Question

Two forces act on a hook. The magnitudes are F1= 300N and F2= 125N. The angle between the two forces is 48. Find the direction(with respect to F1) and the magnitude of the resultant.

Answer 1

R^2=P^2+Q^2+2PQ Cos x
tany=Q Sinx/(P+Qcosx),where y is the angle between P&R
F^2=300^2+125^2+2(300)(125)COS48
F^2=25^2[12^2+5^2+2*12*5Cos48]
F^2=625[144+25+120Cos48]=625(169+120cos48)
F=25 sqrt(169+120cos48)
If y is the angle between F1 &F
tan y=125 sin48/(300+125cos48)
i have no calculator,you can solve cos48& sin48

Answer 2

TakeF=R

Answer 3

Thank You so much!