Consider the object subjected to the loading shown in the figure. Take the value of MD to be equal to 6 kN.m. 10 kN B 2 kN E 2 kN 2 m MD 2 m 2 m x Determine an equivalent force system at point A, using a vector approach. The value of F R of the equivalent force system is Ĵ) KN. The value of M RA of the equivalent force system is + k) kN.m. The magnitude of M RA is KN-m. (Round the final answer to four decimal places.)

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Chapter1: Units, Trigonometry. And Vectors
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Consider the object subjected to the loading shown in the figure. Take the value of MD to be equal to 6 kN.m.
10 kN
B
2 kN
E
2 kN
2 m
MD
2 m
2 m
x
Determine an equivalent force system at point A, using a vector approach.
The value of F
R
of the equivalent force system is
Ĵ) KN.
The value of M RA of the equivalent force system is
+
k) kN.m.
The magnitude of M RA is
KN-m. (Round the final answer to four decimal places.)
Transcribed Image Text:Consider the object subjected to the loading shown in the figure. Take the value of MD to be equal to 6 kN.m. 10 kN B 2 kN E 2 kN 2 m MD 2 m 2 m x Determine an equivalent force system at point A, using a vector approach. The value of F R of the equivalent force system is Ĵ) KN. The value of M RA of the equivalent force system is + k) kN.m. The magnitude of M RA is KN-m. (Round the final answer to four decimal places.)
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