Example:The resultant of the concurrent forces shown in figureis 300 N pointing up along the y-axis. Compute thevalue of F and e required to give this resultant.50ON30240No Since the resultant is up along y-axis then EF = 0o HW.o F = Fcos 0 + 240 cos 30 – 500 = 0o Fcos 0- 292.15 = 0o Repeat this problem if theresultant is 40ON down to theright at 60° with the x-axis.o Also repeat the problem ifthe body is in equilibrium.o Fcos 0 = 292.15... ....(1)o F, = R = Fsin 0 - 240 sin 30 = 300Fsin 0 = 300+ 120 = 420 ..... (2)Divide eqn. (2) by (1) yieldsF sin 0420%3DF cos e292.15o tan 0 = 1.4376,0 = tan 1.4376 = 55.18°420F =sin 55.18= 511.6N

Solution: The above diagram shows the position of the given resultant with x -axis. From the above…

Answered: 500N Example: The resultant of the…

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

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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Concept explainers

Types of loading

In mechanical terminology, the term load indicates that external force/weight, either constant or variable, acts on a mechanical member/structure. The weight of an object of a specific material is also an external load, and the external force that works on a structure comes under the categories of loading. The external load generates its effects as stress and strain in the structure.

Question

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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