Forces on a slope. Static and kinetic friction.

Version: 0.1 10-19-08
To do:
(i) Figure out arbitrary motion (initial velocity)

P = pitch of slope; m_s = coefficient of static friction; m_k = coefficient of kinetic friction; m = mass of block; g = acceleration due to gravity.

m_s=0.7,m_k=0.6,m=3,g=1

o=slider([0,1])

P=o*pi/2

Basis vectors, rotated basis vectors and rotated coordinates:

O=vector(0,0),I=vector(1,0),J=vector(0,1),I_p=vector(cos(P),sin(P)),J_p=vector(-sin(P),cos(P))

function(X,x,y)=x*cos(P)+y*sin(P),function(Y,x,y)=-(x*sin(P))+y*cos(P)

Size of block; distance uphill:

a=slider([0,2,40])

b=slider([0,1,40])

d=slider([0,20])

D=d+1/2*function(A,h)*n^2

Block with center at X=d and Y=b:

[(function(X,x,y)-D)/a]^40+[(function(Y,x,y)-b)/b]^40=1

C=vector(D*cos(P)-(b*sin(P)),D*sin(P)+b*cos(P))

Slope & ground:

y=x*tan(P),x>0

y=0,x>0

Gravitaional force and components parallel and perpendicular to incline:

F_g=vector(0,-(m*g))

F_(g*X)=I_p*[-(m*g*sin(P))]

F_(g*Y)=J_p*[-(m*g*cos(P))]

C,C+F_g

C+F_(g*Y),C+F_(g*Y)+F_(g*X)

C,C+F_(g*Y)

Normal force and its magnitude:

F_N=-F_(g*Y)

C,C+F_N

N=m*g*cos(P)

F_A = applied force; F_F = friction force; F_s/k = static/kinetic friction: F_Q = net X force save friction:

f=slider([-8,8])

F_A=I_p*f

C,C+F_A

F_Q=f-(m*g*sin(P))

Direction of kinetic friction hinky because it depends on velocity, not net applied force –
UNLESS you begin from rest (so force/acceleration tracks initial slip velocity)

F_F=branch(if(-F_Q,abs(F_Q)<m_s*N),if(-(m_k*N),F_Q>m_s*N),if(m_k*N,F_Q<-(m_s*N)))

C,C+I_p*F_F

Net force:

C,C+I_p*[F_Q+F_F]

Block accelerates when net applied force exceeds maximum static friction. a=1 to disable:

function(A,h)=branch(if(0,abs(F_Q)<m_s*N),if([F_Q+F_F]/m,abs(F_Q)>m_s*N),if(0,a=1))

a=slider([0,1,1])



Author: David A. Craig <http://web.lemoyne.edu/~craigda/>


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