- #1
Oblivion77
- 113
- 0
Homework Statement
Here is the problem
Homework Equations
Sum of the forces in x, y and moments
The Attempt at a Solution
I am having troubles drawing a good FBD.
Doc Al said:What forces act on the rod?
Hint: Find the spring stretch as a function of angle.
Doc Al said:Which way do "normals" usually act?
Exactly.Oblivion77 said:perpendicular, so the normal at the bottom is vertical and the normal at the top is horizontal?
You tell me. How is the spring oriented? Which way does it push?Oblivion77 said:Ok, so where does the force in the spring come along in the FBD?
Yep. You should have a good handle on that FBD now.Oblivion77 said:Would the force in the spring be tension going straight up?
Doc Al said:Yep. You should have a good handle on that FBD now.
The unstretched length of the spring is unknown and unneeded. What you do need is how much the spring is stretched when the bar is at an angle. When the bar is vertical the spring is unstretched, thus the the amount of stretch is zero.Oblivion77 said:I am having some trouble seeing the stretched and un-stretched lengths of the spring. When the bar is vertical it looks like the un-stretched length is 0.
Something like that--but not that.Oblivion77 said:Is it something like 5sinα?
No, it's the height of the spring end measured from the horizontal fixed bar. Compare that to where the spring was when the angled bar was vertical (and the spring unstretched). The difference in height will tell you the amount the spring was stretched.Oblivion77 said:So is 5Cosα the stretched?
You need L-Lo, but not Lo.Oblivion77 said:I know F=k(L-Lo). but earlier you said the un-stretched length was not needed.
Doc Al said:You need L-Lo, but not Lo.
Do this. Find the height of spring end above the fixed horizontal bar when the 5m bar is vertical. (That should be easy!) Then find its height when the 5m bar is at an angle. (You've already done that.) The difference between those two heights is the amount of stretch in the spring (which is all that L-Lo is) when the bar moves from vertical to some angle.
A 2D equilibrium problem involves analyzing the forces acting on a stationary object in two dimensions. It requires the use of vector addition and trigonometry to determine the magnitude and direction of the forces involved.
The main difference is the number of dimensions being analyzed. In 2D equilibrium, forces are only acting in two dimensions, while in 3D equilibrium, forces can act in three dimensions. This makes the math and calculations more complex in 3D equilibrium problems.
The first step is to draw a free body diagram, identifying all the forces acting on the object. Then, break down each force into its x and y components. Next, use vector addition and trigonometry to find the resultant force in each direction. Finally, set up and solve equations to find the unknown forces or angles.
The key concepts include understanding vector addition, trigonometry, and the concept of equilibrium, which means the sum of all forces acting on the object equals zero. It is also important to understand how to break down forces into their components and how to set up and solve equations to find the unknowns.
2D equilibrium problems can be applied to many real-life situations, such as analyzing the forces acting on a bridge, calculating the tension in cables of a suspension bridge, determining the forces on a crane lifting a heavy object, or understanding the forces involved in an object at rest on an incline. It is also used in mechanical engineering, architecture, and physics.