This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 91* Determine the horizontal
force F required to start
moving the 250lb block
shown in Fig. P91 up the
inclined surface. The
coefficient of friction
between the inclined surface and the block is
u = 0.30. S SOLUTION .a A free—body diagram for the
block is shown at the right. For impending motion: Ff = HF“ = 0.30Fn + T 2F = F cos 30°  F sin 300 — H
y n f = F cos 30°  0.3017n sin 30° — 250 = o n = 349.15 lb = P  Fn sin 300  Fr cos 300
P  349.15 sin 30°  0.30(349.15) cos 30° = 0 265.29 lb 3 265 lb 8J6 “INEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Two blocks with masses
mA = 20 kg and m8 = 80 kg are connected with a
flexible cable that passes
over a frictionless pulley
as shown in Fig. P94. The
coefficient of friction
between the blocks is 0.25.
If motion of the blocks is
impending. determine the
coefficient of friction
between block B and the
inclined surface and the
tension in the cable
between the two blocks. SOLUTION A freebody diagram for block
A is shown at the right. ' mAg = 20(9.807) = 196.14 N For impending motion
of block A: + /” BE? A — 196.14 cos 35° = o  = 160.67 N n + \y 2Fx —T + 0.25(160.67) + 196.14 sin 350 = 0 = 152.67 N
i ‘3 52.47 N 784.54 N ,r' A freebody diagram for block 35° )LO.(.'7 N
B is shown at the right.
‘4c>.)7 )1 W3 = mag = 80(9.807) = 784.56 N For impending motion
of block B: + /” ZFY = B  160.67  784.56 cos 35° = 0 303.34 N n
+ \y ZFX Br — 40.17  152.67 + 784.56 sin 350 = 0 257.17 N B .
_ r _ 257.17 _
‘ B ' 303.34 ' 0'320 Ans' n ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Two blocks with masses
mA = 50 kg and m8 = 100 kg are connected with a
flexible cable that passes
over a frictionless pulley
as shown in Fig. P96.
Determine the force F
required to start moving
the blocks if the
coefficient of friction
between the blocks and the surface is 0.10. SOLUTION ‘h freebody diagram for block
A is shown at the right. WA = mAg + 50(9.807) = 490.4 N For impending motion: + “\ Epy F  490.4 cos 30° = o n
+ /” 2F, _ T  Fr — w sin 30° T  0.10(424.7)  490.4 sin 30° = o A free—body diagram for block
B is shown at the right. W8 = mBg + 100(9.807) = 980.7 N For impending motion: T + EFY = 9 sin 25° + Fn — 980.7 = o —a + 2Fx P cos 25°  Ff  T 9 cos 25°  0.10Fn  287.7 = 0
Solving yields: P = 406.7 N E 407 N ENGINEERING MECHANICS  STATICS, 2nd. Ed. A homogeneous, triangular block of weight W has height h and base width b as shown in Fig. P97. Develop an expression for the coefficient of friction between the block and the surface for which impending motion by slipping and tipping occur simultaneously if (a) Force F acts in the direction
shown on the figure. (b) The direction of force F is
reversed. SOLUTION (a; From the freebody diagram
for the triangular block: +T2F =
Y +—)2F :
x For slipping: = Af(max) = HA“ 5 b For tipping: + C 2MB = Pth  Wtﬂ  For simultaneous slipping and tipping:
From which: (b) From the freebody diagram
for the triangular block: + T 2F
y + e 2F
X
For slipping: For tipping: For simultaneous slipping and tipping: From which: 82?. wt F. RILEY AND L. D. STURGES ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES The masses of blocks A and B of Fig. P924 are mA 100 kg and mB = 50 kg. If the coefficient
of friction is 0.15 for block A
and 0.25 for block B, determine
the maximum angle 6 for equilibrium of the blocks.
SOLUTION From a freebody diagram for block
B when motion is impending: = 50(9.807) 490.35 = Bn — WB cos 6 B  490.35 cos 6 = 0 n 490.35 cos 6 0.25Bn = 0.25(490.35 cos 6) = 122.59 cos 6 + /' 2F. T  wB sin 6 + 131‘ = T  490.35 sin 6 + 122.59 cos 6 = o T = 122.59 cos 6  490.35 sin 6 From a free—body diagram for block wﬁ=980.7N
A when motion is impending: = 100(9.807) = 980. Ar(max) = uAAn = A  WA cos 9 n A  980.7 cos 6 n 980.7 cos 6 = 0.15A = 0.15(980.7 cos 9) = 147.11 cos 6 n
= T + Ar  WA sin 6
= 122.59 cos 6 — 490.35 sin 6 + 147.11 cos 6  980.7 sin 6 = 269.70 cos 6  1471.05 sin 9 = 0 1 260.70
n _______ 0 0
1471.05  10.389 E 10.39 ...
View
Full Document
 Spring '08
 SWIM

Click to edit the document details