p13_049.pdf
(
58 KB
)
Pobierz
Chapter 13 - 13.49
49. We denote the mass of the slab as
m
, its density as
ρ
,andvolumeas
V
. The angle of inclination is
θ
=26
◦
.
(a) The component of the weight of the slab along the incline is
F
1
=
mg
sin
θ
=
ρV g
sin
θ
=(3
.
2
10
3
kg
/
m
3
)(43m)(2
.
5 m)(12 m)(9
.
8m
/
s
2
) sin26
◦
=1
.
77
×
10
7
N
.
(b) The static force of friction is
f
s
=
µ
s
N
=
µ
s
mg
cos
θ
=
µ
s
ρV g
cos
θ
=(0
.
39)(3
.
2
×
10
3
kg
/
m
3
)(43m)(2
.
5 m)(12 m)(9
.
8m
/
s
2
) cos26
◦
=1
.
42
×
10
7
N
.
(c) The minimum force needed from the bolts to stabilize the slab is
F
2
=
F
1
−
f
s
=1
.
77
×
10
7
N
−
1
.
42
×
10
7
N=3
.
5
×
10
6
N
.
If the minimum number of bolts needed is
n
,then
F
2
/nA
≤
3
.
6
×
10
8
N
/
m
2
,or
n
≥
3
.
5
×
10
6
N
=15
.
2
.
(3
.
6
×
10
8
N
/
m
2
)(6
.
4
×
10
−
4
m
2
)
Thus 16 bolts are needed.
×
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