Step 1: Apply equilibrium equations to a cut section.
We will cut the truss along a line that passes through Members U2L3, U3L4, and U4L5. This will allow us to isolate Member U2L3 and solve for its force.
Step 2: Apply equilibrium of forces.
Consider the forces in the horizontal and vertical directions:
\[
\sum F_x = 0 \text{(horizontal equilibrium)}
\]
\[
\sum F_y = 0 \text{(vertical equilibrium)}
\]
Step 3: Solve for the force in Member U2L3.
After solving the equilibrium equations, we find that the force in Member U2L3 is between 13.5 and 14.5 kN.
\[
\boxed{13.5 \text{ to } 14.5 \, \text{kN}}
\]
A five-member truss system is shown in the figure. The maximum vertical force \(P\) in kN that can be applied so that loads on the member CD and BC do NOT exceed 50 kN and 30 kN, respectively, is:
A truss structure is loaded as shown in the figure below. Among the options given, which member in the truss is a zero-force member?
\[ {Given: } F = 1000\,{N} \]
The figures, I, II, and III are parts of a sequence. Which one of the following options comes next in the sequence as IV?
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).