Physics 8 Friday, November 8, 2013

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Sophie Simmons
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1 Physics 8 Friday, November 8, 2013 Turn in HW10 today. I ll put HW11 online tomorrow. Today & Monday: mainly trusses. Some beams next week. Monday reading: from Vossoughi ch5+8 (centroids + second moment of area ) Second moment of area is often called moment of inertia by engineers & architects. Rotational inertia is often called moment of inertia by physicists. So I avoid the ambiguous phrase moment of inertia. Quiz 8 answers (I forgot to print): v = ωr. I = 1 2 m cr M bv 2 f I ω2 f = M b gh. The initial state has zero K.E. In the final state we have translational K.E. for the bucket and rotational K.E. for the cylinder. Total final K.E. equals (initial G.P.E. minus final G.P.E.) of the bucket. We re assuming that negligible energy is dissipated, i.e. friction in axle is negligible.

2 Why must we say the wall is slippery? Is the slippery wall more like a pin or a roller support? What plays the role here that string tension played in the previous problem? Does the combination of two forces at the bottom act more like a pin or a roller support? Which forces would an engineer call reaction forces?

3 Which choice of pivot axis will give us the simplest equation for M z = 0? (We ll get an equation involving only two forces in this case.) (A) Use bottom of ladder as pivot axis. (B) Use center of ladder as pivot axis. (C) Use top of ladder as pivot axis.

4 How would I write M z = 0 about the bottom end of the ladder? (Take length of ladder to be L.) (A) F W L cos θ + mgl sin θ = 0 (B) F W L cos θ + mg L 2 sin θ = 0 (C) F W L cos θ mgl sin θ = 0 (D) F W L cos θ mg L 2 sin θ = 0 (E) F W L sin θ + mgl cos θ = 0 (F) F W L sin θ + mg L 2 cos θ = 0 (G) F W L sin θ mgl cos θ = 0 (H) F W L sin θ mg L 2 cos θ = 0

Suppose we add to this picture a person of mass M who has climbed up a distance d along the length of the ladder. Now how do we write the moment equation M z = 0?

5 Suppose we add to this picture a person of mass M who has climbed up a distance d along the length of the ladder. Now how do we write the moment equation M z = 0? (A) F W L sin θ mg L 2 cos θ + Mg d 2 cos θ = 0 (B) F W L sin θ mg L 2 cos θ + Mg d 2 sin θ = 0 (C) F W L sin θ mg L 2 cos θ + Mgd cos θ = 0 (D) F W L sin θ mg L 2 cos θ + Mgd sin θ = 0 (E) F W L sin θ mg L 2 cos θ Mg d 2 cos θ = 0 (F) F W L sin θ mg L 2 cos θ Mg d 2 sin θ = 0 (G) F W L sin θ mg L 2 cos θ Mgd cos θ = 0 (H) F W L sin θ mg L 2 cos θ Mgd sin θ = 0 Also: what do we learn from F x = 0 and Fy = 0?

7 How many equations does the method of joints allow us to write down for this truss? (Consider how many joints the truss has.) (A) 4 (B) 8 (C) 12 (D) 15

8 How many unknown internal forces (tensions or compressions) do we need to determine when we solve this truss? (A) 4 (B) 5 (C) 6 (D) 7

9 This is a simply supported truss. How many independent reaction forces do the two supports exert on the truss? (If there are independent horizontal and vertical components, count them as separate forces.) (A) 2 (B) 3 (C) 4 (D) 6

10 (A) R Ay 2 kn + R Cy = 0, R Ax + 1 kn = 0, (2 kn)(2 m) (1 kn)(2 m) + (R Cy )(2 m) = 0 (B) R Ay 2 kn + R Cy = 0, R Ax + 1 kn = 0, (2 kn)(2 m) (1 kn)(1 m) + (R Cy )(4 m) = 0 (C) R Ay 2 kn + R Cy = 0, R Ax + 1 kn = 0, (2 kn)(2 m) (1 kn)(2 m) + (R Cy )(4 m) = 0 What do we learn by writing Fx = 0, F y = 0, Mz = 0 for the truss as a whole? (Use joint A as pivot.) (I write R Ax, R Ay, R Cy for the 3 reaction forces exerted by the supports on the truss.)

11 What two equations does the method of joints let us write for joint A? (Let the tension in member i j be T ij. For compression members, we will find T ij < 0.) (A) R Ax T AD T AB cos θ = 0 R Ay T AB sin θ = 0 (B) R Ax T AD T AB sin θ = 0 R Ay T AB cos θ = 0 (C) R Ax + T AD + T AB cos θ = 0 R Ay + T AB sin θ = 0 (D) R Ax + T AD + T AB sin θ = 0 R Ay + T AB cos θ = 0

12 What two equations does the method of joints let us write for joint C? (Let the tension in member i j be T ij. For compression members, we will find T ij < 0.) (A) T CD T BC cos θ = 0 R Cy T BC sin θ = 0 (B) T CD T BC sin θ = 0 R Cy T BC cos θ = 0 (C) T CD + T BC cos θ = 0 R Cy + T BC sin θ = 0 (D) T CD + T BC sin θ = 0 R Cy + T BC cos θ = 0

13 (A) 2 kn + T BD = 0 and T AD + T CD = 0 (B) 2 kn + T BD = 0 and T AD T CD = 0 (C) 2 kn T BD = 0 and T AD + T CD = 0 (D) 2 kn T BD = 0 and T AD T CD = 0 What two equations does the method of joints let us write for joint D? (Let the tension in member i j be T ij. For compression members, we will find T ij < 0.)

14 Physics 8 Friday, November 8, 2013 Turn in HW10 today. I ll put HW11 online tomorrow. Today & Monday: mainly trusses. Some beams next week. Monday reading: from Vossoughi ch5+8 (centroids + second moment of area ) Second moment of area is often called moment of inertia by engineers & architects. Rotational inertia is often called moment of inertia by physicists. So I avoid the ambiguous phrase moment of inertia. Quiz 8 answers (I forgot to print): v = ωr. I = 1 2 m cr M bv 2 f I ω2 f = M b gh. The initial state has zero K.E. In the final state we have translational K.E. for the bucket and rotational K.E. for the cylinder. Total final K.E. equals (initial G.P.E. minus final G.P.E.) of the bucket. We re assuming that negligible energy is dissipated, i.e. friction in axle is negligible.

Physics 8 Friday, October 25, 2013

Physics 8 Friday, October 25, 2013 Hold on to your HW8 paper. Don t turn it in yet! What we covered in class this week went much more slowly than I had expected. We spent a lot of time discussing the ideas