# Thread: a more difficult statics problem (three force member)

1. ## a more difficult statics problem (three force member)

I really hate my statics book (hibbeler 12th edition). There are no example solutions of any three force members, and the author only spends one sentence describing parallel three force members! I was able to figure out the numerical method with trial and error, but I would appreciate it if someone could explain it to me in words.

The answer in the back of the book says the angle is 1.02 degrees. So I took the tangent of that angle and tried combinations of the numbers given in order to find something that worked. I assumed that the spring at A is compressed $\left(\frac{800}{5000}\right) = 0.16$ meters since the applied force is only a meter away. dividing .16 by 3 (length of the beam) and doing tan inverse gives me
$\arctan\left(\frac{0.16}{3}\right) = 3.0528825$. Dividing by 3 again gives me the answer in the back of the book, or what they most certainly had approximated, 1.0176275 degrees.

If this is incorrect, I would really appreciate some guidance. And even if my method is correct, the physics going on is not obvious to me. Wouldn't the applied force create a larger torque in the opposite direction about B? Does the larger moment at B essentially assist in raising up the opposite end of the beam (so that the deflection is only 0.053333 meters, that would elimiate the second division by 3 after the arctan operation)?

I've searched google, wikipedia, and google books, and I can't find any simple "general case" equations that govern three force members of parallel forces.

thanks guys

2. ## Physics

The key is to balance the (vertical) forces and torques. You have the downward force of the weight on the beam, of 800 N. This must be balanced by the upward forces the springs exert on the beam, $F_A$ and $F_B$, so that we have $F_A+F_B=800\;\text{N}$. Now, consider the torques about the center of the beam. The load acts 0.5 m from the beam center, and so exerts a torque of (800 N)(0.5 m)=400 N•m counterclockwise (in our diagram). Spring B, in turn, exerts a torque in the same direction of $F_B\cdot(1.5\;\text{m})$. Lastly, spring A exerts a torque in the opposite direction of $F_A\cdot(1.5\;\text{m})$, thus
$F_A\cdot(1.5\;\text{m})=F_B\cdot(1.5\;\text{m})+40 0\;\text{N}\cdot\text{m}$
$(1.5\;\text{m})(F_A-F_B)=400\;\text{N}\cdot\text{m}$
$F_A-F_B=267\;\text{N}$
And so you have a system of two equations for your two unknown forces; solving for these forces, you use the fact that the force exerted by each spring will be the stiffness k=5 kN/m times the compression of the spring; you can then find how much each spring is compressed. The difference in these is the difference in the heights of the ends of the beams, which is 3 m (the length of the beam) times the sine of the angle you want to find.

--Kevin C.

3. got it, thanks