Problem Solving Steps & drawing: trajectory, vectors axes free-body diagram table of known and unknown quantities, implied data”. s ( with reasoning comments ! ), ion in algebraic form, and nswers in algebraic form !!! l calculations and answers. imensional, functional, scale, sign, … analy ers and solution. ,... , , , j F a v r , , , z y x
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Problem Solving Steps 1. Geometry & drawing: trajectory, vectors, coordinate axes free-body diagram, … 2. Data: a table of known and unknown quantities,
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Data: m = 200 kg Find: (a) a force Fy to keep scaffold in rest;(b) an acceleration ay if Fy = - 400 N;(c) a length of rope in a scaffold that would allow it to go downward by 10 m
SolutionNewton’s second law: WTam
5
(a) Newton’s third law: Fy = - Ty , in rest ay = 0→ F(a=0)= W/5= mg/5 =392 N
(b) ay= (5T-mg)/m = 5 (-Fy)/m – g = 0.2 m/s2
(c) L = 5·10 m = 50 m (pulley’s geometry)
Data: L, β Find: (a) tension force F;(b) speed v;(c) period T.
Solution:Newton’s second law
j
cj amF
Centripetal force along x: RmvmaF c /sin 2Equilibrium along y: cos/)(cos mgFamgF
cos/sincos/sin
sin,tansin)/()(
2
2
LgLgv
LRRgmFRvb
g
LTLgLvRTc
cos2cos//2/2)(
Two equations with two unknowns: F,v
The conical pendulum (example 5.20)
or a bead sliding on a vertical hoop (problem 5.115)
Exam Example 12:
R
FF
gm
gm ca
ca
Exam Example 13: Stopping Distance (problems 6.29, 7.29)
x
v
0a
Data: v0 = 50 mph, m = 1000 kg, μk = 0.5 Find: (a) kinetic friction force fkx ;(b)work done by friction W for stopping a car;(c)stopping distance d ;(d)stopping time T;(e) friction power P at x=0 and at x=d/2;(f) stopping distance d’ if v0’ = 2v0 .
2 /2 , P(x=d/2) = P(x=0)/21/2 = -μkmgv0 /21/2 .(f) According to (c), d depends quadratically on v0 → d’ = (2v0)2/(2μkg) = 4d
Exam Example 14: Swing (example 6.8)Find the work done by each force if(a) F supports quasi-equilibrium or(b) F = const ,as well as the final kinetic energy K.
Solution:
(a) Σ Fx = 0 → F = T sinθ , Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .
WT =0 always since ldT
Rddsdl
0000
)cos1(sincostancos wRdwRRdwdsFldFWF
0
0 0
)cos1(sin
)sin(
wRdwR
RdwldwWgrav
0 0
sincos)( FRRdFldFWb F
Data: m, R, θ
)cossin( wwFRWWWWK TgravF
Exam Example 15: Riding loop-the-loop (problem 7.46)
Data: R= 20 m, v0=0, m=100 kg
Find: (a) min h such that a car does not fall off at point B,(b) kinetic energies for that hmin at the points B, C, and D,(c) if h = 3.5 R, compute velocity and acceleration at C.
D
Solution: v
rada
tana
a
(a)To avoid falling off, centripetal acceleration v2/R > g → v2 > gR.Conservation of energy: KB+2mgR=mgh → (1/2)mvB
2=mg(h-2R) . Thus, 2g(h-2R) > gR → h > 5R/2 , that is hmin = 5R/2.
Find: (a) kinetic energy and speed at the 1st and 2nd passages of y=0, (b) the lowest position ys and friction energy losses on a way to ys, (c) the highest position yf after rebound. m
Exam Example 17: Proton Bombardment (problem 6.76)
Data: mass m, potential energy U=α/x,initial position x0>0 and velocity v0x<0.
Find: (a) Speed v(x) at point x.(b) How close to the repulsive uranium nucleus 238U does the proton get?(c) What is the speed of the proton when it is again at initial position x0?
Solution: Proton is repelled by 238U with a force
Newton’s 2nd law, ax=Fx/m, allows one to find trajectory x(t) as a solution
of the second order differential equation: (a)Easier way: conservation of energy
(b)Turning point: v(xmin)=0
(c)It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)