7 - roofner(bar784 Homework 7 Weathers(17104 This print-out...

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

View Full Document Right Arrow Icon
roofner (bar784) – Homework 7 – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A force of 34 . 3 N pushes and pulls to blocks as shown in the figure below. The vertical contact surfaces between the two blocks are frictionless. The contact between the blocks and the horizontal surface has a coefficient of friction of 0 . 17. The acceleration of gravity is 9 . 8 m / s 2 . 4 . 9 kg 9 . 8 kg F F 50 μ k = 0 . 17 a What is the magnitude a of the acceleration of the blocks? Correct answer: 2 . 47103 m / s 2 . Explanation: Given : M = 4 . 9 kg , 2 M = 9 . 8 kg , θ = 50 , and μ k = 0 . 17 . Basic Concept: vector F net = mvectora Solution: The frictional forces are f M = μ k N M = μ k M g. f 2 M = μ k N 2 M = μ k (2 M g F sin θ ) . F net = F + F cos θ f M f 2 M = 3 M a Therefore a = F + F cos θ f M f 2 M 3 M = F (1 + cos θ + μ k sin θ ) 3 M 3 μ k M g 3 M = (34 . 3 N)(1 + cos 50 + 0 . 17 sin 50 ) 3 (4 . 9 kg) 3 (0 . 17) (4 . 9 kg) (9 . 8 m / s 2 ) 3 (4 . 9 kg) = 2 . 47103 m / s 2 . 002 10.0 points Three blocks in contact with each other are pushed across a rough horizontal surface by a 72 N force as shown. The acceleration of gravity is 9 . 8 m / s 2 . 2 kg 5 . 2 kg 2 . 5 kg F μ = 0 . 19 If the coefficient of kinetic friction between each of the blocks and the surface is 0 . 19, find the magnitude of the force exerted on the 5 . 2 kg block by the 2 . 5 kg block. Correct answer: 18 . 5567 N. Explanation: M 1 M 2 M 3 F μ = 0 . 19 Given : F = 72 N , μ = 0 . 19 , M 1 = 2 kg , M 2 = 5 . 2 kg , M 3 = 2 . 5 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction. m 1 + m 2 + m 3 N g F f μ ( m 1 + m 2 + m 3 ) g If we treat the blocks as one object of mass M 1 + M 2 + M 3 , the frictional force opposing the applied force will be f μ = μ ( M 1 + M 2 + M 3 ) g
Image of page 1

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

View Full Document Right Arrow Icon
roofner (bar784) – Homework 7 – Weathers – (17104) 2 and the net force equals M total a is F f μ = ( M 1 + M 2 + M 3 ) a . Since M 1 + M 2 + M 3 = 2 kg + 5 . 2 kg + 2 . 5 kg = 9 . 7 kg , a = F μ ( M 1 + M 2 + M 3 ) g M 1 + M 2 + M 3 = 72 N 9 . 7 kg (0 . 19) (9 . 7 kg) (9 . 8 m / s 2 ) 9 . 7 kg = 5 . 56068 m / s 2 , Now that we know the acceleration, the force exerted on M 3 by M 2 is F 3 = M 3 a + μ M 3 g = (2 . 5 kg) (5 . 56068 m / s 2 ) + (0 . 19) (2 . 5 kg) (9 . 8 m / s 2 ) = 18 . 5567 N . According to Newton’s third law, this is also the force exerted on M 2 by M 3 . Using the free-body diagram for the third block, we have F 32 μ M 3 g = M 3 a F 32 = M 3 bracketleftbigg F M 1 + M 2 + M 3 μ g + μ g bracketrightbigg = bracketleftbigg M 3 M 1 + M 2 + M 3 bracketrightbigg F (1) = bracketleftbigg (2 . 5 kg) (2 kg) + (5 . 2 kg) + (2 . 5 kg) bracketrightbigg (72 N) = bracketleftbigg 2 . 5 kg 9 . 7 kg bracketrightbigg (72 N) = 18 . 5567 N . Using the free-body diagram for the first and second blocks, we have F F 23 μ [ M 1 + M 2 ] g = [ M 1 + M 2 ] a F 23 = [ M 1 + M 2 ] bracketleftbigg F M 1 + M 2 + M 3 μ g + μ g bracketrightbigg + F = bracketleftbigg M 1 + M 2 M 1 + M 2 + M 3 + 1 bracketrightbigg F = bracketleftbigg M 3 M 1 + M 2 + M 3 bracketrightbigg F (1 ) which is the same as Eq. 1 above, since
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