1. Digression Toolbox contains: Englishmathematics kinematicsforce problems* conservation of energy**conservation of momentum *includes torque, uniform circular motion **so far, just mechanical energy; includes KE of rotation  kinematics is a mathematical description of motion  force problems are solved using Newton’s laws  conservation of energy and conservation of momentum are truly fundamental

2. If I have a choice, I solve problems using energy methods. Why? How about “simpler?”  only scalars  no vectors! This web page is pertinent after the “previous” lecture, which I think is going to actually be the “next” lecture (based on room availability): http://auto.howstuffworks.com/clutch.htm

3. Bodies in Equilibrium In order for an object to remain motionless (be static), it must neither translate nor rotate: F x = 0F y = 0 z = 0 Remember to choose your axes, including the rotation axis, and explicitly indicate the positive directions! You already know how to solve force and torque problems. There is no new physics here. Just applications.

4. In this problem, you will be asked to calculate the tension in the cable, but the beam is the object of interest. Choose any convenient point O about which to calculate torques. You may or may not need to calculate sums of forces. Here’s an example of a statics problem. To solve statics problems, you must first identify the object of interest.

5. Test-Level Problem on Statics (from a Test given in Winter 1996). Physics 23 Supplementary Problem Winter/Spring 2003 A stationary ladder of total length L=13 m and weight W=216 N leans against a vertical smooth wall, while its bottom legs rest on a rough horizontal surface at a distance d=5 m from the base of the wall. Its top is located a distance H=12 m above the floor. A rope is attached to the ladder at a distance d=2m from its base. A worker pulls horizontally on the rope, producing a tension of T=260 N.

6. In the diagram to the right, sketch all the forces acting on the ladder, and the location where they act. Weight. W Weight. At L/2 up the ladder. L/2 Tension, of course. T Normal force exerted by wall. N The floor clearly exerts an upward normal force, to “balance” W. If TN then a friction force acts on the ladder due to the floor. We can’t tell the direction of this friction force from the data given. The total floor force (call it P) is the sum of the floor normal and friction forces (assumed in + directions). P

7. W L/2 T N P What are the vertical and horiz- ontal components of the force applied on the ladder by the floor if the ladder does not slip? First think. We have been studying torques. Do I need to use torques? I will have two sum of force equations (x and y components). I have three unknowns, P x, P y, and N. I need a third equation, so yes, I will have to use the sum of torques equation.

8. W L/2 T N P Think some more. Where is a good place to put the rotation axis? Therefore, I choose to put the origin at the floor contact point, where I have two unknown torques. I think force equations are easier than torque equations (don’t need to do R  F or RF  or RF sin). Answer: at the point of application of one of the forces. Therefore, I choose to put the origin at the floor contact point, where I have two unknown torques. I’ll pick clockwise for positive rotations. +

9. W L/2 T N P + OSE: F y =ma y N y + W y + T y + P y = ma y 000 -W + P y = 0 P y = +W OSE: F x =ma x N x + W x + T x + P x = ma x 00 -N + T + P x = 0 P x = N - T Don’t know N, don’t know P x. Need another equation.

10. OSE:  τ z = I z W L/2 T N P + τ Nz + τ Wz + τ Tz + τ Pz = I z 00 -NH + (D/2)W + Td  = 0 Dang! Forgot to show D/2 and d  on the diagram! Better do it now! D/2 dd Dang again! d  is not a system parameter! Better figure it out now! There are several ways to do it. I see two similar triangles. Click twice to see them. d  /d = (H/2) / (L/2)  d  = d H/L

11. W L/2 T N P + D/2 dd Continuing… -NH + (D/2)W + Td  = 0 -NH + (D/2)W + T d H/L = 0 Two equations, two unknowns, solve for P x. P x = N - T -NH + (D/2)W + T d H/L = 0 NH = (D/2)W + T d H/L N = DW/2H + T d /L P x = DW/2H + T d / L - T Numerically, P x = -175 N, so it is directed in the –x direction. You are invited to check my algebra.

12. An "art" sculpture (in the shape of a "greater than" mathematics sign) is made of two uniform segments, each of length L and weight W. The structure is attached to a pivot P at its base and to a guy-wire at point A which is directly above P. The lower member makes angle θ with the horizontal, while the wire makes angle  to the horizontal. D = 2Lsinθ is the distance between A and P. What is the tension in the wire? What is the force at the pivot P?