Physics 1 Topic 4: . . . Reminders from Topic 3: put the scenario info on a diagram, at the place the info is about; label the arrows!
. . . you want to immediately split each Force into components parallel to (and perpendicular to) the acceleration (Δp) direction
gravity Force: proportional to the object's mass m ... points straight down!
... also depends on the gravity field (intensity and direction) at the place that the object's mass is:
=> Fgrav. = m g . . . loosely called the object's weight.
Tension Force: in rope , cord , string ... ropes pull inward at both ends , with equal strength , along the rope
= > FTension = T = parallel to rope .
spring Force: proportional to the spring's stiffness ks ... we will usually treat this as uniform
. . . (having the same value, regardless of how far the spring has been stretched or compressed)
. . . also depends on the distance and direction that the spring has been stretched : s = ΔL :
= > Fspring = − k s . . . opposite the stretch or compression
Pressure Force: proportional to the pressure at the contact surface ... solid surfaces "react" to being pushed.
. . . also depends on the contact Area ... pressure Force pierces through (perpendicular to, "Normal" to) the Area:
=> F | = P A ... usually strong enough that the surface is not deformed
the vector SUM of all external Forces applied to an object (thru the object's system-bag)
. . . causes the object's total mass to accelerate . . . or its momentum to change over time .
You write : ΣF|| = m a|| ... and : Σ F_|_ = m a_|_
. . . A. Banked turn with friction ... 4 arrows, but Forces are only 2-d (momentum is ⊥ Force triangle)
Scenario 1a . . . too fast for the bank design angle
1500 kg race-car going 75 m/s around a horizontal curve, with center 800 m to right and bank angle 30° ... what friction coefficient is needed?
a) Force diagram: Fgravity is down, road Normal is diagonal up&right, road friction is diagonal downramp ... ma is rightward.
. . . Fgrav,z = mg = 1500kg(−9.8N/kg) = −14700N (↓)
Froad,z + Ffrict,z (negative) = +14700N (↑)
. . . Σ Fx = m ax = m v²/R = 10550 N (→).
notice that the car needs more than mg tan θ = 8487 N, so friction is necessary.
. . . not all of the excess 2063 N comes from friction ... Froad,N increases as Froad,f increases.
it is the hypotenuse of the excess Froad,⊥ + Froad,|| that needs to be the 2063 N.
... so the friction Force required is 2063 N cos 30° = 1787 N
... the road Normal is m g cos 30° + 2063 N sin 30° = 12731 N + 1031 N = 11698 N
=> the friction coefficient μ = Ff/FN = 0.153 ... unitless, quite small and easily obtained.
. . . B. cause for (planetary) gravity
Gravity : every mass in the universe pulls on every other mass in the universe
. . . each source mass contributes to the gravity field g at any place:
the gravitational flux emanating from a source mass "M" influences its local environment ~ making "Tension" in space ... pointing toward the mass.
total is 4π GM ... (maybe visualize it as the gradient of the source's "mass aura") it G = 6.67E−11[Nm²/kg²] = 66.7[m/s² · (Mkm)²/(1E30kg)] ,
or . . . G = 5.931E−3[m/s² · (au)²/Mass_Sun] <= astronomical units
. . . at large distances this influence has spread, 2 ways in 3-d space, so its contribution to g is much weaker far from the source.
the contribution to g from every source mass (reasonably nearby) must be added as a vector .
=> gby M = G M / r²M-to-point (toward M) . . . then use Fgrav to m = m g ; NOTICE : m ≠ M !
Knowing that Earth's radius is 6400 000 m, and gravity's intensity at its surface is 9.8 N/kg (↓), what is Earth's mass?
. . . gby Earth = G M / r² = 9.8 N/kg . . . solve this for MEarth = g REarth²/G
... hoping that other masses (Moon & Sun) are far enough away that we can ignore their contributions to gravity here
=> MEarth = (9.81 N/kg)(6.4E6 m)² /(6.67E−11 Nm²/kg²) = 6.02E24 kg ... like it says inside the book cover.
. . . C. planets orbit the Sun along "almost circular" paths: angular velocity ω = v/r = Δθ / Δt , in radians/second
. . . Every orbit travels all the way around ... 2π radians ... in one orbit time period => data tables list orbit times
if we also know the orbit distance, we can figure out the mass of the thing being orbited.
. . . in ac = v² /r , replace both v's with r ω
=> ac = r ω² . . . greater acceleration needed at greater distances, for the same angular speed rotation.
We know Earth's orbit distance is 149.6E9 m from the Sun ... what is the Sun's mass?
. . . our orbit time is 1 year ... 365.2564 days (star-to-star) ... = 365.2564 × 24 × 60 × 60 = 31 558 153 seconds
ω for Earth orbit = 2 π / orbit time = 0.1990986 radian/Ms ... Mega = "million" ... = .199 μrad/s ... μ = "micro" = "millionth"
. . . ac = r ω² = (149.6E9 m)(0.1990986E−6 rad/s)² = 0.005930 m/s² ... caused by Sun's gravity g .
re-arrange the formula for calculating gravity ... solve it for the Sun's mass:
. . . MSun = g r² /G = .0054(149E9)²/6.67E-11 = 1.9898 kg.
