October 24, 2017 Physics 102 Inclined Plane Problem 2 Bodies Consider the two-bo

Question

October 24, 2017 Physics 102 Inclined Plane Problem 2 Bodies Consider the two-body situation at the right. A 3.50 x103 kg crate (m1) rests on an inclined plane and is connected by a cable to a 1.00 x 103 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 mI In order to solve this problem we must determine the direction of motion of the bodies. If 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 the incline plane and the hanging mass (m2) will accelerate upward. a. Prove mathematically which way the mass on the incline will accelerate b. Determine the acceleration of the system (ans 0.247 m/s2) c. Determine the tension in the cable. (ans 1.00 x 104 N) Once the direction of motion is determined the solution of (a) and (b) incorporates our 6 rules using the approach of free-body diagrams, component tables and Newton's second law analysis of each individual mass.

Explanation / Answer

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 sin() = 34300 N x sin(30°) = 17150 N

Fperpendicular = m1 g cos() = 34300 N x cos(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:

17150 N - 6237.98 N - Ftens = (3500 kg) a ..........(1)

For mass m2:

Fgrav = m2 g = (1000 kg) (9.8 N/kg) = 9800 N

Applying Newton's second law to m2:

Ftens - 9800 N - = (1000 kg) a .........(2)

Combining Equations 1 and 2 leads to the answers:

a = 0.24712 m/s2

Ftens = 10047 N

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