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 1 
Physics 112  HW #11 Solutions
Spring 2005
1110
[Leaning Ladder]
(a)
Maximum frictional force is f
s,max
=
µ
s
N
g
Apply N’s 2
nd
law in vertical direction:
Σ
F
y
=
N
g
– W
L
 W
p
=
0
so N
g
= W
L
+ W
p
= 160 N + 740 N = 900 N
f
s,max
=
µ
s
N
g
=
(0.40)(900 N)
=
360 N
(b) and (c):
Consider man on ladder,
s
meters along it.
To find friction at bottom, take torques about top of ladder
Σ
τ
top
=
(W
L
)(D/2) + W
p
(D – x )
 N
g
D
+
f
s
H
=
0
so
f
s
H
=
 (W
L
)(D/2)
 W
p
D
L  s
L
+ N
g
D
(f
s
)(4.0 m)
=
 (160 N)(1.5 m)
 (740 N)(3.0 m)
5.0 m

s
5.0 m
+
(900N)(3.0 m)
f
s
= 60 N +
(111 N/m)
s
(1)
For part (b),
let s = 1.0 m in Eq (1) :
Then f
s
= 60 N +
(111 N/m)(1.0 m) =
171 N
For part (c), put f
s
= f
s,max
=
36
0 N in Eq (1) :
360 N = 60 N
+ (111 N/m)
s
Solving for
s
gives
s
= 2.70 m
w
p
w
L
N
w
N
g
f
s
L
D
H
x
s
D = 3.0 m, L = 5.0 m, so H = 4.0 m
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1111
[Diving Board]
Freebody diagram showing forces acting on Person + Board:
(a) Torques about left end of board:
Σ
τ
=
F
e
(0) + F
s
(1.00 m)  w
b
(1.50 m)  w
p
(3.00 m)
=
0
⇒
F
s
=
w
b
(1.50 m) + w
p
(3.00 m)
1.00 m
=
(280 N)(1.50 m) + (500 N)(3.00 m)
1.00 m
=
1920 N
↑
(b)
Σ
F
y
=
F
s
 F
e
 w
b
 w
p
=
0
⇒
F
e
=
F
s
 w
b
 w
p
=
1920 N  280 N  500 N
=
1140 N
↓
1152
[Truck on Drawbridge]
(a) Torques about left end of drawbridge:
Σ
τ
= (T sin 70
°
)(40.0 m) – w
T
(30.0 m)cos 30
°
– w
B
(20.0 m)cos 30
°
=
0
T
=
{ m
B
(20.0 m) + m
T
(30.0 m) } g cos 30
°
(sin 70
°
)(40.0 m)
T
=
{ (30,000 kg )(30.0 m) + (12,000 kg)(20.0 m) } (9.81 m/s
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 LECLAIR,A
 mechanics, Force, Friction

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