fisica1-youn-e-freedman-exercicios-resolvidos

10.1: Equation (10.2) or Eq. (10.3) is used for all parts.
a) ( 4.00 m)(10.0 N) sin 90° = 40.00 N ⋅ m, out of the page.
b) ( 4.00 m)(10.0 N) sin 120° = 34.6 N ⋅ m, out of the page.
c) ( 4.00 m)(10.0 N) sin 30° = 20.0 N ⋅ m, out of the page.
d) ( 2.00 m)(10.00 N) sin 60° = 17.3 N ⋅ m, into the page.
e) The force is applied at the origin, so τ = 0.
f) ( 4.00 m)(10.0 N) sin 180°
= 0.
10.2: τ
1
= −(8.00 N)(5.00 m) = −40.0 N ⋅ m,
τ = (12.0 N)(2.00 m) sin 30°
= 12.0 N
2
⋅
where positive torques are taken counterclockwise, so the net torque is − 28.0 N ⋅ m, with
the minus sign indicating a clockwise torque, or a torque into the page.
m,
10.3: Taking positive torques to be counterclockwise (out of the page),
τ = − 0.090 m) × (180.0 N) = −1.62 N ⋅ m, τ = (0.09 m)(26.0 N) = 2.34 N
1
(
2
⋅
3
= ( 2)
(0.090 m) (14.0 N) = 1.78 N ⋅ m,
.50 N
τ so the net torque is 2 ⋅ m, with the
direction counterclockwise (out of the page). Note that for τ
3
the applied force is
perpendicular to the lever arm.
m,
10.4: τ
1
+ τ2
= −F1
R + F2
R = ( F2
− F1
) R
= ( 5.30 N − 7.50 N)(0.330 m) = −0.726 N ⋅ m.
10.5: a)
b) Into the plane of the page.
r
c) F
r
× = [( −0.450 m) iˆ
+ (0.150 m) ˆ] j × [( −5.00 n)ˆ i + (4.00 N)ˆ i ]
= ( −0.450 m) (4.00 N) − (0.150 m)( −5.00 N) kˆ
= ( −1.05 N ⋅ m) kˆ