Answer :
Answer:
Fa=774 N
Fb=346 N
Explanation:
We will solve this problem by equating forces on each axis.
- On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
- On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative
While towing we know that car is mot moving in y-direction so net force in y-axis must be zero
⇒∑Fy=0
⇒[tex]Fa*sin(50)-Fb*sin(20)=0[/tex]
⇒[tex]Fa*sin(50)=Fb*sin(20)[/tex]
⇒[tex]Fa=2.24Fb[/tex]
Given that resultant force on car is 950N in positive x-direction
⇒∑Fx=950
⇒[tex]Fa*cos(20)+Fb*cos(50)=950[/tex]
⇒[tex]2.24*Fb*cos(20)+Fb(50)=950[/tex]
⇒[tex]Fb*(2.24*cos(20)+cos(50))=950[/tex]
⇒[tex]Fb=\frac{950}{2.24*cos(20)+cos(50)}[/tex]
⇒[tex]Fb=\frac{950}{2.24*0.94+0.64}[/tex]
⇒ [tex]Fb=\frac{950}{2.75}=345.5[/tex]
⇒[tex]Fa=2.24*Fb[/tex]
[tex]=2.24*345.5[/tex]
[tex]=773.93[/tex]
Therefore approximately, Fa=774 N and Fb=346 N