The figure shows an idealized plane truss. If a horizontal force of 300 N is applied at point A, then the magnitude of the force produced in member CD is ________N.

To find the magnitude of the force in member CD, we can analyze the joints of the truss using the method of joints and the properties of zero-force members.

Step 1: Identify Collinear Members and Angles
From the given figure:

• The outer members on the left side, $AB$, $BD$, and $DF$, are collinear and lie along a single straight line.
• The outer members on the right side, $AC$, $CE$, and $EG$, are also collinear and lie along a single straight line.
• Members $BC$ and $DE$ are horizontal.

Step 2: Analyze Joint B
Let us consider the equilibrium of joint B. The members meeting at joint B are:

• $AB$ and $BD$ (which are collinear).
• $BC$ (which is horizontal and not collinear with $AB$ and $BD$).

There is no external load acting at joint B.

By taking the sum of forces perpendicular to the line of members $ABD$:

$F_{BC}·\sin \left(\theta \right)=0$

where $\theta$ is the angle between the member $BC$ and the line $ABD$ (which is non-zero). Since $\sin \left(\theta \right)\ne 0$, we must have:

$F_{BC}=0$

Step 3: Analyze Joint C
Now, let us analyze joint C. The members meeting at joint C are:

• $AC$ and $CE$ (which are collinear).
• $BC$ (horizontal).
• $CD$ (diagonal).

Since we established in Step 2 that $F_{BC}=0$, member $BC$ exerts no force on joint C.

Thus, the active forces acting at joint C are only from the collinear members $AC$ and $CE$, and the diagonal member $CD$.

For the joint to be in equilibrium, the sum of forces perpendicular to the collinear line $ACE$ must be zero:

$F_{CD}·\sin \left(\phi \right)=0$

where $\phi$ is the angle between the member $CD$ and the line $ACE$. Because $\phi \ne 0$, it follows that:

$F_{CD}=0$

Thus, the force produced in member CD is 0 N.