Find the magnitude and direction of the resultant force for the given forces F1 = 11 kN, F2 = 14 kN and F3 = 24 kN acting on hook fixed to a wall as shown in figure. Take e = 35° & e2 = 43°. -- 3 Marks. F1 'F3 F2 (NOTE: ENTER ONLY THE VALUES BY REFERRING TO THE UNITS GIVEN IN THE BRACKET. ALSO, SUBMIT THE HANDWRITTEN ANSWER SHEET IN THE LINK PROVIDED) The sum of horizontal force component for the given system (unit is in kN), E, : (1 Mark) The sum of vertical force component for the given system (unit is in kN), E, =. (1 Mark) The resultant of the given force system is (unit in kN), R = (0.5 Mark) The direction of the resultant force (in degrees) = (0.5 Mark) %3D Windows bu

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Question 12
Find the magnitude and direction of the resultant force for the given forces F1 = 11 kN, F, = 14 kN and F3 = 24 kN acting on hook fixed to a wall as shown in
figure. Take e, = 35° & e2 = 43°, -- 3 Marks.
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answered
Marked out of
3.00
P Flag question
F1
F3
F2
(NOTE: ENTER ONLY THE VALUES BY REFERRING TO THE UNITS GIVEN IN THE BRACKET. ALSO, SUBMIT THE HANDWRITTEN ANSWER SHEET
IN THE LINK PROVIDED)
The sum of horizontal force component for the given system (unit is in kN), E,
(1 Mark)
The sum of vertical force component for the given system (unit is in kN), E., =
(1 Mark)
The resultant of the given force system is (unit in kN), R
- (0.5 Mark)
The direction of the resultant force (in degrees) =
(0.5 Mark)
Windows byw
.Windows b U Ulw Jout
Transcribed Image Text:Question 12 Find the magnitude and direction of the resultant force for the given forces F1 = 11 kN, F, = 14 kN and F3 = 24 kN acting on hook fixed to a wall as shown in figure. Take e, = 35° & e2 = 43°, -- 3 Marks. Not yet answered Marked out of 3.00 P Flag question F1 F3 F2 (NOTE: ENTER ONLY THE VALUES BY REFERRING TO THE UNITS GIVEN IN THE BRACKET. ALSO, SUBMIT THE HANDWRITTEN ANSWER SHEET IN THE LINK PROVIDED) The sum of horizontal force component for the given system (unit is in kN), E, (1 Mark) The sum of vertical force component for the given system (unit is in kN), E., = (1 Mark) The resultant of the given force system is (unit in kN), R - (0.5 Mark) The direction of the resultant force (in degrees) = (0.5 Mark) Windows byw .Windows b U Ulw Jout
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