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.
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This note was uploaded on 03/19/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics

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