University:
Missouri State University
Course:
PHY 203 | Foundations of Physics I
Academic year:

2019
Views:

419

Pages:

12
Author:

ck5o8fauc

AP Mechanics Chapter 6 Review (Friction, Drag, and Circular Motion) All Problems and Questions below can be found on textbook pages 139-148 Circular Motion Banked Turns: Sample Problem 6-06 from chapter & P#'s 53,51 Horizontal Circles: P# 87 Vertical Circles: Pr# 47 Conical Pendulum: P# 70 Basic Friction: P#'s 1,2 Forces @ Angles: P# 30a Inclined Planes: P# 17 Pulleys: Old HW problems Objects in Contact: What maximum force can be applied to the bottom block so that the top block will not slide? Frictionless 5 kg 55 Friction 10 kg F Drag 2000M2. A rubber ball of mass m is dropped from a cliff. As the ball falls, it is subject to air drag (a resistive force caused by the air). The drag force on the ball has magnitude by, where b is a constant drag, coefficient and v is the instantaneous speed of the ball. The drag coefficient & is directly proportional to the cross-sectional area of the ball and the density of the air and does not depend on the mass of the ball. As the ball falls, its speed approaches a constant value called the terminal speed. a. On the figure below, draw and label all the forces on the ball at some instant before it reaches terminal speed. b. State whether the magnitude of the acceleration of the ball of mass m increases, decreases, or remains the same as the ball approaches terminal speed. Explain. c. Write, but do NOT solve, a differential equation for the instantaneous speed of the ball in terms of time 1, the given quantities, and fundamental constants. d. Determine the terminal speed v, in terms of the given quantities and fundamental constants. e. Sketch a gogh of the objects velocity. VS. time from its initial release from rest to terminal velocity. Include any Knowr velves in the axes, P.115 (F) TEXTBOOK r = 4.1m F_{5} = m*f_{5} C F n y =F T cot infty F_{T} F T x =F T sndG a_{r} = (s*r_{r})/m F_{T} * cos theta = F_{g} F_{T} * cos Delta = ng o = (F_{r} * cos beta - F_{f})/m F_{T} = (ng)/(cos alpha*g) J a_{c} = (F_{r}*s*omega*Q)/M (V ^ 2)/r = F T zeta-4 emptyset mu r^ 2 tilde r g = ln(d*theta) a_{H} = (d*f_{H})/kappa Tr A ^ m [ matrix (L_{1}*mu * mu ^ k)/((9, 1)(4mu)) ]=Q (v ^ 2)/r = mdQ T A overline d' * (v ^ 2)/(rg) = Q 125 deg = 0 deg C Y = 16 16/11 * (1n)/(1k) * (1.116k)/3440.5 = 4.44 * n/7 B = objects in contact top box F_{N} f V 49N f smax = mu_{5}*Fw = (0.55)(49N) = 27N Sigma*F_{A} = max 27 = 5ax a_{x} = 5.4sqrt^2 s So acceleration of whole thing must be <= 5.4n / (s ^ 2) for "whole thing" 2F_{x} = max F = (10 + 5)(5.4) = boxed 81N Tor Rx a = (SF)/m a = F f forex 5 a = (Sigma*F)/M a = F/15 a = a F fhnw 5 = F/15 27/5 = F/15 IF = 81N drag force b * v ^ 2 decreases: gets Sigma*F = ma V gets bigger, buz gets bigger so SF bv bu²-my z ma b * v ^ 2 - mg = m * d/dt (v) terminal velocity occurs. when F_{D} = F_{g} b * v ^ 2 = mg v = sqrt((mg)/b) t