# Wright State University Lake Campus/2017-9/Phy1110/log

## Journal

### 12:43, 30 August 2017 (UTC)

13:09, 30 August 2017 (UTC) is wikitext!

Syllabus is under construction. But we need to start. 85% tests and 15% labs*.

• Roughly 100 "homework" problems. Your exams will be based this homework. I will not see or grade your homework. All solutions are posted. Pilot might be involved. For, now everything is on Google {Wikiversity Lake Campus} (w/o quotation marks)
• Tentative attendance policy: Roughly 10% absent w/ no penalty. Graduated penalty.
• Lab policy: I need lab reports that you will usually write in class. Computer labs are OK, but must be handed in on paper

How I do lab and lecture attendance: Labs are by report. Lecture is by one piece of paper (perhaps small).

### 6 Sep Lecture

On a x versus t graph v is the slope
On a v versus t graph a is the slope


The "opposite" of slope is area under the curve.

On a v versus t graph, area is the distance travelled.  This is obvious for constant velocity and for uniform acceleration.


All this is nicely explained at

### Friday 9:00 am, 8 September 2017 (UTC)

We will probably go to a computer lab (190) today and work on this:

### Monday 9:00 11 September: Relative and 2D motion allpresent

• Relative_velocity wikipedia:Special:Permalink/791470005. This explains the notation and introduces the idea using a man walking on top of a train. It also does 2D motion, but not very well because it's missing The river needs a cork. It also does 1-D relativistic motion, but unfortunately in the wrong units: Best to let the speed of light be unity (Just ignore whatever the units are here; it is common to normalize so that fundamental constants are equal to one -- if you insist on actual units, measure time in years and distance in light-years).

In the case where two objects are traveling in parallel directions, the relativistic formula for relative velocity is similar in form to the formula for addition of relativistic velocities. (From Wikipedia, letting the speed of light equal one)

${\displaystyle {\vec {v}}_{\mathrm {B|A} }={\frac {{\vec {v}}_{\mathrm {B} }-{\vec {v}}_{\mathrm {A} }}{1-{\vec {v}}_{\mathrm {A} }{\vec {v}}_{\mathrm {B} }}}}$

### Lab Fri 22 Sep: Force Probe Project

done by wsul 14,22,23,35,36:  must record

### M 25 September 2017

working on forces quiz. Started #Formal group report #1

### W 27 Sep

#### 10 am

Look at solutions for test 2

### Wed Oct 11 Lab Newton's cannon

• Write report as you work. Begin with task and drawing that explains what v and d are:
• Find d for v = 1000m/s and for 4000m/s. Graph the points on Excel, from that graph, draw a straight line and attempt to predict hwat would appen at 2000m/s. Check your prediction and find the % error.

### 25 October Derivation of a=v^2/r

• ${\displaystyle \Delta \ell =|{\vec {r}}_{2}-{\vec {r}}_{1}|}$
• ${\displaystyle \Delta v=|{\vec {v}}_{2}-{\vec {v}}_{1}|}$S

## Reports

### Getting force probe to work[1]

1. All five present already know how to use LabQuest from previous (motion detector lab).
2. On import of spreadsheet data into lab computer: The LabQuest has a USB to ??? THE ??? was either Micro-B plug, UC-E6[b], Mini-B plug, type-A receptacle[c] ,type-A plug , or type-B plug. We think its Mini-B plug. Looks like a small version of the USB port. (Obtained hint from wikipedia:USB.
3. All present knows how to use the force probe. Easy: We plugged adaptor into 120V AC. Go to menue > New File (play). The LabQuest apparently detects that it is a force probe.
4. First Data: Attached 1kg mass to steel spring approx 5 cm by 2 cm. The stretch due to the weight was about 1 cm. Made it oscillate w/ peak to peak amplitude of about 1 cm. We estimated the period to be one second (± 0.2 sec)
5. Classroom computer was unable to recognize the Labquest device through the USB port. I will try and restart computer. After class left, Labquest file was saved as 15.qmbl (an overwrite).
6. Found Labquest2 on classroom computer, Control Panel>Hardware and Sound > Devices and Printers

### friction static with downhill pull on block [2]

${\displaystyle T+mg\cos \theta =f}$
${\displaystyle N=mg\cos \theta }$
${\displaystyle f\leq \mu N}$,
where T is tension, m is mass, θ is incline angle, N is normal force, f is static friction, and μ is the coefficient of static friction.

angle =
0 deg    .9 N
5 deg    1.0 N
15 deg    .4
force to start (downhill)
null force on scale 0.1 already accounted for
Details in reports

 there was no "error" boy somebody left as usual


### first written report by student in front of class

We believe that the block weighed .285 Kg. We placed a block on an inclined plane and attached a string attached to force sensor. We had to subtract an offset of 0.1 Newtons. (We added a data point at 11 degrees with 0.7 Newtons.)

