Biologia

Cap 6
1. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person’s push F in the +x direction). Applying
Newton’s second law to the x and y axes, we obtain
, max
0
s
N
F f ma
F mg
−
− respectively. The second equation yields the normal force FN = mg, whereupon the maximum static friction is found to be (from Eq. 6-1) fs,max s mg . Thus, the first equation becomes
F mg ma s −0 where we have set a = 0 to be consistent with the fact that the static friction is still (just barely) able to prevent the bureau from moving.
(a) With  s 0.45 and m = 45 kg, the equation above leads to F = 198 N. To bring the bureau into a state of motion, the person should push with any force greater than this value. Rounding to two significant figures, we can therefore say the minimum required push is F = 2.0 102 N.
(b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly
F 1.2102 N.
2. To maintain the stone’s motion, a horizontal force (in the +x direction) is needed that cancels the retarding effect due to kinetic friction. Applying Newton’s second to the x and y axes, we obtain
0
k
N
F f ma
F mg
−
− respectively. The second equation yields the normal force FN = mg, so that (using Eq. 6-2) the kinetic friction becomes fk =  k mg. Thus, the first equation becomes
F −kmg ma 0 where we have set a = 0 to be consistent with the idea that the horizontal velocity of the stone should remain constant. With m = 20 kg and  k = 0.80, we find F = 1.6 102 N.
3. We denote F as the horizontal force of the person exerted on the crate (in the +x direction), fk is the force of kinetic friction (in the –x direction), N F is the vertical normal force exerted by the floor (in the +y direction), and mg is the force of gravity.
The magnitude of the force of friction is given by fk =  kFN (Eq. 6-2). Applying Newton’s
second