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A particle is acted upon by two forces F1 and F2 given by F1 =(-4i + 7j) N and F2 = (2ai - aj) N, where a is a positive constant. The resultant of F1 and F2 is R.
Given that R is parallel to 3i + j find the value of a. (4 marks)​


A Particle Is Acted Upon By Two Forces F1 And F2 Given By F1 4i 7j N And F2 2ai Aj N Where A Is A Positive Constant The Resultant Of F1 And F2 Is RGiven That R class=

Sagot :

The two given forces,

[tex]\mathbf F_1 = (-4 \, \mathbf i + 7 \, \mathbf j) \, \mathrm N[/tex]

[tex]\mathbf F_2 = (2a \, \mathbf i - a \, \mathbf j) \, \mathrm N[/tex]

have resultant

[tex]\mathbf R = \mathbf F_1 + \mathbf F_2 = ((2a-4) \, \mathbf i + (7 - a) \, \mathbf j) \, \mathrm N[/tex]

R is parallel to the vector 3i + j, which means the angle between these two vectors is either θ = 0° or θ = 180°. Then

[tex]\mathbf R \cdot (3 \, \mathbf i + \mathbf j) = \|\mathbf R\| \|3\,\mathbf i+\mathbf j\| \cos(\theta)[/tex]

[tex]\implies 3(2a-4)+(7-a) = \pm \sqrt{10} \, \sqrt{(2a-4)^2+(7-a)^2}[/tex]

[tex]\implies 5a - 5 = \pm \sqrt{50} \, \sqrt{a^2 - 6a + 13}[/tex]

[tex]\implies a - 1 = \pm \sqrt2 \, \sqrt{a^2 - 6a + 13}[/tex]

Solve for a :

[tex]a - 1 = \pm \sqrt2 \, \sqrt{a^2 - 6a + 13}[/tex]

[tex]\implies (a - 1)^2 = \left(\pm \sqrt2 \, \sqrt{a^2-6a+13}\right)^2[/tex]

[tex]\implies a^2 - 2a + 1 = 2a^2 - 12a + 26[/tex]

[tex]\implies 0 = a^2 - 10a + 25[/tex]

[tex]\implies 0 = (a - 5)^2[/tex]

[tex]\implies \boxed{a = 5}[/tex]

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