcq option1=”2 kgf, 3 kgf” option2=”3 kgf, 4 kgf” option3=”4 kgf, 5 kgf” option4=”5 kgf, 3 kgf” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B. 3 kgf, 4 kgf}}$.<\/p>\n
Let $F_1$ and $F_2$ be the magnitudes of the two forces. We know that $F_1^2 + F_2^2 = 5^2 = 25$ and $(F_1 + F_2)^2 = 37^2 = 1369$. Solving these two equations, we get $F_1 = 3$ and $F_2 = 4$.<\/p>\n
The two forces can be represented by two vectors, $F_1$ and $F_2$, that are at right angles to each other. The resultant force, $R$, is the diagonal of the rectangle formed by the two vectors. The magnitude of $R$ is given by the Pythagorean theorem:<\/p>\n
$$R^2 = F_1^2 + F_2^2$$<\/p>\n
In this case, $R = 5$, so $F_1^2 + F_2^2 = 25$.<\/p>\n
If the two forces are at an angle of 60\u00c2\u00b0 to each other, the resultant force can be found using the law of cosines:<\/p>\n
$$R^2 = F_1^2 + F_2^2 – 2F_1F_2\\cos(60^\\circ)$$<\/p>\n
In this case, $R = 37$, so $F_1^2 + F_2^2 – 2F_1F_2\\cos(60^\\circ) = 1369$.<\/p>\n
Solving this equation for $F_1$ and $F_2$, we get $F_1 = 3$ and $F_2 = 4$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”2 kgf, 3 kgf” option2=”3 kgf, 4 kgf” option3=”4 kgf, 5 kgf” option4=”5 kgf, 3 kgf” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[645],"tags":[],"class_list":["post-6767","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n