# Homework Help: [tips for doing static equilibrium questions]

1. Feb 20, 2005

### F|234K

i have found that static equilibrium questions (which require the sum of all forces and torques to be zero) are the most challenging to me.

i don't know why, maybe there are a lot of different type of static equilibrium questions (they always ask the same questions but with different situations...).

hence, i would really appreciate it if you guys will tell me some of ur "strategies" when it comes to doing these questions, or some cautious things people need to pay close attention to when doing them, or just some mental notes you made to yourself when you were trying to learn how to do them.

2. Feb 20, 2005

### Davorak

Well it has more then likely been told to you before but I will say it again, draw a freed body diagram. Draw arrows where the forces are. Then break these force arrows up into their x and y directions. Practice is also important as in anything else. Some things will not be obvious on first glance, but will seem so on your third or fourth time on the same type of problem.

3. Feb 20, 2005

### F|234K

thanks.

ok, but then sometimes, there are those weird situations (like ladders) which are hard to analyse.

4. Feb 20, 2005

### Davorak

$$\begin{picture}(100,100)(0,0) \put(0,0){\line(1,0){100}} \put(0,0){\line(0,1){100}} \put(0,75){\line(1,-1){75}} \put(40,5){{\bf b}} \end{picture}$$

b is the angle between the ladder and the ground. If a weight is place on the ladder like so:
$$\begin{picture}(100,100)(0,0) \put(0,0){\line(1,0){100}} \put(0,0){\line(0,1){100}} \put(0,75){\line(1,-1){75}} \put(37.5,37.5){\circle{5}} \put(37.5,37.5){\line(0,-1){20}} \put(37.5,17.5){\line(1,1){10}} \put(37.5,17.5){\line(-1,1){10}} \put(07.5,17.5){{\bf Fa}} \put(40,5){{\bf b}} \end{picture}$$

The wall can not provide any upward force so the upward force must be provided by the part of the ladder touching the ground so:?

$$\begin{picture}(100,100)(0,0) \put(0,0){\line(1,0){100}} \put(0,0){\line(0,1){100}} \put(0,75){\line(1,-1){75}} \put(37.5,37.5){\circle{5}} \put(37.5,37.5){\line(0,-1){20}} \put(37.5,17.5){\line(1,1){10}} \put(37.5,17.5){\line(-1,1){10}} \put(07.5,17.5){{\bf Fa}} \put(75.5,0){\line(0,11){20}} \put(75,20){\line(-1,-1){10}} \put(75,20){\line(1,-1){10}} \put(95,20){{\bf Fb}} \put(40,5){{\bf b}} \end{picture}$$

Fa therefore must equal Fb

Now to find the force on the wall.

$$\begin{picture}(100,100)(50,0) \put(50,0){\line(1,0){100}} \put(50,0){\line(0,1){100}} \put(50,75){\line(1,-1){75}} \put(87.5,37.5){\circle{5}} \put(87.5,37.5){\line(0,-1){20}} \put(87.5,17.5){\line(1,1){10}} \put(87.5,17.5){\line(-1,1){10}} \put(57.5,17.5){{\bf Fa}} \put(125.5,0){\line(0,1){20}} \put(125,20){\line(-1,-1){10}} \put(125,20){\line(1,-1){10}} \put(145,20){{\bf Fb}} \put(50,75){\line(-1,0){20}} \put(30,75){\line(1,1){10}} \put(30,75){\line(1,-1){10}} \put(30,55){{\bf Fc}} \put(90,5){{\bf b}} \end{picture}$$

The angle between the ladder and the force Fb is 90-b.
Now we can use the triangle made by Fb and Fc
Fb Tan(90-b) = Fa

5. Feb 22, 2005

### F|234K

ok. thanks a lot.