# oldhw 11 - alexander(jra2623 – oldhomework 11 – Turner...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: alexander (jra2623) – oldhomework 11 – Turner – (92510) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points The suspended 2 . 3 kg mass on the right is moving up, the 1 . 3 kg mass slides down the ramp, and the suspended 8 . 4 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 17 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 3 k g μ = . 1 7 24 ◦ 8 . 4 kg 2 . 3 kg What is the acceleration of the three block system? Correct answer: 5 . 24861 m / s 2 . Explanation: Let : m 1 = 2 . 3 kg , m 2 = 1 . 3 kg , m 3 = 8 . 4 kg , and θ = 24 ◦ . Basic Concept: F net = ma negationslash = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di- rected upward F net 1 = m 1 a = T 1 − m 1 g (1) For the mass on the table, the parallel compo- nent of its weight is mg sin θ and the perpen- dicular component of its weight is mg cos θ . ( N = mg cos θ from equilibrium). The accel- eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μ m 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ − T 1 − μ m 2 g cos θ . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di- rected downward F net 3 = m 3 a = m 3 g − T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin θ − μ m 2 g cos θ − m 1 g . Solving for a , we have a = [ m 2 sin θ − μ m 2 cos θ + ( m 3 − m 1 )] g m 1 + m 2 + m 3 = (1 . 3 kg) (9 . 8 m / s 2 ) sin 24 ◦ 2 . 3 kg + 1 . 3 kg + 8 . 4 kg − (0 . 17) (1 . 3 kg) (9 . 8 m / s 2 ) cos 24 ◦ 2 . 3 kg + 1 . 3 kg + 8 . 4 kg + (8 . 4 kg − 2 . 3 kg) (9 . 8 m / s 2 ) 2 . 3 kg + 1 . 3 kg + 8 . 4 kg = 5 . 24861 m / s 2 . alexander (jra2623) – oldhomework 11 – Turner – (92510) 2 002 (part 2 of 3) 10.0 points(part 2 of 3) 10....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

oldhw 11 - alexander(jra2623 – oldhomework 11 – Turner...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online