solution_6_pdf (1) - rodriguez (fr4537) – Homework 6...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rodriguez (fr4537) – Homework 6 Aplications of Newton’s Laws – catala – (20104) 1 This print-out should have 31 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 A 16 kg block with a pulley attached slides along a frictionless surface. It is connected by a massless string to a 5 . 3 kg block via the arrangement shown. The acceleration of gravity is 9 . 81 m / s 2 . 16 kg 5 . 3 kg Find the horizontal distance the 16 kg block moves when the 5 . 3 kg block descends a dis- tance of 9 . 2 cm . The pulleys are massless and frictionless. Correct answer: 4 . 6 cm. Explanation: Let : m 1 = 16 kg , m 2 = 5 . 3 kg , Δ x 2 = 9 . 2 cm , and g = 9 . 81 m / s 2 . Choose a coordinate system in which the positive x direction is to the right and consider the forces acting on each body. m 1 m 2 a 1 2 T N m 1 g a 2 = 2 a 1 T m 2 g Because of the pulley, the mass m 1 moves half the distance as the suspended mass: Δ x 1 = 1 2 Δ x 2 = 1 2 (9 . 2 cm) = 4 . 6 cm . 002 (part 2 of 3) 10.0 points Find the acceleration of the 16 kg block. Correct answer: 2 . 79532 m / s 2 . Explanation: The pulley causes the same relationship on the accelerations: a 1 = 1 2 a 2 = a . Applying summationdisplay F 1 = m 1 a 1 to the m 1 block, 2 T = m 1 a . (1) Applying summationdisplay F 2 = m 2 a 2 to the m 2 block, m 2 g- T = m 2 (2 a ) . (2) Double eq (2) and add to eq (1): 2 m 2 g = m 1 a + 4 m 2 a a = 2 m 2 g m 1 + 4 m 2 (4) = 2 (5 . 3 kg) ( 9 . 81 m / s 2 ) 16 kg + 4 (5 . 3 kg) = 2 . 79532 m / s 2 . 003 (part 3 of 3) 10.0 points Find the tension in the connecting string. Correct answer: 22 . 3626 N. Explanation: Using Eq (1), T = m 1 a 2 = (16 kg) ( 2 . 79532 m / s 2 ) 2 = 22 . 3626 N . 004 (part 1 of 4) 10.0 points Two 125 kg boxes are dragged along a fric- tionless surface with a constant acceleration of 1 . 22 m / s 2 , as shown in the figure. rodriguez (fr4537) – Homework 6 Aplications of Newton’s Laws – catala – (20104) 2 Each rope has a mass of 1 . 03 kg. 125 kg 125 kg 1 . 03 kg 1 . 03 kg F Find the force F . Correct answer: 307 . 513 N. Explanation: Let : m 1 = 125 kg , m 2 = 1 . 03 kg , m 3 = 125 kg , m 4 = 1 . 03 kg , and a = 1 . 22 m / s 2 . The free-body diagrams for the boxes and the ropes are below. F 12 m 1 F 21 F 23 m 2 F 32 F 34 m 3 F 43 F m 4 Because the vertical forces have no bearing on the problem they have not been included. vector F 34 = tension force exerted by m 4 on m 3 , vector F 43 = tension force exerted by m 3 on m 4 , vector F 23 = tension force exerted by m 3 on m 2 , vector F 32 = tension force exerted by m 2 on m 3 , vector F 12 = tension force exerted by m 2 on m 1 , vector F 21 = tension force exerted by m 1 on m 2 , The equal and opposite pairs of forces are vector F 21 =- vector F 12 , vector F 32 =- vector F 23 , vector F 43 =- vector F 34 ....
View Full Document

This note was uploaded on 04/15/2011 for the course PHY 2048 taught by Professor Unkown during the Spring '06 term at Miami Dade College, Miami.

Page1 / 13

solution_6_pdf (1) - rodriguez (fr4537) – Homework 6...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online