Three masses (12 kg, 20 kg and 61 kg) are connected by strings. The 61 kg mass s

Question

Three masses (12 kg, 20 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 20 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 coecient of sliding
friction, µ, between the 61 kg mass and the
table top is 0.102.
The acceleration of gravity is 9.8 m/s.Which equation represents T13 in the 12 kg
and 20 kg mass system?
1. T13 = (61 kg) (g a)
2. T13 = (12 kg) (g a)
3. T13 = (32 kg) (g a)
4. T13 = (12 kg) (g + a)
5. T13 = (20 kg) (g + a)
6. T13 = (61 kg) (g + a)
7. T13 = (20 kg) (g a)
8. T13 = (93 kg) (g + a)
9. T13 = (32 kg) (g + a)
10. T13 = (93 kg) (g a)

Which equation represents T13 in the 61 kg
mass system?
1. T13 = (g a) (61 kg)
2. T13 = (µ g + a) (61 kg)
3. T13 T12 = (61 kg) (g 2)
4. T13 = a (61 kg)
5. T13 T12 = [20 kg + µ (61 kg)] (g a)
6. T13 = (a g) (61 kg)
7. T13 T12 = [20 kg + µ (61 kg)] g
8. T13 = (µ g a) (61 kg)
9. T13 T12 = µ g (61 kg)
10. T13 = (a µ g) (61 kg)

Find the magnitude of acceleration of the system.
Answer in units of m/s2

Which equation correctly represents T12 in
the 12 kg and/or 20 kg mass system?
1. T12 = (20 kg) (g a)
2. T12 = (12 kg) (g a)
3. T12 = (12 kg + 20 kg) (g a)
4. T12 = (20 kg) g (12 kg) a
5. T12 = (20 kg) g

Explanation / Answer

1) 7. T13 = (20 kg) (g a)

2) 2. T13 = (µ g + a) (61 kg)

3) 1. T12 = (20 kg) (g a)

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