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Physics I Enhancement (Phy.261) 2022 Fall

Topic Three - Vector Components


Readings for Topic 4 :

centripetal components : Ch.4 §5 , Ch.6 §3 - §4
elastic Force : Ch.6 §2
Gravitation Source : Ch.9 §1 - §4

elastic material : Ch.14 §4


Topic 3 HomeWork practice (not for grading)

HomeWork 7 (classes 11,12,13) for grading , due Friday Sep.17 ... Monday's Quiz is about this ...

  1. A tower for radio antenna is 30[m] tall ; at its top, three guy wires apply Forces (3-d problem!)
        F1 = 500[N] Eastward + −700[N] Upward (pulling "downward", toward its anchor on the ground)
        F2 = 1000[N] , in the direction 40° downward from horizontally North ;
        F3 = |F3| at a 45° angle below horizontal ... toward an anchor which you need to locate.
          0) Draw a "side view" for F2 (seen from the East so that F2 is along your paper)
          ... what is F2 's North component?
          a) Draw a "top view" for all 3 Forces applied to the tower top
          ... What is the "East" component , and the "North" component for F3 , if the tower top has a = 0?
          b) Use Pythagoras to find the total "horizontal" component for F3 .
          c) Draw a "side view" for the tower, with guy wire 3 along the paper ... how far from the base of the tower should guy wire 3 be anchored?
          d) Draw a "side view" for F3 , with that Force along the page ... what is F3 's "Upward" component?
          e) What Force from the ground would make all 4 Forces "add up" to zero? (ignore weight)
  2. A 1.5[Mg] log rests on a 15% grade (that rises 15[m] for 100[m] horizontal distance) ;
      the surface is very slippery mud. A rope from the log is parallel to the slope and passes over a pulley, attaching to a 0.5[Mg] counterweight that hangs freely.
          a), b) find the acceleration of the log, and the Tension in the rope.
          c) Find the Normal (Pressure) Force by the muddy slope applied to the log.
          d) If the mud is NOT perfectly slippery, but has μ = 0.14 , use the friction Force to find the log's new acceleration.
  3. Last week a 2[ton] device was removed from the Science building roof ; as it hung from the crane by a cable, a rope pulled sideways (+x) so the cable was 7° from vertical.
          a), b) Find the Tension in the cable, and the Tension in the rope.
          c) If the rope Tension suddenly doubled, what would be the initial acceleration for the device?
  4. A punter stands in a narrow flatboat (a punt) and pushes a pole against the river bottom. If
          the total mass being punted is 400[kg] , and you punt with 150[N] along (parallel to) the pole
          while the pole is 30° from vertical , calculate :
          a) horizontal & vertical Force components applied to boat+people system by 3 main causes
          b) the boat system's initial acceleration components (ignore viscous friction , −bv ≈ 0)
          c) the change in velocity during a 5[s] stroke   . . . d) what happens just as you stop pushing?
  5. A 50[kg] ice skater travels counter-clockwise around a 3m radius circle at constant speed ...
      ... with the circle center at origin : at t=0 she is 19.5° past the x-axis
      ... 1.335 seconds later she still has to go 19.5° to get to the y-axis . Find vector components for :
          a) initial location (t=0)   b) later location (t=1.335[s])   c) average velocity ... keep several digits .
          d) Determine the constant speed, from the fraction of the circumference traveled (51°/360°)
          e) how does this compare with magnitude of the average vavg .
  6. . . . same ice-skater scenario . . . use the constant speed , along the circle , to determine components for
          a) initial velocity   b) velocity at t=1.335[s]  
          c) average acceleration   d) average Force [from a]   ... e) what external object applied that Force?
  7. A 1200[kg] bison has been roped by 2 bison ranchers :
        rope one pulls the bison directly along x with 400[N] Force toward rancher 1;
        rope two pulls the bison toward rancher 2 , at 53° from −x toward +y , with Tension 800[N] .
          a) compute the net Force (vector!) applied to the bison by these two ropes
          b) compute the expected acceleration of the bison , in magnitude and direction .
      The bison is seen accelerating 2× that quickly, but at a 30° angle counterclockwise from that direction .
          c) What other Force (components, or strength & direction) is being applied to the bison?
          . . . what might be causing that Force? (speculate!)
  8. In the FirstMars Olympics (2112), a 0.8[kg] javelin, thrown with speed 30.5[m/s] at 41° above horizontal,
      hits the scoreboard with its velocity 17° below horizontal. Using gmars = 3.72[N/kg] ,
          a) find both velocity components for initial condition , and for final condition
          b) find the time duration for javelin flight before hitting scoreboard (Mars!) . . . {hint: what change do we know? what made it change?}
          c) find the displacement components : Δx , and Δz
          d) If you had been given Δx and Δt instead of the initial velocity vector, outline a scheme to deduce v.
  9. Wiley E. Coyote (20[kg]) lights his rocket-powered roller-skates just as Roadrunner passes him at 15[m/s] ; one rocket thrusts forward 60[N] and one thrusts upward 60[N] ... < oops, Wiley did not fasten rocket #2's orientation > .
          (a) How long (time) does it take for Wiley's speed to match Roadrunner's , on this level road?
          . . . (you should be able to verify that Wiley does stay in contact with the road ... what is FN ? )
          (b) How far does Wiley need on level straight road before he catches Roadrunner? ... going how fast, by then?
      ... the road curves sharply moments (t − tcatch ≈0) before then ; Wiley goes straight instead, off a 400[m]-high-cliff, with rockets blazing.
          (c) How far from the cliff edge should we look , to see his puff of dust at the ground ? {show all steps!}
          (d) What would Wiley's velocity (vector components) be, just before he lands (if rockets keep blazing)?

