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500N Example: The resultant of the concurrent forces shown in figure is 300 N pointing up along the y-axis. Compute the value of F and e required to give this resultant. 30 240N

500N Example: The resultant of the concurrent forces shown in figure is 300 N pointing up along the y-axis. Compute the value of F and e required to give this resultant. 30 240N

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Example: The resultant of the concurrent forces shown in figure is 300 N pointing up along the y-axis. Compute the value of F and e required to give this resultant. 50ON 30 240N o Since the resultant is up along y- axis then EF = 0 o HW. o F = Fcos 0 + 240 cos 30 – 500 = 0 o Fcos 0- 292.15 = 0 o Repeat this problem if the resultant is 40ON down to the right at 60° with the x-axis. o Also repeat the problem if the body is in equilibrium. o Fcos 0 = 292.15... .... (1) o F, = R = Fsin 0 - 240 sin 30 = 300 Fsin 0 = 300+ 120 = 420 ..... (2) Divide eqn. (2) by (1) yields F sin 0 420 %3D F cos e 292.15 o tan 0 = 1.4376, 0 = tan 1.4376 = 55.18° 420 F = sin 55.18 = 511.6N
Transcribed Image Text:Example: The resultant of the concurrent forces shown in figure is 300 N pointing up along the y-axis. Compute the value of F and e required to give this resultant. 50ON 30 240N o Since the resultant is up along y- axis then EF = 0 o HW. o F = Fcos 0 + 240 cos 30 – 500 = 0 o Fcos 0- 292.15 = 0 o Repeat this problem if the resultant is 40ON down to the right at 60° with the x-axis. o Also repeat the problem if the body is in equilibrium. o Fcos 0 = 292.15... .... (1) o F, = R = Fsin 0 - 240 sin 30 = 300 Fsin 0 = 300+ 120 = 420 ..... (2) Divide eqn. (2) by (1) yields F sin 0 420 %3D F cos e 292.15 o tan 0 = 1.4376, 0 = tan 1.4376 = 55.18° 420 F = sin 55.18 = 511.6N
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