Question #3024

Three force F1 , F2, F3 act on a 5.00 kg object . The angle between F1 and F2 is 60 degree the angle between F2 and F3 is 30 degree .Taking F1 = 20.0 N and F2 = 15.0 and F3 = 10.0 N
A )- Find the magnitude and direction of the net force acting on the object ?
B)- Find the magnitude and direction of the acceleration of the object ?

Expert's answer

Let F2 has x direction. Then

F_{x}=20*cos(60)+15+10*cos(30)=25+5√(3)

F_{y}=20*sin(60)-10*sin(30)=10√(3)-5

Net force:

F=sqrt(F_{x}^{2}+F_{y}^{2})=5sqrt(41+6*√(3))

Direction:

the angle between net force and F_{2 }is

arctg(F_{y/}F_{x})=arctg((-1+√(3))/2)

acceleration has the same direction. its magnitude

a=F/m=√(41+6*√(3))

F

F

Net force:

F=sqrt(F

Direction:

the angle between net force and F

arctg(F

acceleration has the same direction. its magnitude

a=F/m=√(41+6*√(3))

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