Question

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop a resultant force of 950 N directed along the positive x axis. Set θ = 50°.

Answer 1

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
2. 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

⇒$Fa*sin(50)-Fb*sin(20)=0$

⇒$Fa*sin(50)=Fb*sin(20)$

⇒$Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒$Fa*cos(20)+Fb*cos(50)=950$

⇒$2.24*Fb*cos(20)+Fb(50)=950$

⇒$Fb*(2.24*cos(20)+cos(50))=950$

⇒$Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒$Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒$Fa=2.24*Fb$

      $=2.24*345.5$

      $=773.93$

Therefore approximately, Fa=774 N and Fb=346 N