posted by ally .
In the figure the coefficient of static friction between mass (MA) and the table is 0.40, whereas the coefficient of kinetic friction is 0.28 ?
part a) What minimum value of (MA) will keep the system from starting to move?
part b) What value of(MA) will keep the system moving at constant speed?
physics - Elena, Tuesday, May 8, 2012 at 1:35pm
Block on the table m(A) = m1,
block on the cord m2,
the coefficient of static friction is k1=0.4,
the coefficient of kinetic friction is k2 =0.28
T = F(fr) = k1 •N = k(s) • m1 •g,
Block B: T = m2•g.
k1 • m1 •g= m2•g,
m1 = m2/k(s) = m2/0.4.
T = F(fr) = k2 •N = k2 • m1 •g,
T = m2•g.
k2• m1 •g= m2•g,
m1 = m2/k2 = m2/0.28.
- thanks for the help, i am having a lot of trouble with this problem, what is m2 ?
nvm i got the answer, i was just cofused on how you explain it. Thank you . I got 5 for part a and for part b got 7.14. thank you