Question

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.

Answer 1

Answer: a = 0.247 m/s² and F_tens = 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:
F_grav = m₁•g = (3500 kg)•(9.8 N/kg) = 34300 N
F_parallel = m₁•g•sine(θ) = 34300 N • sine(30°) = 17150 N
F_{perpendicular} = m₁•g•cosine(θ) = 34300 N • cosine(30°) = 29704.67 N
F_norm = F_{perpendicular} = 29704.67 N
Ffrict = µ•F_norm = (.210)•(29704.67 N) = 6237.98 N

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

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

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

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

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