This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 93 workers are pulling a 4001b 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 F that the workers must
exert to start sliding
the crate up the incline.
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 freebody diagram for the
block is shown at the right. (a) For impending motion of
the block up the incline: + z” EFx p cos 15°  400 sin 15°  0.2An o n + “\ 2F? A  P sin 15°  400 cos 15° = 0
Solving yields: A = 437.6 lb P = 19?.?8 lb 8 19?.8 lb TI For impending motion of the
block down the incline: + /’ EFx 9 cos 15°  400 sin 15° + 0.2An = o + “\ EFY A — P sin 15°  400 cos 15° = 0 :1 Solving yields: A = 393.0 lb P = 25.30 lb The masses of blocks A and B of Fig. P916 are mA = 50
kg and '3 = 75 kg. If the
coefficient of friction is
0.25 for both surfaces,
determine the force F
required to cause impending
notion of block B. SOLUTION A freebody diagram for block
A is shown at the right. HA mhg + 50(9.807l = 490.35 N For impending motion: + » 2F 2 A  T cos 45°
3 f 0.25.4n  T cos 45° = o T = 0.3536An + T 2F = A + T sin 45° H
y n A = An + 0.3536 An sin 450  490.35 = 0 = 392.27 N A freebody diagram for block
B is shown at the right. ”B = nag + 75(9.807} = 735.53 N For impending motion: + T SE = B  W  A
y B 1'! h B  735.53  392.27 = 0 1127.30 N P  AF — BF
P  0.25{392.27I  0.25(1127.BO} = 0 380.02 N 8 380 N 920 A 30kg box is sitting on
an inclined surface as
shown in Fig. PQZO. If
the coefficient of friction
between the box and the
surface is 0550 and the
angle a = 60 . determine
the maximum force T for
which the box will be in
equilibrium. Is impending
motion by tipping or by
slipping? SOLUTION W mg = 30(9.807} = 294.2 N From a freebody diagram for the rectangular block
when motion is impending: For sliding:
A = Aftmax) = “A“ = 0.50An f
+ /” 2Fx 0.5mn  T cos 60° ~ 294.2 sin 20° = 0 + “\ EFY A + T sin 60°  294.2 cos 20° = 0 Solving yields: 241.55 N 40.31 N 8 40.3 N (for sliding} For tipping: + C Sun = T sin 60° (0.91 + T cos 60° {0.6}
+ 294.2 sin 20° {0.31  294.2 cos 20° (0.45) = 0 T = 87.29 N 3 87.3 N [for tipping) Therefore: T = 40.3 N max Impending motion is by slipping ...
View
Full Document
 Fall '06
 Anderson

Click to edit the document details