  1. A 1.5[kg] lab cart on a ramp, inclined 22° from horizontal, is pushed (by hand)
      with a 10[N] Force up-wardly along (parallel to) the ramp .
          a) compute the Force caused by Earth gravity applied to the lab cart ... what direction is this?
          b) compute the two components of weight that are parallel to the ramp, and perpendicular to the ramp;
          c) calculate the total Force component along the ramp ... what is the cart's acceleration?
          d) does the cart accelerate into (or out from) the ramp surface? ... so, what's the (Normal) Pressure Force by the ramp?
  2. a 0.058 [kg] tennis ball was moving −20[m/s] x + −2[m/s] z ; then Venus hits it with her racquet, which is tilted upward 8° , and squishes the ball 12[mm] before the ball's x-velocity component momentarily stops (soon after, it bounces off with positive x-velocity!)
          a) What was the ball's average acceleration x-component during this "squish" half of the collision?
          b) What was the average Force x-component during the "squish" ?
          c) If the racquet's Force is thru the surface (along A, at 8° upward) , what is the vertical Force component by the racquet?
          d) including the weight of the tennis ball, what is the total Force applied to the ball?
          e) If the same average Force is applied for the "un-squish" half of the collision, what would the ball's velocity components end as?
  3. The 80[kg] hero drags a 102[kg] unconscious alien 12000[m] across the (horizontal) sand in 5 [hours],
      by pulling with 400[N] on parachute cord that is 37° from horizontal.
          a) compute the horizontal & vertical Force components applied to the alien by the cord ;
          b) compute the Force components applied to the alien by Earth's gravity
          c) calculate the two Force components applied to the alien by the sand {hint: what is the alien's accel?}
      . . . the component along the sand is usually called sliding (dry) friction.
          d) compare the Force component along surface (|| v) to Force component thru surface ( | v) ... [divide Ffr by FN]
      . . . that ratio (called friction co-efficient , μ ) depends mostly on the surface condition (roughness)
  4. . . . same hero-with-captured-alien scenario . . .
          a) use Newton#3 to write the horizontal & vertical Force components applied to the hero by the cord ;
          b) compute the Force components applied to the hero by Earth's gravity
          c) calculate the two Force components applied to the hero by the sand
      . . . notice that the hero's Friction-to-Normal ratio is greater than the alien's (or else hero's feet would slip)

Previous Topic Reminders :

USE UNITS all thru your answer ... don't just append them onto the final number !

DRAW a DIAGRAM for each scenario . . . label the diagram with SYMBOLS , as you read
. . . condition points , words like   "start" , "end" , "turn-around" . . . with vector arrows
. . . process spans as brackets or lines , words like   "average" , "moved" , "change" , "Δ" , "t=0→2" ...

Write statements as symbols first ; manipulate symbols before plugging numbers.
. . . each statement should have a SUBJECT . . .
. . . keep track of adjectives such as   "initial" , "average" , "final" , "stopped" . . . "change"

subscripts for "total" or "system" , "part A" ...

keep track of the SOURCE for the external Forces applied to the (passive) object

Recognize whether you are predicting based on theory . . . reasoning from causes to their effects ... (what should it do?)
      or whether you are deducing from observation . . . (what does this imply about it?)

