Given the truss supported by cable AB and pin C – (1) draw the free-body diagram (FBD) of the truss (2) find the forces (FCx, Fcy) at point C and FAB in the cable to keep the structure at equilibrium. Given: a = 2 m b = 2 m F = 3 kN F, = 4 kN 0 = 30°

Given the truss supported by cable AB and pin C – (1) draw the free-body diagram (FBD) of the truss (2) find the forces (FCx, Fcy) at point C and FAB in the cable to keep the structure at equilibrium. Given: a = 2 m b = 2 m F = 3 kN F, = 4 kN 0 = 30°

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Chemical and Phase Equilibrium

Phase diagrams or equilibrium phase diagrams are the graphical representations that show the phase transformations of the constituting elements with respect to the pressure, temperature, and compositions. The diagrams show the conditions that limit the different phases, namely solid, liquid, and gas. The diagram can also be utilized to verify the relations between the state variables. Most of the phase diagrams are temperature-pressure phase diagrams, with the variable parameter as the composition. Usually other variables such as solubility and pressure parameters are not indicated in the phase diagram. For a one-component system (unary phase diagram), the phase diagram is generally a function of temperature and pressure as the state variables, or temperature and volume. However, for a binary system (two-phase) having two components, the phase diagram is a function of temperature and composition as the state variables only. Coexisting of the different phases of the constituting elements is indicated in the phase diagram. A ternary phase diagram can be characterized by consisting of three phases in equilibrium.

Question

Problem 1
Given the truss supported by cable AB and pin C –
(1) draw the free-body diagram (FBD) of the truss
(2) find the forces (Fcx, Fcy) at point C and FAB in the cable to keep the structure at equilibrium.
Given:
a = 2 m
b= 2 m
F = 3 kN
F = 4 kN
0 = 30°
a

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