I have been bothered with a problem that I know should not be monopolizing as much time as it has. It just annoys me how I keep missing the little thing that would make the problem more understandable.(adsbygoogle = window.adsbygoogle || []).push({});

Anyways, here is the problem, with my attempt of a solution:

a tow truck pulls on a car with a 2500N force directed along the cable used for towing. The purpose of the problem is to find the force in the members of the 'towing crane' used, assuming that both the car and the truck are not moving. (the attached FBD should make more sense.. sorry)

Here are the calculations and results I got:

(sum of forces F) F = ma = 0 // at the point C where the three forces are joined

--> Fx= 0 and Fy=0

F1= -2500* cos(40)i- 2500*sin(40)j// finding the vector corresponding to each force

Fa= - Faj

Fb= Fb*cos(50)i- Fb*sin(50)j

Fx= [ (-2500* cos(30) + Fb*cos(50)]= 0 //applying the equilibrium equation

--> Fb = 2500* cos(30) / *cos(50)

Fb= 1632 N

Fy= [ - 2500*sin(30) - Fa - Fb*sin(50)] = 0

--> Fa = -(2500*sin(30) + Fb*sin(50))

Fa = -(2500*sin(30) + 2979*sin(50))

Fa= 3214 N

It seems logic to me that it'd be the correct resolution path, but according to the textbook, the correct values for Fa and Fb are

Fa= 3.83kN and Fb=3.37kN

Quite different from what I got indeed.

If anyone could help me with this problem, it'd be great. I tried. I really did. :(

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Laws or motions and a truck that won't move.

**Physics Forums | Science Articles, Homework Help, Discussion**