Solving a Statics Problem with F=80N

  • Thread starter Thread starter ebinesos
  • Start date Start date
  • Tags Tags
    Statics
AI Thread Summary
The discussion centers on solving a statics problem involving a force of 80 N directed downward, with components along bars AB and AC positioned at specific angles. The user struggles with applying the dot product and correctly determining the components of the force. It is clarified that the equation used for the components is correct, but adding the components together does not equal the original force due to their directional nature. The components of the force should be treated separately, as they are not in static equilibrium when summed directly. Understanding the relationship between the components and their angles is crucial for solving the problem accurately.
ebinesos
Messages
1
Reaction score
0
hey all. I'm having problems with what should be a simple statics problem.

"The Magnitude of force F is 80.0 N. What are the madnitude of its components a) F\ab and b) F\ac that are parallel to the bars AB and AC respectively."

The picture: Force F pointing straight down. bar AB pointing 30 left of straight up (NW), and and bar AC pointing 70 degrees left of straight up (NW also).

I know i need to use the dot product somehwhere, but I can't determine what i need to do it with respect to. But I did rotate the problem 90 degrees CCW, and had the resulting equation:

|80*F\ac|cos110 + |80*F\ab|cos150=0

additional question: will F\ab + F\ac = 80. When adding up all those forces this problem doesn't seem to be in static equallibrium...

Thanks
 
Physics news on Phys.org
No, adding the force onto each bar will not equal the original force.

Each component part of your equation is right, but I do not understand why you are trying to add the components. When you see a force written as F=#i+#j, the 'i' and 'j' are 90 degrees apart.
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top