solution_pdf7

# solution_pdf7 - magalhaes(bam2734 – Homework#5 –...

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: magalhaes (bam2734) – Homework #5 – Erskine – (58200) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points There is friction between the block and the table. The suspended 4 kg mass on the left is moving up, the 3 kg mass slides to the right on the table, and the suspended mass 7 kg on the right is moving down. The acceleration of gravity is 9 . 8 m / s 2 . 4 kg 3 kg 7 kg μ = 0 . 2 What is the magnitude of the acceleration of the system? Correct answer: 1 . 68 m / s 2 . Explanation: m 1 m 2 m 3 μ a Let : m 1 = 4 kg , m 2 = 3 kg , m 3 = 7 kg , and μ = 0 . 2 . Basic Concepts: The acceleration a of each mass is the same, but the tensions in the two strings will be different. F net = ma negationslash = 0 Solution: Let T 1 be the tension in the left string and T 2 be the tension in the right string. Consider the free body diagrams for each mass T 1 m 1 g a T 2 m 3 g a T 1 T 2 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, so F net 1 = m 1 a = T 1 − m 1 g . (1) For the mass on the table, a is directed to the right, T 2 acts to the right, T 1 acts to the left, and the motion is to the right so that the frictional force μm 2 g acts to the left and F net 2 = m 2 a = T 2 − T 1 − μm 2 g . (2) For the mass m 3 , T 2 acts up and the weight m 3 g acts down, with the acceleration a di- rected downward, so F net 3 = m 3 a = m 3 g − T 2 . (3) Adding these equations yields ( m 1 + m 2 + m 3 ) a = m 3 g − μm 2 g − m 1 g a = m 3 − μm 2 − m 1 m 1 + m 2 + m 3 g = 7 kg − (0 . 2) (3 kg) − 4 kg 4 kg + 3 kg + 7 kg × (9 . 8 m / s 2 ) = 1 . 68 m / s 2 . 002 10.0 points The suspended 2 . 9 kg mass on the right is moving up, the 2 . 5 kg mass slides down the ramp, and the suspended 8 . 1 kg mass on the magalhaes (bam2734) – Homework #5 – Erskine – (58200) 2 left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 5 k g μ = . 1 5 18 ◦ 8 . 1 kg 2 . 9 kg What is the tension in the cord connected to the 8 . 1 kg block? Correct answer: 46 . 3585 N. Explanation: Let : m 1 = 2 . 9 kg , m 2 = 2 . 5 kg , m 3 = 8 . 1 kg , and θ = 18 ◦ . 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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

solution_pdf7 - magalhaes(bam2734 – Homework#5 –...

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

View Full Document
Ask a homework question - tutors are online