fisica1-youn-e-freedman-exercicios-resolvidos

11.74: a) The center of gravity of top block can be as far out as the edge of the lower
block. The center of gravity of this combination is then 3L 4 from the right edge of the
upper block, so the overhang is 3L 4.
b) Take the two-block combination from part (a), and place it on the third block
such that the overhang of 3L 4 is from the right edge of the third block; that is, the center
of gravity of the first two blocks is above the right edge of the third block. The center of
mass of the three-block combination, measured from the right end of the bottom block, is
− L 6 and so the largest possible overhang is ( 3L 4) + ( L 6) = 11L
12.
Similarly, placing this three-block combination with its center of gravity over the right
edge of the fourth block allows an extra overhang of L 8,
for a total of 25L 24.
c) As
the result of part (b) shows, with only four blocks, the overhang can be larger than the
length of a single block.
11.75: a)
F
B
= 2w
= 1.47 N
sinθ
= R
τ =
C
0, axis at
2R
so θ = 30°
P
F (2Rcosθ)
− wR = 0
mg
FC
= = 0.424 N
2cos30°
FA
= FC
= 0.424 N
b) Consider the forces on the bottom marble. The horizontal forces must sum to
zero, so
F = nsinθ
A
FA
n = = 0.848 N
sin 30°
Could use instead that the vertical forces sum to zero
F − mg − ncosθ
= 0
B
FB
− mg
n = =
cos30°
0.848 N, which checks.