Question

Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb which acts up to the left, in the same direction as from B towards A. Take θ = 30 ∘.

Answer 1

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = $\sqrt{400^2 + 650^2 - 2(400)(650)cos30}$

      = $\sqrt{160000 + 422500 - 80210}$

      = 708.72 Ib

using the sine law to determine the design angle

$\frac{sin}{400} = \frac{sin 30}{Fac}$

hence ∅ = $sin^{-1} (\frac{sin 30 *400}{708.72} )$

              = $sin^{-1} 0.0871$ =  4.9968 ≈ 5⁰

Answer 2

The design angle of the struts system given is;

ϕ = 38.26°

Parallelogram law of vector addition

The diagram showing the struts system is missing and so i have attached it.

• Now, from the forces given to us acting on the system attached, I have drawn a force diagram in form of a parallelogram showing the direction of the forces action. The second attached image shows this force diagram called diagram a.

From the diagram, we can see that it follows parallelogram law of vector addition.

Now, I have narrowed down the force system to the upper part of the strut as depicted in the third image attached.

• From the third image attached, we can use the concept of parallelogram law of vector addition and the law of cosine to get the force on the strut AC which is;

F_AC = √(400² + 600² - 2(400 × 600)cos 30)

F_AC = 322.97 lb

• Now, to get the design angle, we will solve as;

(sinϕ)/(sin 30) = 400/322.97

Where; ϕ is the design angle.

Thus;

(sinϕ) = (400/322.97) × sin 30

(sinϕ) = 1.2385 × 0.5

(sinϕ) = 0.61925

ϕ = sin^(-1) 0.61925

ϕ = 38.26°

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