oldquiz 2 answers - Overton Mays Oldquiz 2 Due Mar 8 2005...

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

View Full Document Right Arrow Icon
Overton, Mays – Oldquiz 2 – Due: Mar 8 2005, noon – Inst: Turner 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points The suspended 2 . 6 kg mass on the right is moving up, the 1 . 4 kg mass slides down the ramp, and the suspended 7 . 2 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 15 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 4 kg μ = 0 . 15 34 7 . 2 kg 2 . 6 kg What is the acceleration of the three block system? Correct answer: 4 . 55768 m / s 2 . Explanation: Let : m 1 = 2 . 6 kg , m 2 = 1 . 4 kg , m 3 = 7 . 2 kg , and θ = 34 . Basic Concept: F net = m a 6 = 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 m g sin θ and the perpen- dicular component of its weight is m g cos θ . ( N = m g 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 . 4 kg) (9 . 8 m / s 2 ) sin 34 2 . 6 kg + 1 . 4 kg + 7 . 2 kg - (0 . 15) (1 . 4 kg) (9 . 8 m / s 2 ) cos 34 2 . 6 kg + 1 . 4 kg + 7 . 2 kg + (7 . 2 kg - 2 . 6 kg) (9 . 8 m / s 2 ) 2 . 6 kg + 1 . 4 kg + 7 . 2 kg = 4 . 55768 m / s 2 .
Image of page 1

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

View Full Document Right Arrow Icon
Overton, Mays – Oldquiz 2 – Due: Mar 8 2005, noon – Inst: Turner 2 002 (part 2 of 3) 10 points What is the tension in the cord connected to the 2 . 6 kg block? Correct answer: 37 . 33 N. Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 6 kg) (9 . 8 m / s 2 + 4 . 55768 m / s 2 ) = 37 . 33 N . 003 (part 3 of 3) 10 points What is the tension in the cord connected to the 7 . 2 kg block? Correct answer: 37 . 7447 N. Explanation: Using Eq. 3, we have T 3 = m 3 g - m 3 a = (7 . 2 kg) (9 . 8 m / s 2 - 4 . 55768 m / s 2 ) = 37 . 7447 N . 004 (part 1 of 2) 10 points Given: The coefficient of static friction be- tween the person and the wall is 0 . 68 , the mass of the person is 78 kg , the radius of the cylinder is 8 . 7 m , and g = 9 . 8 m / s 2 . A barrel of fun consists of a large vertical cylinder that spins about the vertical axis.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern