Question

the truss section shown here is in equilibrium. it is being acted on by four forces, with directions as shown. the angle 0 between the horizontal and diagonal members is 36 degrees.
note: figure may not be drawn to scale.
vectors a and b represent known forces. the magnitude of force a is 50 kn, and the magnitude of force b is 20 kn.
find the magnitude of vector c in kn. include two digits after the decimal point in your answer.
(hint: first find the components of vector d.)
answer

Explanation:

Step1: Find the horizontal component of vector D

Since the truss is in equilibrium, the sum of horizontal forces is zero. Let the magnitude of vector D be \(D\). The horizontal component of \(D\) is \(D\cos\theta\). We know that \(A = D\cos\theta\) (from horizontal force balance, \(A\) is the horizontal force and \(D\cos\theta\) is the horizontal component of \(D\)). Given \(A = 50\space kN\) and \(\theta=36^{\circ}\), we can solve for \(D\): \(D=\frac{A}{\cos\theta}=\frac{50}{\cos(36^{\circ})}\).

Step2: Find the vertical component of vector D

The vertical component of \(D\) is \(D\sin\theta\). Substituting \(D = \frac{50}{\cos(36^{\circ})}\) into \(D\sin\theta\), we get \(D\sin\theta=\frac{50\sin(36^{\circ})}{\cos(36^{\circ})}=50\tan(36^{\circ})\)

Step3: Find the magnitude of vector C

Since the truss is in equilibrium, the sum of vertical forces is zero. So \(C + B=D\sin\theta\). Given \(B = 20\space kN\), then \(C=D\sin\theta - B\). Substituting \(D\sin\theta = 50\tan(36^{\circ})\) into the equation: \(C = 50\tan(36^{\circ})- 20\)

We know that \(\tan(36^{\circ})\approx0.7265\)

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Answer:

\(16.33\)