This preview shows page 1. Sign up to view the full content.
65. (a) To hold the crate at equilibrium in the final situation,
G
F
must have the same
magnitude as the horizontal component of the rope’s tension
T
sin
θ
, where
is the
angle between the rope (in the final position) and vertical:
=
F
H
G
I
K
J
=°
−
sin
.
.
..
1
400
12 0
19 5
But the vertical component of the tension supports against the weight:
T
cos
=
mg
.
Thus, the tension is
T
= (230 kg)(9.80 m/s
2
)/cos 19.5° = 2391 N
and
F
= (2391 N) sin 19.5° = 797 N.
An alternative approach based on drawing a vector triangle (of forces) in the final
situation provides a quick solution.
(b) Since there is no change in kinetic energy, the net work on it is zero.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 NAJAFZADEH
 mechanics

Click to edit the document details