## Lab: Force Probe (from a sandbox)

a long group lab done in Special:Permalink/1744498
From a laboratory manual in physics, to accompany Black and Davis' "Practical physics for secondary schools," (1913) (14764785362).jpg

### Abstract

1. Entire class attempted an abstract.

This lab's objective is to verify that a free body diagram describes a static system consisting of a weight suspended from two strings. Force probes (spring scales) were used to measure tension. This lab generated three equations. These equations involved tensions, mass, and angles. Each equation can be verified by comparing the LHS (left-hand-side) to the RHS (right-hand-side).

The first equation is that the tensions multiplied by the cosines of the angles are equal.
The angle sine (trying to describe equation in word..."degree of the sin..."
Um, so we are just describing equations, or what


### Guy writes equantions

${\displaystyle T_{A}\cos \theta _{A}=T_{B}\cos \theta _{B}}$

### outline

1. Calibrate the force probe with a 5 newton weight.
2. The yellow string interferes slightly, so we used a binder clip to fix the issue. Therefore, we null the force probe by .5 newtons on the right side. And also rotated the right force probe.
3. Calculations
${\displaystyle T_{1}=.8}$
${\displaystyle T_{3}=1.4}$
${\displaystyle \theta _{1}=40deg}$
${\displaystyle \theta _{3}=70deg}$
${\displaystyle \theta _{3}+90\deg =160\deg }$
${\displaystyle \theta _{1}+90\deg =130\deg }$
${\displaystyle \theta _{1}+\theta _{3}+75\deg =185\deg }$

Check:

LHS=0.8sin40deg+1.4sin70deg

off by a factor of

2.45/1.829=1.34 or 1.829/2.45=.7465= 25%error

### Procedure

in order to calibrate the force probes we rotated the right force probe approximately 2 inches so that the angle of the strings attached matched. The calibration was done using a 250g weight where each force probed equaled approximately 1 newton. For greater accuracy the strings were bound together with a binder clip. Each force probe carries the capacity for 5 newtons. Continuing the experiment we will now test the force probes using a force probe meter and a Lab Quest device to record then analyze our data.

### Free body diagram: Beware the physical triangles!

Derive important equations from this diagram

Be able to derive:

${\displaystyle T_{1}\cos \theta _{1}=T_{3}\cos \theta _{3}}$ .. and .. ${\displaystyle mg=T_{1}\sin \theta _{1}+T_{3}\sin \theta _{3}}$

### Practice FBD13:06, 9 October 2017 (UTC)

• Practice the 3-tension fbd until you can do it in your notes. The goal is to be able to derive the fundamental equations. The students wrote an abstract/summary:
• We made a visual (Free Body Diagram) to explain what T2 is and the components were used to solve for T2. Am more advanced version is in my notes to study for the tests.
• Class rewrote: as follows
We made a FBD (Free Body Diagram) for a mass suspended by two strings, one of which was horizontal. The horizontal string had tension T1, while the other string with tension T2 was oriented an angle θ above the horizontal. To solve this for T2 we used the fact that the vertical (y) component was equal to mg, since the other tension exerted no force in that direction.

### Lab Report

Our goal was to verify each side of an equation through math calculations. To do so, we had to calculate the percent error (30%) or off by a factor of 1.34. (Experimental-True Value)/(True Value). In another case we got 180=185.

${\displaystyle 500g\left({\frac {1kg}{1000g}}\right)=.5kg}$
F=mg= ≈ 5N, since g≈10m/2

### 27 October 2017 (UTC)Friday: Uniform circular motion derived

Hints:

1. Know exact meaning of ${\displaystyle {\vec {r}}_{1}}$   ${\displaystyle {\vec {r}}_{2}}$   ${\displaystyle {\vec {v}}_{1}}$   ${\displaystyle {\vec {v}}_{2}}$   ${\displaystyle r}$   ${\displaystyle v}$   ${\displaystyle \Delta \ell }$   ${\displaystyle \Delta v}$  ${\displaystyle \Delta \theta }$   ${\displaystyle \Delta t}$
2. Draw two similar triangles to represent velocity and displacement vectors. Identify the lengths of all sides of the triangle using the variables shown above (note that we have no need for scalars r1 and r2 because they both equal r.)

## End notes

1. by user:Guy vandegrift with input from class
2. Drawing and equations by user:Guy vandegrift, data and text by class members.