HW 11 - Morby Grant Homework 11 Due noon Inst Drummond This...

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

View Full Document Right Arrow Icon
Morby, Grant – Homework 11 – Due: Feb 15 2006, noon – Inst: Drummond 1 This print-out should have 11 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 1) 10 points The force required to stretch a spring varies directly with the amount the spring is stretched. A force of 11 pounds is needed to stretch a spring 66 inches, as shown in the right-hand figure below. 57 00 66 00 W 11 lbs How much force is required to stretch the spring 57 inches? Correct answer: 9 . 5 lbs. Explanation: Given : F 2 = 11 lbs , x 2 = 66 inches , and x 1 = 57 inches . F x , so F = k x = k = F x . Thus F 1 x 1 = F 2 x 2 = F 2 x 1 x 2 = (11 lbs) (57 inches) 66 inches = 9 . 5 lbs . 002 (part 1 of 3) 10 points The suspended 2 . 4 kg mass on the right is moving up, the 2 . 2 kg mass slides down the ramp, and the suspended 7 . 6 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 13 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 2 kg μ = 0 . 13 32 7 . 6 kg 2 . 4 kg What is the acceleration of the three block system? Correct answer: 4 . 9187 m / s 2 . Explanation: Let : m 1 = 2 . 4 kg , m 2 = 2 . 2 kg , m 3 = 7 . 6 kg , and θ = 32 . 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
Image of page 1

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

View Full Document Right Arrow Icon
Morby, Grant – Homework 11 – Due: Feb 15 2006, noon – Inst: Drummond 2 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 = (2 . 2 kg) (9 . 8 m / s 2 ) sin 32 2 . 4 kg + 2 . 2 kg + 7 . 6 kg - (0 . 13) (2 . 2 kg) (9 . 8 m / s 2 ) cos 32 2 . 4 kg + 2 . 2 kg + 7 . 6 kg + (7 . 6 kg - 2 . 4 kg) (9 . 8 m / s 2 ) 2 . 4 kg + 2 . 2 kg + 7 . 6 kg = 4 . 9187 m / s 2 . 003 (part 2 of 3) 10 points What is the tension in the cord connected to the 2 . 4 kg block? Correct answer: 35 . 3249 N. Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 4 kg) (9 . 8 m / s 2 + 4 . 9187 m / s 2 ) = 35 . 3249 N .
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