Spring 2010 - HW07_Solution - banuelos (ojb93) Homework 07...

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: banuelos (ojb93) Homework 07 florin (58140) 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 A person weighing 0 . 6 kN rides in an elevator that has a downward acceleration of 2 . 3 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the magnitude of the force of the elevator floor on the person? Correct answer: 0 . 459184 kN. Explanation: Basic Concepts: summationdisplay vector F = mvectora vector W = mvectorg Solution: Since W = mg , F net = ma = W f f = W ma = W parenleftbigg 1 a g parenrightbigg = (0 . 6 kN) parenleftbigg 1 2 . 3 m / s 2 9 . 8 m / s 2 parenrightbigg = 0 . 459184 kN . 002 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 2 kg, 2 kg, and 6 kg masses are suspended as in the figure. 2 . 1 m 24 . 3 cm 2 kg 2 kg 6 kg T 2 T 1 T 3 What is the tension T 1 in the string be- tween the two blocks on the left-hand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 23 . 52 N. Explanation: Let : R = 24 . 3 cm , m 1 = 2 kg , m 2 = 2 kg , m 3 = 6 kg , and h = 2 . 1 m . Consider the free body diagrams 2 kg 2 kg 6 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a For each mass in the system vector F net = mvectora. Since the string changes direction around the pulley, the forces due to the tensions T 2 and T 3 are in the same direction (up). The acceleration of the system will be down to the right ( m 3 > m 1 + m 2 ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this acceleration rate be a and the tension over the pulley be T T 2 = T 3 . For the lower left-hand mass m 1 the accel- eration is up and T 1 m 1 g = m 1 a. (1) For the upper left-hand mass m 2 the acceler- ation is up and T T 1 m 2 g = m 2 a. (2) banuelos (ojb93) Homework 07 florin (58140) 2 For the right-hand mass m 3 the acceleration is down and T + m 3 g = m 3 a. (3) Adding Eqs. (1), (2), and (3), we have ( m 3 m 1 m 2 ) g = ( m 1 + m 2 + m 3 ) a (4) a = m 3 m 1 m 2 m 1 + m 2 + m 3 g (5) = 6 kg 2 kg 2 kg 2 kg + 2 kg + 6 kg (9 . 8 m / s 2 ) = 1 . 96 m / s 2 . The tension in the string between block m 1 and block m 2 (on the left-hand side of the pulley) can be determined from Eq. (1). T 1 = m 1 ( a + g ) (6) = (2 kg) (1 . 96 m / s 2 + 9 . 8 m / s 2 ) = 23 . 52 N . 003 (part 2 of 2) 10.0 points What is the magnitude of the acceleration of the lower left-hand block? Correct answer: 1 . 96 m / s 2 . Explanation: The acceleration is the same for every mass, since the string is inextensible. keywords: 004 10.0 points A 7 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar....
View Full Document

Page1 / 10

Spring 2010 - HW07_Solution - banuelos (ojb93) Homework 07...

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