ENGINEERING MECHANICSPLS DO STEP BY STEP SOLUTION THIS FOR MY EXAM REVISION THANK YOU FOR HELPING ME APPRECIATED A LOT ! Determine each force in Figure7mX35°as a Cartesian Vector FormFAB=155 N25 mBZ15 mFBC = 105 N8 m35 m13 mBy referring to the figure, calculate the magnitude and coordinate direction angles of the resultantforce

Given: Force→FAB=155 N ; →FBC=105 N From the given figure for force along AB: Component of that…

Answered: Determine each force in Figure as a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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ENGINEERING MECHANICS

PLS DO STEP BY STEP SOLUTION THIS FOR MY EXAM REVISION THANK YOU FOR HELPING ME APPRECIATED A LOT !

Determine each force in Figure
7m
X
35°
as a Cartesian Vector Form
FAB=155 N
25 m
B
Z
15 m
FBC = 105 N
8 m
35 m
13 m
By referring to the figure, calculate the magnitude and coordinate direction angles of the resultant
force

Expert Solution

Step 1

Given:

Force $\to F_{A B} = 155 N; \to F_{B C} = 105 N$

From the given figure for force along AB:

Component of that force in z direction:

$\to F_{z 1} = - F_{A B} \times \cos (\tan^{- 1} (\frac{25}{35 - 7})) = - 155 \times \cos (\tan^{- 1} (\frac{25}{35 - 7})) = - 115.62 \hat{k} N$

Component of that force in x direction:

$\to F_{x 1} = F_{A B} \times \sin (\tan^{- 1} (\frac{25}{35 - 7})) \times \cos 35 ° = 155 \times \sin (\tan^{- 1} (\frac{25}{35 - 7})) \times \cos 35 ° = 84.56 \hat{i} N$

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