Statics

1. In order for an object to be in equilibrium when concurrent forces are acting upon it, the components of the force vectors must sum to 0. Thus net force x = 0, net force y = 0, and net force in all = 0.

2. For an object to be in equilibrium when parallel forces are acting on it, the net torques must sum to 0. Sum of torque clockwise will equal sum of torque counterclockwise. It is thus important to choose a fulcrum.

3. The three types of equilibrium are stable, unstable, and neutral. If an object is said to be in equilibrium and to be displaced slightly, the objecting returning to its original position is in stable equilibrium. If the object moves farther from its original position, then it is unstable equilibrium. If the object remains in its new position, then it is said to be in neutral equilibrium.

4. Elongation can be considered the change of length of an object. This is used in the definition of stress, the change in length divided by the original length. Stress is defined as force divided by the cross sectional area the force is acting on. This is in N/m^2, also called pressure. Strain is the ratio of the change in length to the original length. It has no unit, so it’s the fractional change in the length of the object. E, Elastic modulus is a constant of proportionality of a material, stress/strain. Also known as Young’s modulus. Shear modulus can be recognized as G. /\L = 1/G * F/A * L0. This is when the change in length is perpendicular to the original length. (think of shearing scissors or sears tower) Bulk modulus is the proportionality of the change in pressure times the original volume divided by the change in volume. This only applies to liquids and gases.

5. Fracture is the term used to describe what happens to an object when the stress is too great. Ultimate strength is the maximum force that can be applied to an object without breaking. N/m^2. This value depends on the material the object is composed of as well. There are Tensile Strength, Compressive Strength, and Shear strength.

Problem 1
A uniform 7.0 m long ladder of mass 15kg leans against a smooth wall so the force exerted by the wall F(wall) is perpindicular to the wall. The ladder makes an angle of 20 degrees with the vertical wall, and the ground is rough. a) Calculate the components of the force exerted by the ground on the ladder at its base, and b) determine what the coefficient of friction at the base of the ladder must be if the ladder is not to slip when a 70kg duck stands 3/4 of the way up the ladder

A) That's a big problem! Place the fulcrum where the ladder contacts the ground.
First of all, determine the net forces acting horizontally, and vertically.
Remember, torque = r _|_ F
Sum Fx = 0. F wallx = F groundx
Sum Fy = 0. F groundy = mg
Sum T = 0 Torque CW = Torque CCW
Fgroundy = 15*9.8 => 17N
This equation is found by Torque CW - Torque CCW = 0-> 7 sin 70*(Fwall) - 1.2(mg) = 0. (m is the mass of the ladder)
Fwall = 27N
Fgroundx = Fwall. F groundx = 27N

B) This is a lot of use of torque to start it off.
F cg = mg = 15(9.8)
F duck = mg = 70(9.8)
7 sin 70(F wall) - 1.2(F cg) -(3/4)(7) cos 70(F duck) = 0
Solve for F wall.
F wall = 214N.
Sum Fx = F friction - F wall = 0.
F friction = 214N
F friction = mu*Fn.
214N = mu*(70+15)(9.8) sin 70
mu = .273

Problem 2
The leaning Tower of Pisa is 55m high and 7m in diameter. The top of the tower is displaced 4.5m from the vertical. Treat the tower as a uniform, circular cylinder. What additional displacement, measured at the top, will bring the tower to the verge of toppling?

Start it off by drawing a picture in 2d, (I'm lazy right now, fill in picture later)
It'll look like a rectangle rotated about its corner on the ground. Oddly enough, when you draw a perpendicular to the ground from the corner that touches the ground to the opposite corner, the net torque is 0(a condition that brings the tower to the verge of toppling).
Now we have a nice idea of what the situation looks like. Since we want to find the additional displacement, we want to find distance from the vertical that of which will cause this to occur. So we use tan theta = diameter/length(vertical angles are equivalent).
theta = 7.3 degrees
distance from vertical = diameter/2 + (length/2)sin theta = 7m.
So, now we do 7m - 4.5m to find the additional displacement, 2.5m.

Problem 3 (thanks SBTM)
A non-uniform bar of length 1.3 m is hanging from two scales which support it so that it is horizontal and at rest. The scale which supports the bar at the very left reads 78kg. The second scale, 0.25m from the right end reads 92kg. How far is the center of mass from the mid-point of the bar?

let x be the distance from the left support to the center of mass.
let clockwise be positive. net torque = 0. 0 = x*78*g - ((1.3-.25)-x)*92g
x = 0.568
The midpoint of the bar lies at 1.3/2. So find the difference between that and x to get 0.082m.