Sarthaks Test
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Consider the two-body situation at the right. A 3.50x103-kg crate (m1) rests on an inclined plane and is connected by a cable to a 1.00x103-kg mass (m2). This second mass (m2) is suspended over a pulley. The incline angle is 30.0° and the surface has a coefficient of friction of 0.210. Determine the acceleration of the system and the tension in the cable.

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by (50.6k points)

Answer: a = 0.247 m/s2 and Ftens = 1.00x104 N

The solution here will use the approach of a free-body diagram and Newton's second law analysis of each individual mass. The free-body diagrams for the two objects are shown below.

Because the parallel component of gravity on m1 exceeds the sum of the force of gravity on m2 and the force of friction, the mass on the inclined plane (m1) will accelerate down it and the hanging mass (m2) will accelerate upward.

For mass m1:
Fgrav = m1•g = (3500 kg)•(9.8 N/kg) = 34300 N
Fparallel = m1•g•sine(θ) = 34300 N • sine(30°) = 17150 N
Fperpendicular = m1•g•cosine(θ) = 34300 N • cosine(30°) = 29704.67 N
Fnorm = Fperpendicular = 29704.67 N
Ffrict = µ•Fnorm = (.210)•(29704.67 N) = 6237.98 N

Applying Newton's second law to m1:
Equation 10: 17150 N - 6237.98 N - Ftens = (3500 kg)•a

For mass m2:
Fgrav = m2•g = (1000 kg)•(9.8 N/kg) = 9800 N

Applying Newton's second law to m2:
Equation 11: Ftens - 9800 N - = (1000 kg)•a

Combining Equations 10 and 11 leads to the answers:
a = 0.24712 m/s2 = ~0.247 m/s2
Ftens = 10047 N = ~1.00x104 N

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