Question

Determine components of 2kN force along oblique axes a and b. Determine projections of F on a and b axes.​

Answer 1

Final Answer: The projections of F on the a and b axes are 1.414kN and 1.932kN respectively.

Explanation: Let the components of 2kN (say F) be F₁ and F₂ along oblique axes a and b and projections of F on a and b axes are P₁ and P₂ respectively.

As shown in the figure, Drawn a line F₂ and F₁.

Ф₁ = 45°Ф₂ = 120° and Ф₃ = 15°

According to the formula

$\frac{F}{sin120} = \frac{F_1}{sin15} =\frac{F_2}{sin45}$

On solving the first two terms,

$F_1 = \frac{2 * sin15}{sin120} \\\\F_1 =\frac{ 2* 0.2588 }{0.8660} \\\\F_1 = 0.5977kN$

Similarly,

$F_2 = \frac{2 * sin45}{sin120} \\\\F_2 =\frac{ 2* 0.7071 }{0.8660} \\\\F_2 = 1.63239kN$

Now,

P₁ = Fcos45°

P₁ = 2 ₓ 0.7071 =1,4142kN

P₂ = Fcos15°

P₂ = 2 ₓ 0.9659 = 1.93182kN.