10. In addition to the forces already shown in Fig. 6-22, a free-body diagram would include an upward
normal force
Nexerted by the floor on the block, a downward mgrepresenting the gravitational pull
exerted by Earth, and an assumed-leftward
ffor the kinetic or static friction. We choose +xrightwards
and +yupwards. We apply Newton’s second law to these axes:
(6.0N)−f=ma
P+N−mg =0
where m=2.5 kg is the mass of the block.
(a) In this case, P=8.0NleadstoN=(2.5)(9.8)−8.0 so that the normal force is N=16.5 N. Using
Eq. 6-1, this implies fs,max =µsN=6.6 N, which is larger than the 6.0 N rightward force – so
the block (which was initially at rest) does not move. Putting a= 0 into the first of our equations
above yields a static friction force of f=P=6.0 N. Since its value is positive, then our assumption
for the direction of
f(leftward) is correct.
(b) In this case, P=10Nleadsto N=(2.5)(9.8) −10 so that the normal force is N=14.5 N. Using
Eq. 6-1, this implies fs,max =µsN=5.8 N, which is less than the 6.0 N rightward force – so the
block does move. Hence, we are dealing not with static but with kinetic friction, which Eq. 6-2
reveals to be fk=µkN=3.6 N. Again, its value is positive, so our assumption for the direction of
f(leftward) is correct.
(c) In this last case, P=12NleadstoN=12.5Nandthustofs,max =µsN=5.0N,which(as
expected) is less than the 6.0 N rightward force – so the block moves. The kinetic friction force,
then, is fk=µkN=3.1 N. Once again, its value is positive, so our assumption for the direction of
f(leftward) is correct.