# Forces and Equilibrium

1. Jan 13, 2010

### Peto

The Crane in P19 (Attached) supports a 20.0-kN load.. to simplify things take the bales to be represented by two separate bundles: Cable-AB and cable-BC. (a) Draw a free-body diagram of point-B putting in the forces exerted by the boom, cable-AB, and cable -BC. (b) determin the horizontal and the vertical components of each force. (c) Using Sigma(Fx) = 0 and Sigma(Fy) = 0, determine the compressive force on the boom.

Since Fy, and Fx = 0, The load must be in Equilibrium.

I'm kind of stuck on this question, I had skipped over it before but now I still want to figure it out.

Any help would be much appreciated!

#### Attached Files:

• ###### Physics 12 P.138 Scan.jpg
File size:
41.2 KB
Views:
84
2. Jan 13, 2010

### PhanthomJay

Take the suggestions in parts a, b, and c. There are 3 forces acting on B, one is known (tension in BC, pulling away from B), the others are not (tension in AB pulling away from B), and compression in the boom (pushing towards point B). Break the forces into components and apply the equilibrium equations in the x and y directions. You'll get 2 equations with 2 unknowns, solve for T_AB and the boom compressive force. Watch your geometry and trig and plus and minus signs!

3. Jan 20, 2010

### Peto

Sorry for the late response I was caught up with other work, anyways,
so I drew the free-body diagram as suggested in part a,
For part b,
F_BCy = 20000N down
F_BCx = 0
F_BAy = 20000sin40 = 12855N down
F_BAx = 20000cos40 = 15320N left
Since in equilibrium, the boom must counteract these forces yielding,
F_boomy = 20000 + 12855 = 32855N up
F_boomx = 0 + 15320 = 15320N right

therefore the compression of the boom would be a^2 + b^2 = c^2,
sqrt(32855^2 + 15320^2) = 36251N

is this method correct ?