Question 3.5.2: If, in Example 3.5-1 (Fig. 3.5-1), the force P is applied at...
If, in Example 3.5-1 (Fig. 3.5-1), the force P is applied at joint F instead of joint E, determine the forces in all members.
By Rule 2, members 4, 5, and 6 are all unstressed (dotted lines, Fig. 3.5-2).
At joint F, where P acts, member 1 is the only member not lying in the horizontal xy plane. Therefore (Rule 1), considering vertical equilibrium of the joint,
S_{1} \cos 50°= P
Hence
S_{1} = \underline{ 1.55P} (compressive)
Again, the force in member 3 is the only force at joint F not lying in the yz plane.
Hence (Rule 1), considering force components normal to that plane, we must have
S_{3} \sin 60°=0 i.e. S_{3}=\underline{0}
whence, by inspection,
S_{2} = S_{1} \sin 50° = \underline{1.19P} (tensile)
S_{4} = S_{5} = S_{6} = \underline{0} as stated above
