minimum value
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1. ( a ) 33.6 N. ( b ) aA = 4.76 m/s^ 2 , aB = 3.08 m/s ^2 , aC = 1.401 m/s ^2
2. ( a ) 0.1904, motion impending downward. ( b ) 0.349, motion impending upward.
1. The coefficients of friction between blocks A and C and the horizontal surfaces are µs =0.24 and µk = 0.20. Knowing that Ma = 5 kg, Mb = 10 kg, and Mc = 10 kg, determine ( a) the tension in the cord, ( b) the acceleration of each block.
2. A small, 0.6- lb collar D can slide on portion AB of a rod which is bent as shown. Knowing that r = 8 in. and that the rod rotates about the vertical AC at a constant rate of 10 rad/ s, determine the smallest allowable value of the coefficient of static friction between the collar and the rod if the collar is not to slide when ( a) a = 15°, ( b) a = 45°. Indicate for each case the direction of the impending motion.
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Solution Summary
This post finds the minimum value. The coefficients of frictions between blocks are determined.
Solution Preview
Problem 1: RA aA aC RC
A T T T T C
fA fC
5g T T 10g
T T
B aB
10g
Forces acting on the system are shown in the fig..
5g = Weight of block A
10g = Weight of block B
10g = Weight of block C
RA = Normal reaction on block A
RC = Normal reaction on block C
fA = Frictional force on block A
fC = Frictional force on block C
T = Tension in the string
As block A has no vertical motion, net vertical force on it must be zero. Hence, RA = 5g = 5x9.8 = 49 N
Similarly, RC = 10g = 10x9.8 = 98 N
As we know, frictional force f = μR. Hence, fA = μkRA = 0.2 x 49 = 9.8 N
fC = μkRC = 0.2 x 98 = 19.6 N
Net horizontal force on block A = FA = T - fA = T - 9.8 ...
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