Get un-stuck (often) by wondering "why isn't it the same as it used to be?"

Topic 3 Summary :

Quantities which have direction are called vectors. Vectors are drawn as arrows (with labels, of course).
... we will pattern our treatment for all object vectors on Displacement .
We will always use coordinates with axes that are perpendicular to each other ; then,
motion in any direction is independant of motion in all other directions
. . . an event condition relates these directions by its time
. . . algebraically, the key is to keep each component (x) separate from the others (y and z) !
      then , a "big & scary" 2-dimensional scenario becomes 2 "small" 1-dimensional scenarios ... connected via time .

Vector representations : first choose an origin and coordinate system (showing each positive direction).
drawing : a location vector is drawn as an arrow with its tail at the origin, its tip at that location.
      this represents a location as if it is a radially-pointing range at some angle from the coordinate axes ... hence the abbreviation r .
. . . vectors of other quantities can have their tail at the object being described,
      but must point in the direction of the quantity (e.g, v in velocity's direction).

bold italic letters represent vector quantities !
component : The ordered set of the location coordinates ( x , y , z ) . . . = (right , forward , up)   or   (East , North , Zenith)
      is the algebraic way to write this location vector (Cartesian component form)
. . . a velocity vector is written as the set of velocity components ( vx , vy , vz ) in the same order ... .


Use Trig (right) triangle formulas to obtain components ... and/or   extract a diagonal from a component
. . . soh cah toa   ... for right triangles
. . . Pythagoras : diagonal   d² = a² + b² + c²   ... for orthogonal components

if NOT a right triangle, a·b = a² + b² + 2ab cos(θa,b)


In almost every case, it is best to :
0) diagram the "Before" scenario ... include arrows (and labels!) for the vector condition quantities.
½) notice which vector quantities might be different in the "Meanwhile" and the "After" diagram
1) draw and label the important process quantities , from Before to After ,along the path taken.
2) choose a coordinate system ; often parallel (and perp.to) the path v or p ... except in free-fall .
3) split important vector quantities into (parallel , perpendicular) or ( horizontal , vertical ) components.
4) write how the each direction's vector quantities relate to one another ... or to TIME.

Addition : drawing : usually done on a small coordinate system , separate from the main diagram (so it doesn't get cluttered).
      the first arrow tail is at the origin of the addition coordinate axes (x=0,y=0,z=0).
      its tip points the same direction from its tail as it does in your sytory-book diagram ... to ( x1 , y1 , z1 ).
. . . the vector being added to the previous has its tail placed at the previous tip
  its tip points the same direction from its tail as it does in your story-book diagram ... to (x1 + x2 , y1 + y2 , z1 + z2 ).
. . . vector arrows are appended tail-to-previous-tip in sequence, keeping each aligned with its arrow on the first diagram.
The Sum ("Result for the addition") of the vectors is the arrow from first tail (origin) to last tip .
algebraically : the Result of the addition is the list of components : (x1+x2+x3... , y1+y2+y3... , z1+z2+z3... ) .

Adding a vector to itself results in a vector twice as long , in the same direction ; i.e, v + v = 2 v ...
. . . the negative of a vector points in the opposite direction
... a vector added to its opposite ... Δx + ( − Δx ) ... makes the "zero length vector" ... = 0 = (0, 0, 0) .


spring Force : depends on the stiffness konstant k of the spring , which is a scalar property (almost) intrinsic to a particular spring
      . . . (although a real spring's stiffness does change with age or abuse)
. . . and also depends on the distance and direction which that end of the spring has been stretched : s .
      the spring's end pulls in the opposite direction from its stretch vector (or pushes opposite its compression vector)
      . . . this is parallel to the length of the spring !     neglect the mass of ideal (textbook) springs
=> Fspring = − k s .

Two springs hooked end-to-end in series (anchor...spring1...spring2...Force , as : |-v^v^v-·-^v^v^v→ )
. . . will stretch farther than either of them ...the stretch distances add , so stotal = s1 + s2 ,   but their Tensions are the same
=> 1 / kseries = 1 / k1 + 1 / k2 .
this implies that longer springs have smaller stiffness k than otherwise similar short springs .

