.Three masses (12 kg, 27 kg and 61 kg) are connected by strings. The 61 kg mass
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Question
.Three masses (12 kg, 27 kg and 61 kg) are connected by strings. The 61 kg mass slides on a horizontal surface of a table top and the 12 kg and 27 kg masses hang over the edge of the surface. The string connecting the 12 kg and 61 kg masses runs over a massless and frictionless pulley. The coefficient of sliding friction, µ, between the 61 kg mass and the table top is 0.292. The acceleration of gravity is 9.8 m/s. 61 kg 12 kg 27 kg T12 T13 µ = 0.292 a g Which equation represents T13 in the 12 kg and 27 kg mass system? 1. T13 = (61 kg) (g + a) 2. T13 = (39 kg) (g + a) 3. T13 = (27 kg) (g + a) 4. T13 = (39 kg) (g a) 5. T13 = (61 kg) (g a) 6. T13 = (100 kg) (g a) 7. T13 = (12 kg) (g + a) 8. T13 = (27 kg) (g a) 9. T13 = (12 kg) (g a) 10. T13 = (100 kg) (g + a) 012 (part 2 of 4) 10.0 points Which equation represents T13 in the 61 kg mass system? 1. T13 T12 = µ g (61 kg) 2. T13 T12 = (61 kg) (g 2) 3. T13 = (g a) (61 kg) 4. T13 = (a g) (61 kg) 5. T13 = (µ g a) (61 kg) 6. T13 = (µ g + a) (61 kg) 7. T13 = a (61 kg) 8. T13 T12 = [27 kg + µ (61 kg)] g 9. T13 T12 = [27 kg + µ (61 kg)] (g a) 10. T13 = (a µ g) (61 kg) 013 (part 3 of 4) 10.0 points Find the magnitude of acceleration of the system. Answer in units of m/s 2 . 014 (part 4 of 4) 10.0 points Which equation correctly represents T12 in the 12 kg and/or 27 kg mass system? 1. T12 = (12 kg) (g a) 2. T12 = (12 kg + 27 kg) (g a) 3. T12 = (27 kg) g 4. T12 = (27 kg) g (12 kg) a 5. T12 = (27 kg) (g a)
Explanation / Answer
here
by balancing the forces
T2 - u * 61 * g = 61 * a
T2 - 0.292 * 61 * g = 61 * a
T2 - 174.5 = 61 * a -------------------(i)
then
(12 + 27) * g - T2 = (12 + 27) * a
382.2 - T2 = 39 * a ------------------(ii)
by using both equations (i) + (ii)
207.7 = 100 * a
a = 2.077 m/s
Now 27 * g - T1 = 27 * a
T1 = 27 * (g - a)
T1 = 27 * (9.8 - 2.077)
T1 = 208.5 Newton
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