Question

Four people are pushing a body with forces:

310N in the direction 045

280N in the direction 135

400N in the direction 80. Calculate sum of vertical and sum of horizontal component respectively.?

Answer 1

$F_1$ = 310 N,

$F_2$ = 280 N,

$F_3$ = 400 N.

Sum of vertical component:

$F_y=F_{1y}+F_{2y}+F_{3y}=F_1sin45\degree+F_2sin45\degree+F_3sin80\degree=\newline 310\cdot 0.7+280\cdot 0.7+400\cdot 0.98=811.1\space N$

Sum of horizontal component:

$F_x=F_{1x}-F_{2x}+F_{3x}=F_1cos45\degree-F_2cos45\degree+F_3cos80\degree=\newline 310\cdot 0.7-280\cdot 0.7+400\cdot 0.17=90.7\space N$