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Unformatted text preview: 9~l* Determine the horizontal force P required to start
moving the 250lb block
shown in Fig. PQl up the
inclined surface. The
coefficient of friction
between the inclined surface and the block is # = 0.30. S SOLUTION 3 A freebody diagram for the
block is shown at the right. For impending motion: Fr = HF“ = 0.30Fn + T ZFY = F cos 30o  Fr sin 300 — W n F cos 30° — 0.305n sin 30° — 250 = 0 n 349.15 lb '1‘?)
II + —+ IF = p — Fn sin 30°  Ff cos 30° F  349.15 sin 30°  0.30(349.15) cos 30° = 0 P = 265.29 lb E 265 lb Ans. 93 Workers are pulling a 400lb crate up an incline as shown in Fig.
P93. The coefficient of friction between the crate and the surface
is 0.20, and the rope on which the workers are pulling is horizontal.
(a) Determine the force P
that the workers must
exert to start sliding
the crate up the incline. (b) If one of the workers
lets go of the rope for
a moment, determine the
minimum force the other
workers must exert to
keep the crate from
sliding back down the
incline. SOLUTION A free—body diagram for the
block is shown at the right. (a) For impending motion of
the blo k th ' l‘ :
0 up e inc me O 2A” ﬁn + x” XFX = p cos 15° — 400 sin 15°  0.2An = 0
+ "\ ZFY = An — P sin 15°  400 cos 15° = 0
Solving yields:
A“ = 437.6 lb P = 197.78 lb a 197.8 lb Ans.
. . . 400 it
(b) For impending motion of the ,5.
block down the incline: 3"..’7L
u”
oann
Fin + x” SE = p cos 15° — 400 sin 15° + 0.2An = o + \ ZFy = An — P sin 15° — 400 cos 15° = 0 Solving yields: A = 393.0 lb P = 25.80 lb n 25.8 lb Ans. 921* A 120—lb girl is walking up a 48lb
uniform beam as shown in Fig. P921.
Determine how far up the beam the
girl can walk before the beam
starts to slip 1% (a) The coefficient of friction
is 0.20 at all surfaces. (b) The coefficient of friction at
the bottom end of the beam is
increased to 0.40 by placing
a piece of rubber between the
beam and the floor. SOLUTION (a) From a freebody diagram for the
beam when motion is impending: A = MA = 0.2A f n n
B = MB = 0.28
f n n
¢ = tan‘1 2 = 36.870 +
l
M
'11
II B sin 36.870 — 0.23n cos 36.870 — 0.2An = 0 n + T 2F = B cos 36.870 + 0.28n sin 36.870 + An  120 — 48 = 0 V n Solving yields: A = 118.46 lb B = 53.85 lb n n + C EMA = 48(6) cos 36.870 + 120(x) cos 36.870  53.85(10) = 0 X = 3.209 ft g 3.21 ft , Ans.
0'2 Bn 320 “a
(b) From a freebody diagram for the
beam when motion is impending: ‘3
Ar = HA“ = 0.4An
OvQFI
B = MB = 0.2B ‘n1f""' n
f n n
. o o R
+ 4 53Fh = B Sln 36.87 — 0.2B cos 36.87 — 0.4A = 0 y‘
. n n n
+ T EFV = Bn cos 36.870 + 0.213n sin 36.87° + An — 120 — 48 = 0
Solving yields: A = 91.49 lb B = 83.17 lb n n + C EMA = 48(6) cos 36.870 + 120(x) cos 36.870  83.17(10) = 0 x = 6.264 ft 2 6.26 ft Ans. 936 The masses of blocks A and
B of Fig. P9—36 are mA = 40
kg and mB = 85 kg. If the
coefficient of ffiction is 0.25 for both surfaces, determine the force p required to cause impending motion of block B. SOLUTION From a free—body diagram
for block A when motion is impending: T _ _ x
WA = mAg = 40(9.80/) = 392.28 N Ar = Af(max) = HAAn = 0.25An / Rn + T ZFy = A cos 45° + Af sin 450 — W n A  392.28 = 0 0 A cos 45° + 0.25An sin 4o n 32
ll 443.81 N f 0.25An = 0.25(443.81) = 110.95 N >
II From a freebody diagram
for block B when motion
is impending: I?
II B mBg = 85(9 807) = 833.60 N Bf = Bf(max) = MB“ = 0.25Bn
.17 : 0— ' 0— —
. / _Fx P cos 20 WE Sln 45 Ar Bf
= P cos 20o  833.60 sin 45° — 110.95  0.25Bn= 0
+ K\ :F = p sin 20° — w cos 45° — A + B
y B n n 0 = 9 sin 20 — 833.60 cos 45°  443.81 + Bn = 0 Solving yields: P = 935.14 N E 935 N Ans. 9120* The masses of blocks A and
B of Fig. P9120 are mA =
50 kg and mB = 25 kg. If
the coefficient'of static
friction is 0.15 for both
surfaces, determine the
force ﬁ required to cause impending motion of block B. SOLUTION From a freebody diagram
for block A when motion
is impending: WA = mAg = 50(9.807) 490.4 N A cos 45° — B
x f A cos 45°  0.1513n = o + —+ 2F A = 0.21213n
+ T 2F = B — A sin 45° — w = B — 0.21213 sin 45° — 490.4 = 0
y n A n n
Bn = 576.93 N Br = ya“ = 0.15(576.93) = 86.54 N From a free—body diagram
for block B when motion
is impending: W = mBg = 25(9.807) B 245.2 N + T 2F = C — B — W
n Y n B
= Cn — 576.93  245.2 = 0
Cn = 822.13 N
Cr = uCn = O.15(822.13) = 123.32 N
+ * EFX = P  Br  Cf
= P  86.54  123.32 = 0 P = 209.86 N 3 210 N Ans. 913* The ZOOlb block of Fig. P9—13 is sitting on a 300 inclined surface.
The coefficient of friction between the block and the surface is
0.50. A light, inextensible cord is attached to the block, passes
around a frictionless pulley.
and is attached to a second
block of mass M. Determine
the minimum and maximum masses, M . and M , such
min max that the system is in
equilibrium. Is impending
motion by slipping or by
tipping? SOLUTION From a free—body diagram
for the ZOOlb block: + “\ ZFY = A — 200 cos 30° n O A = 173.21 lb n MA = 0.50(173.21) = 86.61 lb A,(max)
1 n For slipping down the incline:
+ /” 2px = p + Af(max) — 200 sin 30°
= p + 86.61 — 200 sin 30° = 0 P3 = 13.39 lb
For tipping:
+ c 2MB = P(0.5)  200 sin 30° (1) + 200 cos 30° (0.5) = 0
Pt = 26.79 lb > PS (Block tips before it slips as P is reduced)
Therefore: Pmin = 26.79 lb
For slipping up the incline:
+ /” ZFX = P — Af(max) — 200 sin 30°
= P  86.61  200 sin 30° = 0 P = 186.61 lb U) P
_ min _ 26.79 _ . .
Mmin — —E — EETI7— — 0.833 slug (Tipping)
Pmax 186.61 _
M = —— = ——— = 0.80 slug (Slipping) Ans. max g 32.17 ...
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This note was uploaded on 07/05/2010 for the course ENGR 120 taught by Professor Jones during the Spring '10 term at Florida State College.
 Spring '10
 jones
 Statics

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