Question

Determine the resultant of the three forces and the couple C, and show it on a sketch if a coordinate system if (a)C=0 ; and (b)C=90 N.m

Answer 1

$Q_2=tan^{-1}(\frac{4}{5}) \\ Q_2=38.66 \\ Q_3=tan^-(\frac{12}{5}) \\ Q_3=67.38$

$F_1=80N \\ F_1=80 î N \\ F_2=150cosQ_2 î +150sinQ_2 Ĵ \\ =117.13 î +93.70Ĵ \\ F_3= - 130 cosQ_3 î + 130sinQ_3 Ĵ \\ = -50.00 î + 119.99Ĵ$

$r_1= 0î+ 0Ĵ$ (distance between 0 and applicat point of force)

$r_2= 2.8 î + 2.4Ĵ \\ r_3= 0 î + 2.4Ĵ$

$F_t=F_1 +F_2 +F_3 \\ = 80 î +117.13 î +93.70Ĵ -50.00î +120Ĵ \\ F_t=147.13 î +213.7Ĵ$

So, this force F_t will be acting on the point 0.

a. c=o then the torque will be at 0,

r=101.248 in counter clock wise directional

$F_t$ =147.13 î + 213.7Ĵ

|$F_t$ |=259.45 N (at point 0)

b. c=90Nm=90Nm kˆ (conter clock wise)

$r_t$ =90 + 101.248 kˆ (total torque or moment)

$r_t$ =191.248 in counter clock wise directional

$F_t$ =147.13 î +213.7Ĵ

|$F_t$ |=259.45 N