Physics 1 Topic 3: . . . Reminders from Topic 2: draw each Force arrow, label by its cause (source)
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. Some 2-d scenarios are decomposable into separate 1-d scenarios
Scenario 1a . . . Forces applied along diagonals to a flat surface
If the mass stays on the surface, the velocity will be along the surface, so the acceleration will be along the surface.
. . . choose a coordinate axis x along the surface ... the other (y or z or n) will be ⊥ to the surface .
Decompose each Force vector into components: 1) along the surface and 2) perpendicular to the surface
. . . each Force becomes the hypotenuse in a right triangle ... soh cah toa shows the legs as Fsin θ and Fcos θ .
A 1.5 kg block is pulled by a spring stretched 3 N at an angle 30° leftward from straight up, as the block moves leftward with constant speed.
a) What support Force does the table provide? ... less than it used to before the spring's upward help !
. . . Fgrav,z = mg = 1.5kg(−9.8N/kg) = −14.70N (that is, ↓)
Fspring,up = 3N cos30° = +2.60N
. . . Σ Fz = m aup = 0 => Ftable,z = −Fgrav,z − Fspring,z = −(−14.70N) − 2.60N = 12.10N (↑)
b) What's happening horizontally, Force-wise?
. . . Fspring,leftward = 3N sin30° = −1.50N , which would accelerate the block 1 m/s² ... but it seems to be constant velocity (Δv = Δp =0)
=> so, there must be a rightward Force which cancels the spring's leftward component
... this is friction, which comes from the table (contact thru the microscopic surface roughness) into the block
Scenario 1b . . . Cart on a Ramp, with Forces applied along diagonals to an inclined surface
This might look "harder" than scenario 1a, but it isn't any different ... just tilt your coordinate system.
. . . the gravity Force now has a normal component into the ramp: Fgrav, | = m g cosθ (into ramp) ... it is cancelled by the ramp's Force outward (Normal, into the cart)
and it now also has a small component along the ramp: Fgrav,|| = m g sinθ (downramp)
. . . the downramp component all by itself would cause the cart to accelerate downramp: a|| = m g sinθ / m = g sinθ (downramp) ... if no friction and no other pushes or pulls.
each additional Force must be decomposed into its downramp component , and all of them added before setting equal to ma
... their "into-ramp" components will increase or decrease the Force from the ramp.
Scenario 2a . . . Projectile to a wall, ignoring air viscosity and drag Forces
gravity Force (↓) is the only Force, so the acceleration direction is (9.8 m/s²) down ... please choose up as your (+) z-axis ...
Shooting a basketball ... use the horizontal distance to find out what time it becomes "over the basket";
Δx = vx,avgΔt = vdiag,i cos θ Δt ... no acceleration, simple topic 1.
... 7m/s at 30° becomes 6.06m/s (x) and 3.5m/s (z) ... so Δx= 5m takes 0.825 seconds.
. . . then find the vertical displacement during that time span ... including the squared acceleration term.
Δz = vz,avgΔt = vdiag,i sin θ Δt + ½ az,avg(Δt²) .
... Δz = 2.888m + (−)(4.9m/s²)(.681s²) = 2.89m − 3.33m = −0.447m ... below the 2m initial height !
. . . this ball was only rising for 0.35 s (since Δv = a Δt) , and its average vertical speed during the rise was ½(3.5m/s) , so it rose 0.613 meters
I'll bet we make this basket if we use 11m/s (off the backboard)
Scenario 2b . . . Projectile to ceiling or floor
. . . the "ending time" is usually determined by the vertical location, so you solve the quadratic using the quadratic formula.
A t² + B t + C = 0 A = &plusm;½g ; B = vi or vf ; C = Δz .
. . .
. . . B. Friction Force opposes relative motion slide along contact: Ffriction ≈ μ FN
. . . friction depends mostly on how strongly the surfaces are being pushed together
(microscopic contact Area between solids has little to do with macroscopic measurable Area)
. . . different surface materials have different friction coefficients μ ≤ 100% effective ; you look them up in a data table
. . . C. Constant-Speed Motion along a Circular Path: a·r = −v·v
mass on a spring, rotating around a horizontal circle
200 gram = 0.2kg; spring ≈30° from vertical, circle ≈0.2m radius, about once per second
. . . spring Force vertical component cancels gravity Force, so is (0.2kg)(9.8N/kg) ≈ 2N upward.
. . . spring Force inward component makes the mass accelerate inward "centre-pointedly"
mass traveled 2πr = 1.25m each second, so acentr = v²/r ≈ 7.89 m/s² .
(0.2kg)(7.89m/s²) ≈ 1.58N inward.
. . . Pythagoras tells us that the spring diagonal Force should be [sqrt(6.5N²)] = 2.55N
Scenario 1 . . . Banked Curve around horizontal circle, with no friction
. . . gravity's Force is down, ⊥ to the −radius (center-pointing) vector ... provides no acentripetal .
. . . road surface ⊥ (Normal) has one component (↑) that cancels gravity's Force, and one component that provides mAc ;
a car traveling around a 100m radius I-64 entrance ramp ... suppose it is banked at 15°.
. . . Froad,N cos 15° = − mg ⇒ Froad,N = mg /cos(15°) = m 10.15N/kg .
. . . Froad,N sin 15° = m 2.626 N/kg inward , so v² is supposed to be −a·r = 262.6 m²/s² ⇒ vdesign = 16 m/s ≈ 36 mi/hr .
