**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.

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**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.