Determine the range of mass m over which the system is in equilibrium (a) if the
ID: 1862924 • Letter: D
Question
Determine the range of mass m over which the system is in equilibrium (a) if the coefficient of static friction is 0.20 at all three fixed shafts and (b) if the coefficient of static friction associated with shaft B is increased to 0.50. The answer is (a) 0.304 <= m <= 13.17 and (b) 0.1183 <= m <= 33.8
Explanation / Answer
a) u = 0.2
T1 = Tension in rope between m and A
T2 = Tension in rope between A and B
T3 = Tension in rope between B and C
T4 = Tension in rope between C and 2 kg mass
Angle of wrap, theta = pi rad = 3.14 rad
For motion of m impending downwards:
mg = T1
T1 /T2 = e^(0.2*3.14)
T2 / T3 = e^(0.2*3.14)
T3 / T4 = e^(0.2*3.14)
T4 = 2g = 2*9.81 N = 19.62 N
Solving these eqns we get,
T3 = 36.765 N
T2 = 68.892 N
T1 = 129.095 N
Thus m = 13.17 kg
For motion of m impending upwards:
mg = T1
T2 /T1 = e^(0.2*3.14)
T3 / T2 = e^(0.2*3.14)
T4 / T3 = e^(0.2*3.14)
T4 = 2g = 2*9.81 N = 19.62 N
Solving these eqns we get,
T3 = 10.47 N
T2 = 5.58 N
T1 = 2.98 N
Thus m = 0.304 kg
Thus 0.304 <= m <= 13.17 kg
b)
For motion of m impending downwards:
mg = T1
T1 /T2 = e^(0.2*3.14)
T2 / T3 = e^(0.5*3.14)
T3 / T4 = e^(0.2*3.14)
T4 = 2g = 2*9.81 N = 19.62 N
Solving these eqns we get,
T3 = 36.765 N
T2 = 176.71 N
T1 = 331.14 N
Thus m = 33.8 kg
For motion of m impending upwards:
mg = T1
T2 /T1 = e^(0.2*3.14)
T3 / T2 = e^(0.5*3.14)
T4 / T3 = e^(0.2*3.14)
T4 = 2g = 2*9.81 N = 19.62 N
Solving these eqns we get,
T3 = 10.47 N
T2 = 2.18 N
T1 = 1.16 N
Thus m = 0.118 kg
Thus 0.118 <= m <= 33.8 kg
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.