Answer: a = 0.247 m/s^{2} and F_{tens} = 1.00x10^{4} 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 m_{1} exceeds the sum of the force of gravity on m_{2} and the force of friction, the mass on the inclined plane (m_{1}) will accelerate down it and the hanging mass (m_{2}) will accelerate upward.

**For mass m**_{1}:

F_{grav} = m_{1}•g = (3500 kg)•(9.8 N/kg) = 34300 N

F_{parallel} = m_{1}•g•sine(θ) = 34300 N • sine(30°) = 17150 N

F_{perpendicular} = m_{1}•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 m_{1}:

Equation 10: 17150 N - 6237.98 N - F_{tens} = (3500 kg)•a

For mass m_{2}:

F_{grav} = m_{2}•g = (1000 kg)•(9.8 N/kg) = 9800 N

Applying Newton's second law to m_{2}:

Equation 11: F_{tens} - 9800 N - = (1000 kg)•a

Combining Equations 10 and 11 leads to the answers:

a = 0.24712 m/s^{2} = ~0.247 m/s^{2}

F_{tens} = 10047 N = ~1.00x10^{4} N