An inverted Howe truss supports three forces at joints along the top edge, as shown in the figure below. The forces have magnitudes F₁ = 70 kN, F₂ = 85 kN, and F3 = 20 kN. The lengths for all horizontal members is s = 3 m and for all vertical members is h = 9 m. The truss is held in equilibrium by two vertical cables at points A and G. Use the method of sections to determine the force in members BC, CK, JK, DE, EJ, and IJ and indicate whether they are in tension or compression. Note: Express tension forces as positive (+) and compression forces as negative (-). F₁ F₂ F3 B C D E F A G S S S K S S PBC = number (rtol=0.01, atol=1e-05) kN Рск = number (rtol=0.01, atol=1e-05) kN PJK = number (rtol=0.01, atol=1e-05) kN PDE = number (rtol=0.01, atol=1e-05) kN PEJ = number (rtol=0.01, atol=1e-05) kN PIJ = number (rtol=0.01, atol=1e-05) kN S S h

International Edition---engineering Mechanics: Statics, 4th Edition
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ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
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Chapter4: Coplanar Equilibrium Analysis
Section: Chapter Questions
Problem 4.97P: The figure shows a wire cutter. Determine the cutting force on the wire at A when the 75-N forces...
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An inverted Howe truss supports three forces at joints along the top edge, as shown in the figure below. The
forces have magnitudes F₁ = 70 kN, F₂ = 85 kN, and F3 = 20 kN. The lengths for all horizontal members is
s = 3 m and for all vertical members is h = 9 m. The truss is held in equilibrium by two vertical cables at
points A and G. Use the method of sections to determine the force in members BC, CK, JK, DE, EJ, and IJ
and indicate whether they are in tension or compression.
Note: Express tension forces as positive (+) and compression forces as negative (-).
F₁
F₂
F3
B
C
D
E
F
A
G
S
S
S
K
S
S
PBC =
number (rtol=0.01, atol=1e-05)
kN
Рск
=
number (rtol=0.01, atol=1e-05)
kN
PJK =
number (rtol=0.01, atol=1e-05)
kN
PDE
=
number (rtol=0.01, atol=1e-05)
kN
PEJ =
number (rtol=0.01, atol=1e-05)
kN
PIJ
=
number (rtol=0.01, atol=1e-05)
kN
S
S
h
Transcribed Image Text:An inverted Howe truss supports three forces at joints along the top edge, as shown in the figure below. The forces have magnitudes F₁ = 70 kN, F₂ = 85 kN, and F3 = 20 kN. The lengths for all horizontal members is s = 3 m and for all vertical members is h = 9 m. The truss is held in equilibrium by two vertical cables at points A and G. Use the method of sections to determine the force in members BC, CK, JK, DE, EJ, and IJ and indicate whether they are in tension or compression. Note: Express tension forces as positive (+) and compression forces as negative (-). F₁ F₂ F3 B C D E F A G S S S K S S PBC = number (rtol=0.01, atol=1e-05) kN Рск = number (rtol=0.01, atol=1e-05) kN PJK = number (rtol=0.01, atol=1e-05) kN PDE = number (rtol=0.01, atol=1e-05) kN PEJ = number (rtol=0.01, atol=1e-05) kN PIJ = number (rtol=0.01, atol=1e-05) kN S S h
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