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Solution, Prob. 1_3201
a. Choosing the positive direction of the x axis to the left, we have for mass A:
 F
x
A
 0  mB g  F cos   f s
Notice that the static friction was selected as positive. Its direction is actually
determined by the competition between the other forces in the problem, such that-
f s  F cos   mB g
may be positive (friction pointing to the left) or negative (friction pointing to the
right).
b. We now demand that the friction is indeed in the static regime:
  N  F cos   mB g   N
mB   mA
mB   mA
gF
g
cos    cos 
cos    cos 
Where here we substituted N  mA g  F sin  from the y axis equation.
CHECK UNITS AND LIMITING CASES!!
c. We now consider the case where F  0 . Newton's law states:
 F
x
A
 mAa  mB g  f k
m

a  B g
 mA

Where here we substituted f k   mA g .
Forces on m:
 Fx( m )  ma  F  f s  mg sin 

(m)
 Fy  0  N m  M  mg cos 
Forces on M:
 Fx( M )  Ma  f s  f k  Mg sin 

(M )
 Fy  0  N surface  N m  M  Mg cos 
And in addition it is known that f k  k N surface . Solve for f s and demand the condition
f s   s N m  M to get-
m

F  Fmax    1 mg cos    s  k 
M
