A plate is supported by the bracket at O. For the figure shown, determine: (a) the moment about O of the force exerted at A, (b) the magnitude and direction of a vertical force that would be applied at B that would make the same moment about O, (c) the magnitude and direction of the smallest force that could be applied at C which would create the same moment about O. FBD hints: Think about what the force in each would need to look like graphically in each part so that the forces in the FBD can be drawn in the correct orientation. No “fancy math" (i.e. Calculus) is required for portion of this problem. any 400 lb 30° D 1 ft 3 ft -3 ft 7 ft B C

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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A plate is supported by the bracket at O. For the figure shown, determine:
(a) the moment about O of the force exerted at A,
(b) the magnitude and direction of a vertical force that would be applied at B that would make the same
moment about O,
(c) the magnitude and direction of the smallest force that could be applied at C which would create the same
moment about O.
FBD hints: Think about what the force in each would need to look like graphically in each part so that the
forces in the FBD can be drawn in the correct orientation. No “fancy math" (i.e. Calculus) is required for
portion of this problem.
any
400 lb
30°
D
1 ft
-3 ft-
-3 ft
7 ft
В
Transcribed Image Text:A plate is supported by the bracket at O. For the figure shown, determine: (a) the moment about O of the force exerted at A, (b) the magnitude and direction of a vertical force that would be applied at B that would make the same moment about O, (c) the magnitude and direction of the smallest force that could be applied at C which would create the same moment about O. FBD hints: Think about what the force in each would need to look like graphically in each part so that the forces in the FBD can be drawn in the correct orientation. No “fancy math" (i.e. Calculus) is required for portion of this problem. any 400 lb 30° D 1 ft -3 ft- -3 ft 7 ft В
Expert Solution
Step 1

Solution:

The force 400 lb is resolved into two components :

Mechanical Engineering homework question answer, step 1, image 1

in x direction:Fx=400 cos 30°=346.4 lbFy=400sin30°=200 lbin vector form:F=-346.4i+200j

The position vector AO is given as:

rA/O=A-O=(-3i+j)=-3i+j

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