Force Diagram

The force in the top member 6-7 needs to be determined when the 1,500 kg car is at joint 2.

First, solve for reactions R₁ and R₅ by applying both the moment and force equilibrium equations for the whole truss.

     ΣM₁ = 0
     -4 (14.72) + 16 R₅ = 0
     R₅ = 3.68 kN

     ΣF_y = 0
     R₁ + R₅ - 14.72 = 0
     R₁ = 11.04 kN

Cut 1 Diagram

Since, the member 6-7 needs to be determined, that member must be included in the cut. However, there is is usually more than one location to make a cut.

It is easiest to make a cut where only three or less members are cut. Thus, a vertical cut through members 6-7, 6-3 and 2-3 was done as shown at the left. Notice that there are only three unknowns, which can be solved for with the three equilibrium equations.

     ΣF_y = 0
     11.04 - 14.72 - F₆₃ cos45 = 0
     F₆₃ = -5.204 kN

     ΣM₁ = 0
     -4 (14.72) - 4cos45 F₆₃ - 4sin45 F₆₃ - 4 F₆₇ = 0
      -58.88 - 2 [(4) (0.7071) ( -5.204)] = 4 F₆₇
     F₆₇ = -7.360 kN  compression

For this cut location, you need to use at least two of the three equilibrium equations. The number of equations may be reduced by other cut locations.

Cut 2 Diagram

By carefully choosing where the cut is made, the number of calculations can be reduced.

For example, if a cut is made through members 6-7, 3-7, 3-8, and 3-4 it is possible to solve for the member force F₆₇ with one equilibrium equation even though four members are cut.

Since all unknowns, except F₆₇, go through
joint 3, the moment about joint 3 has only one unknown.

     ΣM₃ = 0
     -8 (11.04) - 4 F₆₇ + 4 (14.72) = 0
     F₆₇ = -7.360 kN  compression

The solution is identical to the solution for the previous cut.