University:
Angelo State University
Course:
ENGR 2302 | Engineering Mechanics - Dynamics
Academic year:

2024
Views:

302

Pages:

4
Author:

sommatosmby

SOLUTIONS TO HOMEWORK PROBLEMS CHAPTER 14 Planar Kinetics of Rigid Bodies; HW#9 14-6, 14, 18, 26 Problem 14-6 The crate of mass M is subjected to forces F_{1} and F_{2} as shown. If it is originally at rest, determine the distance it slides in order to attain a speed v. The coefficient of kinetic friction between the crate and the surface is D_{k} Units Used: kN = 10 ^ 3 * N Given: F_{1} theta_{t} F_{2} theta_{2} M = 100kg F_{J} = 800N F_{2} = 1.5kN theta_{f} = 30 deg theta_{2} = 20 deg v = 6 * m/s g = 9.81 * m/(s ^ 2) mu_{k} = 0.2 Mg F_{4} F_{2} theta perp theta_{2} mu_{k}*N_{k} N_{C} N_{C} = 867.97N Solution: N_{C} - F_{I} * sin(theta_{I}) - Mg + F_{Z} * sin(theta_{2}) = 0 N_{C} = F_{I} * sin(theta_{I}) + Mg - F_{2} * sin(theta_{2}) T_{1} + U_{12} = T_{2} F_{1} * cos(theta_{1}) * s - mu_{k}*N_{c}*s + F_{2} * cos(theta_{2}) * s = 1/2 * M * v ^ 2 s = (M * v ^ 2)/(2(F_{1} * cos(theta_{1}) - mu_{k}*N_{C} + F_{2} * cos(theta_{2}))) s = 0.933m Problem 14-14 Determine the velocity of the block 4 of weight W_{A} if the two blocks are released from rest and the block B of weight W_{B} moves a distance d up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is mu_{k} Given: W_{A} = 60lb W_{\mathcal{B}} = 40lb theta_{j} = 60 deg theta_{2} = 30 deg d = 2f mu_{k} = 0.1 Solution: L = 2s_{A} + 5B 0 = 2v_{A} + v_{B} Guesses v_{A} = 1 * (ft)/s Given W_{f} theta_{12} 27 F_{A} N_{A} W_{2} theta_{2} A theta_{1} 15 theta_{2} F_{B} N_{k} v_{B} = - Pi/s 0 = 2v_{A} + v_{B} W_{\mathcal{A}}(d/2) * sin(theta_{1}) - W_{\mathcal{B}} * d * sin(theta_{2}) - mu mathcal E W mathcal A cos( theta 2 ) d 2 ...= 1 2g ( W A v A ^ 2 + W B v B ^ 2 ) - mu_{kappa} * W_{B} * cos(theta_{2}) * d VA VB = Find(VA-VB) v_{B} = - 1.543 * (ft)/5 v_{A} = 0.771 * (ft)/s S_{A} S_{H} Problem 14-18 The collar has mass M and rests on the smooth rod. Two springs are attached to it and the ends of the rod as shown. Each spring has an uncompressed length /. If the collar is displaced a distance s = s deg and released from rest, determine its velocity at the instant it returns to the point s = 0 Given: k' = 100 * N/m M = 20 kg x' = 0.5m /= 1 m d = 0.25m Solution: 1/2 * k * s' ^ 2 + 1/2 * k' * s' = 1/2 * M * v_{c} ^ 2 v c = sqrt((k + k')/M) * g^ prime prime v_{c} = 1.37 * m/s k = 50 * N/m M_{B} F_{S} F_{g} ^ * N_{C}