If two springs are both connected to the anchor and to the external Force, in parallel, ( |‾‾‾→ ) , their Forces add
. . . their stretch distance is the same, but Forces add :
=> kparallel = k1 + k2 .
this implies that thicker wire springs have larger stiffness k than otherwise similar thin-wire springs .

elastic modulus : even straight steel wire stretches a little bit if pulled with enough Tension ...
. . . the cause of the stretch is stress : Tension spread across the cross-sectional wire Area , in [N/m²] or [N/atom²]
      the effect of stress is strain : fractional length increase , ΔL /L , due to increased average atom-to-atom distance parallel to T
. . . the ratio : stress/strain is Young's modulus Y , which is a property of the material (not just the object)
      Y describes the Pressure that would cause its length to double (if it didn't break first)
=> P = Y ΔL/L , for Pressure in solids .

for fluids, which have no intrisic shape, we describe the change in Volume as they are compressed :
. . . each atom itself is a little bit compressible , with non-infinite elastic stiffness .
=> P = −B ΔV/V , where B is called the Bulk Modulus for the material.


Pressure : more intense Pressure at deeper depths within a non-accelerating fluid ;
. . . ΣFz=m az = 0 => Pbottom·Aup = Ptop·Adown + mfluidg ;   m = ρ A×δh
      Pressure at bottom of fluid sample = Pressure at top of fluid sample + ρ g δh ...
=> δP = ρ g δh .

Pressure Units are [N/m²] ; this set of units is also called a "Pascal" , abbreviated Pa .
. . . but other pressure units are also used : besides the "psi" = pound-Force per square inch , there are also
      torr : pressure caused by 1[mm] deep mercury column in standard Earth gravity (9.81[N/kg) . . . and
      "inches of water" , P caused by 1" deep column of water in 9.81[N/kg] gravity . . . and
      "atmosphere" , the pressure caused by "standard" depth (to "sea level") dry Earth atmosphere in standard Earth gravity

. . . Force difference : Fon bottom − Fon top is called "Buoyant Force" by the fluid on object ;
=> δF = − ρfluid g Vobject . . . upward, as it used to cancel the weight of fluid that isn't there anymore ... the "displaced fluid"
      typical Bouyant Force for condensed matter (solid or liquid) immersed in sparse matter (gas, plasma) is −0.001 × mg ;
      but condensed matter can float in other condensed matter ; and some gases float in other gases .
=> for many situations bouyant Forces can NOT be ignored .


Important statement #5 about how the Universe works:

Gravity : every mass in the universe pulls on every other mass in the universe

each source mass Ms contributes to the gravity field g at each location.
. . . the total gravitational influence emanating from source mass Ms ~ GMs , pointed toward the source mass ,
      with G = 6.67E−11[Nm²/kg²] = 66.7[N/kg · (Mkm)²/(1E30kg)] ,
      or . . . G = 5.931E−3[N/kg · (au)²/MSolar] <= astronomical units
. . . this influence spreads out, 2 ways in 3-d space, so the contribution to g is weaker far from the source ;
      so strength contributed goes as 1/"distance squared" = 1/r² from that source
      . . . { its contributed g piercing thru the whole surface of any surrounding shell is the same => g·A = 4π GMs }
=> gby M = G M / r²M-to-point (toward M) . . . then use   Fgravity, on m = mg ; NOTICE : m ≠ M !

The contribution to g from every source mass (reasonably nearby) must be added as a vector
      to get the total   g   at that place (the field point) .
. . . so there IS only ONE gravity field vector at any location , after adding all the contributions ... (3 components, one vector)

Since the gravitational influence spreads with distance from the souce Mass ,
      the gravity field is more intense on the "near side" of an object than it is on the "far side" of the object.
. . . relative to the average field at the object , these "excess" and "deficit" Forces (per unit object mass) seem to pull outward
      Tidal Tension (per unit mass in the object) = gneargaverage   and   gaveragegnear   .
=> Δg = g / (1 ± Δr/R
. . . this is almost the same on each side , and if the stressed object is small (compared with source Mass distance ; Δr/R << 1 )
=> Δg ≈ ( d/dr{g(r)}|R ) · Δr   . . . ≈ 2gΔr/R   .

For a "field point" anywhere inside a hollow shell of mass, the gravity field contributions from all the shell source Mass will cancel !
. . . only M located non-isotropically around the field point contribute to g .
=> gravity is weaker at the bottom of a deep hole into Earth ... only the "inside" Mass counts .



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