Chapter 6 Problems

5. (a) Find vertical and horizontal components of applied force at each angle. (b) For chair to slide, F_x must be greater than or equal to f_{s, max}

9. (a) F_min = f_{s, max} (b) F = f_k (c) F_net = f_{s, max} - f_k

11. (a) block will not slide if f < m_sN (b) Force on block is -Ni + fj

16. When block begins to slip, m_s = tanq. For m_k, solve for acceleration in terms of 2nd law and d = 1/2 at²; set two equations equal and solve for m_k

27. (a) Draw free body diagram. Resolve applied force and weight into parallel and perpendicular components. Solve for N in terms of weight and applied force. Write 2nd law equation for acceleration. (b) Use kinematics equation with final velocity = 0 (c) For block to move downwrd, force down plane must be greater than f_{s, max}

29. (a) Before motion, F and f_{s, max} are directed up ramp. This is an equilibrium condition, up and down forces must cancel. (b) For block to move up, f_s is directed down ramp and applied force must be greater than or equal to parallel component of weight plus friction. (c) Constant v means no net force, use f_k

30. Draw free body diagrams for block B and knot. Resolve T at angle into horizontal and vertical components. It's and equilibrium situation so forces up = forces down and forces left = forces right. Write equations relating forces. Tension in cord between block B and knot must be < f_s

51. (a) See sample problem 6-11 (b) From problem 16, just before slipping, m_s = tanq

58. (a) v = d/t (b) centripetal acceleration (c) f_s = ma (d) f_{s, max} = centripetal force

63. At critical speed, tension in cord = 0, weight = centripetal force.