fisica1-youn-e-freedman-exercicios-resolvidos

11.95: Assume that the center of gravity of the loaded girder is at L 2,
and that
the cable is attached a distance x to the right of the pivot. The sine of the angle
2
2
between the lever arm and the cable is then h h + (( L 2) − x)
, and the tension is
obtained from balancing torques about the pivot;
⎡ hx ⎤
T ⎢
⎥ = wL 2,
2
2
⎢⎣
h + (( L 2) − x)
⎥⎦
where w is the total load (the exact value of w and the position of the center of
gravity do not matter for the purposes of this problem). The minimum tension will
occur when the term in square brackets is a maximum; differentiating and setting
the derviative equal to zero gives a maximum, and hence a minimum tension, at
x ( 2
min
= h L)
+ ( L 2). However, if x
min
> L,
which occurs if h > L 2,
the cable must
be attached at L, the furthest point to the right.
11.96: The geometry of the 3-4-5 right triangle simplifies some of the
intermediate algebra. Denote the forces on the ends of the ladders by
F
L
and F R
(left and right). The contact forces at the ground will be vertical, since
the floor is assumed to be frictionless. a) Taking torques about the right end,
F
L( 5.00 m) = (480 N)(3.40 m) + (360 N)(0.90 m), so FL
= 391N. FR
may be found in a
similar manner, or from F
R
= 840 N − FL
= 449 N. b) The tension in the rope may be
found by finding the torque on each ladder, using the point A as the origin. The
lever arm of the rope is 1.50 m. For the left ladder,
T( 1.50 m) = FL (3.20 m) − (480 N)(1.60 m), so T = 322.1 N (322 N to three figures). As a
check, using the torques on the right ladder,
T ( 1.50 m) = F R
(1.80 m) − (360 N)(0.90 m) gives the same result. c) The horizontal
component of the force at A must be equal to the tension found in part (b). The
vertical force must be equal in magnitude to the difference between the weight of
each ladder and the force on the bottom of each ladder, 480 N−391 N = 449
N−360 N = 89 N. The magnitude of the force at A is then
2
2
(322.1 N) + (89 N) = 334 N.
d) The easiest way to do this is to see that the added load will be distributed
at the floor in such a way that
F ′ ( 0.36)(800 N) 679N, and ′
L
= FL
+
= FR
= FR
+ (0.64)(800 N) = 961N. Using these
forces in the form for the tension found in part (b) gives
F′
(3.20 m) − (480 N)(1.60 m) ′(1.80 m) − (360 N)(0.90 m)
= L
FR
T
=
= 936.53 N,
(1.50 m)
(1.50 m)
which is 937 N to three figures.