fisica1-youn-e-freedman-exercicios-resolvidos

11.53: a) Take the torque exerted by F r 2
to be positive; the net torque is then
− F
1( x)sin
φ + F2
( x + l)sin
φ = Flsinφ,
where F is the common magnitude of the forces.
b) τ
1
= −(14.0 N)(3.0 m)sin 37°
= −25.3 N ⋅ m, keeping an extra figure, and
τ
2
= (14.0 N)(4.5 m)sin 37°
= 37.9 N ⋅ m, and the net torque is 12.6 N ⋅ m. About point
P, τ
1
= (14.0 N)(3.0 m)(sin 37°
) = 25.3 N ⋅ m, and
τ
2
= ( −14.0 N)(1.5 m)(sin 37°
) = −12.6 N ⋅ m, and the net torque is 12.6 N ⋅ m. The
result of part (a) predicts ( 14.0 N)(1.5 m)sin 37° , the same result.
11.54: a) Take torques about the pivot. The force that the ground exerts on the ladder is
given to be vertical, and F (6.0 m)sinθ
(250 N)(4.0 m) sinθ
V
=
+ ( 750 N)(1.50 m)sinθ , so F = V
354 N. b) There are no other horizontal forces on the
ladder, so the horizontal pivot force is zero. The vertical force that the pivot exerts on the
ladder must be ( 750 N) + (250 N) − (354 N) = 646 N, up, so the ladder exerts a downward
force of 646 N on the pivot. c) The results in parts (a) and (b) are independent of θ.
11.55: a) V = mg + w and H = T.
To find the tension, take torques about the pivot
point. Then, denoting the length of the strut by L ,
⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ L ⎞
T⎜
L⎟
sinθ
= w⎜
L⎟
cosθ
+ mg⎜
⎟ cosθ,
or
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 6 ⎠
⎛ mg ⎞
T = ⎜ w + ⎟ cotθ.
⎝ 4 ⎠
b) Solving the above for w , and using the maximum tension for T ,
mg
2
w = T tanθ
− = (700 N) tan55.
0° − (5.0 kg)(9.80 m s ) = 951 N.
4
c) Solving the expression obtained in part (a) for tan θ and letting
mg
ω → 0,
tanθ
= = 0.700, so θ = 4.
00°
.
4
T