HW04-solutions - baum (lmb2768) HW04 distler (56295) 1 This...

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: baum (lmb2768) HW04 distler (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A box slides down a ramp inclined at 23 to the horizontal with an acceleration of 2 . 27 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . Determine the coefficient of kinetic friction between the box and the ramp. Correct answer: 0 . 172838. Explanation: W N f x y Let the x axis run down the ramp. Then according to the Second Law of Newton ma = F net x = mg sin - f, 0 = F net y =- mg cos + N, and the kinetic friction coefficient follows from the ratio of the friction force f to the normal force N : f N = mg sin - ma mg cos = 1 cos sin - a g = 1 cos 23 sin 23 - 2 . 27 m / s 2 9 . 8 m / s 2 = . 172838 . 002 10.0 points A force F = 71 . 8 N acts at an angle = 39 with respect to the horizontal on a block of mass m = 27 . 2 kg, which is at rest on a horizontal plane. The acceleration of gravity is 9 . 81 m / s 2 . If the static frictional coefficient is s = . 83, what is the force of static friction? 27 . 2 kg s = 0 . 83 7 1 . 8 N 39 1. 183 . 967 N 2. 46 . 3132 N 3. 37 . 5037 N 4. 221 . 471 N 5. 45 . 1852 N 6. 175 . 157 N 7. 0 N 8. 267 . 784 N 9. 55 . 7991 N correct 10. 258 . 974 N Explanation: Let : m = 27 . 2 kg , F applied = 71 . 8 N , = 39 , s = 0 . 83 , and g = 9 . 81 m / s 2 baum (lmb2768) HW04 distler (56295) 2 m F a p p l i e d mg N F s Basic Concepts: F applied,x = F applied cos F applied,y = F applied sin F y,net = N - F applied,y- mg = 0 F x,net = ma x = F applied,x- F s = 0 when at rest. Solution: F applied,x = (71 . 8 N) cos39 = 55 . 7991 N F applied,y = (71 . 8 N) sin39 = 45 . 1852 N From the horizontal motion, ma x = F applied,x- F s 0 = F applied,x- F s = 55 . 7991 N- F s across the floor. Therefore, F s = 55 . 7991 N 003 10.0 points A 59 kg boy and a 42 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleration of the girl toward the boy is 2 . 2 m / s 2 , determine the magnitude of the acceleration of the boy toward the girl. Correct answer: 1 . 5661 m / s 2 . Explanation: According to Newtons Third Law of Mo- tion the force the boy exerts on the girl is the same as the force the girl exerts on the boy, so F g = F b m g a g = m b a b a b = m g a g m b 004 10.0 points A spherical mass rests upon two wedges, as seen in the figure below. The sphere and the wedges are at rest and stay at rest. There is no friction between the sphere and the wedges. M The following figures show several attempts at drawing free-body diagrams for the sphere....
View Full Document

Page1 / 11

HW04-solutions - baum (lmb2768) HW04 distler (56295) 1 This...

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