THE Moment OF A Force- Diana Mellado PDF

Preview

Summary

NAME:​ Diana Mellado COURSE & YEAR: ​BSME-IIITHE MOMENT OF A FORCE Assuming ​counterclockwise moments as positive, compute the moment of force F = 200 kg and force P = 165 kg about points A, B, C, and D SOLUTION​:FH=F( ) 00( ) 45 = 2 45 FH= 1 60 kg FV =F( ) 00( ) 53 = 2 53 FV = 1 20 kg PH=P(...

Description

NAME: Diana Mellado COURSE & YEAR: BSME-III

THE MOMENT OF A FORCE 1. Assuming counterclockwise  moments as positive, compute the moment of force F = 200 kg and force P = 165 kg about points A, B, C, and D

SOLUTION:

F H = F ( 45 ) = 200( 45 ) F H = 160 kg F V = F ( 35 ) = 200( 35 ) F V = 120 kg PH = P

( ) = 165( 2 √13

2 ) √13

P H = 91.526 kg PV = P

( ) = 165( 3 √13

P V = 137.288 kg

3 ) √13

Moment of force F about points A, B, C, and D: M A = 5(0.3)F V = 5(0.3)(120) M A = 180 kg − m

M B = −6(0.3)F H = − 6(0.3)(160) M B = − 288 kg − m

M C = − 0.3F V − 3(0.3)F H = −0.3(120)−3(0.3)(160)

M C = − 180 kg − m M D = 5(0.3)F V − 6(0.3)F H = 5(0.3)(120)−6(0.3)(160) M D = − 108 kg − m Moment of force P about points A, B, C, and D:

M A = 6(0.3)P H − 4(0.3)P V = 6(0.3)(91.526)−4(0.3)(137.288) MA = 0 M B = 0.3P V = 0.3(137.288) M B = 41.186 kg − m M C = 2(0.3)P V + 3(0.3)P H = 2(0.3)(137.288) + 3(0.3)(91.526) M C = 164.746 kg − m M D = −4(0.3)P V = −4(0.3)(137.288) M D = −164.746 kg − m...