Question

Chapter S2, Problem S2/122 Determine the moment of each individual force about (a) point A and (b) point B. 2.7F 1.3b . 1.5F 2.7b Answers: 1.5F: MA- Mg = 1 즈 2.7F: MA=( 즈 Mg = 즈

Answer 1

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The terms used in the solution of the given problem are as follows:

i = unit vector along x-axis

j = unit vector along y-axis

k = unit vector along z-axis

$\overrightarrow{F}$ = Force

$\overrightarrow{M}$ = Moment of force

Solution:

To find Moment of 1.5F force about A:

$\overrightarrow{M_{A}}$= ($\overrightarrow{AE}$) x $\overrightarrow{F}$ (cross Product)

$\overrightarrow{AE}$ = [0 - (-3.5b)]i = 3.5b i

$\overrightarrow{F}$ = . (1.5F)

EO

• $\widehat{EO}$ = Unit vector along EO
• $\widehat{EO}$ = (2.7b j – 1.3b k) / ​​​​( (2.7)² b² + (1.3)² b² )^1/2

= 1.3515 F j – 0.6507 F k

$\overrightarrow{M_{A}}$ = ( 3.5b i ) x ( 1.3515 F j – 0.6507 F k ) = (2.28 j + 4.73 k ) Fb

$\overrightarrow{M_{A}}$= (0 i + 2.28 j + 4.73 k ) Fb

To find Moment of 1.5F force about B:

$\overrightarrow{M_{B}}$ = ​​​​
BE
x

= 3.5 b i + 1.3 b k

BE

= 1.3515 F j – 0.6507 F k

$\overrightarrow{M_{B}}$ = (3.5 b i + 1.3 b k) x (1.3515 F j – 0.6507 F k)

$\overrightarrow{M_{B}}$= (-1.76 i + 2.28 j + 4.73 k) Fb

To find Moment of 2.7F force about A:

$\overrightarrow{M_{A}}$ =
AD
x

= 2.7 b j

AD

= 2.7 F i

$\overrightarrow{M_{A}}$ = (2.7 b j) x (2.7 F i)

$\overrightarrow{M_{A}}$ = (0 i + 0 j - 7.29 k) Fb

To find Moment of 2.7F force about B:

$\overrightarrow{M_{B}}$ =
BD
x

= 2.7 b j + 1.3 b k

BD

= 2.7 F i

$\overrightarrow{M_{B}}$ = (2.7 b j + 1.3 b k ) x (2.7 F i)

$\overrightarrow{M_{B}}$ = (0 i + 3.51 j - 7.29 k) Fb

Answers:

1.5F :

M_A = (0 i + 2.28 j + 4.73 k ) Fb

M_B = (-1.76 i + 2.28 j + 4.73 k) Fb

2.7F :

M_A = (0 i + 0 j - 7.29 k) Fb

M_B = (0 i + 3.51 j - 7.29 k) Fb