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1 如圖所示,求此力系合力之大小與方向?
A 20N,37°
B 25N,53°
C 25N,127°
D 20N,53°

∑Fx=15×cos53°-26×12/13

=15×3/5-24

=15×3/5-24

-15N

∑Fy=15×sin53°+26×5/13-2

=15×4/5+10-2

=12+8

=20N

R=`sqrt((-15)^2+20^2)`

=25N

方向 = `tan^(-1)(20/(-15))`

用圖解:345的三角形,可知對20N那邊的角度是53°,所以方向就是180°-53°=127°