A simple truss supports a force Fat joint A, as shown in the figure below. The force has a magnitude F and is inclined by an angle = 35° from the horizontal. The lengths for all horizontal members is s = 7 m and for all vertical members is h = 7 m. The truss is held in equilibrium by a pin at point E and a roller at point F. Using the method of joints, determine the largest applied force Fmax so that the force in each truss member does not exceed ±100 kN. Note: Express tension forces as positive (+) and compression forces as negative (-).

International Edition---engineering Mechanics: Statics, 4th Edition
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Author:Andrew Pytel And Jaan Kiusalaas
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Chapter5: Three-dimensional Equilibrium
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Problem 5.42P: The 350-lb homogeneous plate has the shape of an isosceles triangle. The plate is supported by a...
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A simple truss supports a force Fat joint A, as shown in the figure below. The force has a magnitude F and is
inclined by an angle = 35° from the horizontal. The lengths for all horizontal members is s = 7 m and for all
vertical members is h = 7 m. The truss is held in equilibrium by a pin at point E and a roller at point F. Using
the method of joints, determine the largest applied force Fmax so that the force in each truss member
does not exceed +100 kN.
Note: Express tension forces as positive (+) and compression forces as negative (-).
F
0
A
B
S
S
Fmax
= number (rtol=0.01, atol=1e-05)
kN
F
S
E
h
Transcribed Image Text:A simple truss supports a force Fat joint A, as shown in the figure below. The force has a magnitude F and is inclined by an angle = 35° from the horizontal. The lengths for all horizontal members is s = 7 m and for all vertical members is h = 7 m. The truss is held in equilibrium by a pin at point E and a roller at point F. Using the method of joints, determine the largest applied force Fmax so that the force in each truss member does not exceed +100 kN. Note: Express tension forces as positive (+) and compression forces as negative (-). F 0 A B S S Fmax = number (rtol=0.01, atol=1e-05) kN F S E h
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