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

2024
Views:

322

Pages:

7
Author:

Dawson Thornton

SOLUTIONS TO HOMEWORK PROBLEMS HW#4 13-9, 19, 39, 59, 72, 81, 88, 111 Problem 13-9 The elevator E has a mass (M_{E}), and the counterweight at A has a mass (M_{A}). If the motor supplies a constant force F on the cable at B, determine the speed of the elevator at time t starting from rest. Neglect the mass of the pulleys and cable. Units Used: [kN=10^{3}N] Given: [M_{E}=500~kg] [M_{A}=150~kg] [F=5~kN] 1 [t=3~s] Solution: Guesses [T=1~kN] [a=1\frac{m}{s^{2}}] [v=1\frac{m}{s}] Given [T-M_{A}g=-M_{A}a] [F+T-M_{E}g=M_{E}a] [v=at] Find(T,a,v) [T=1.11~kN] [a=2.41\frac{m}{s^{2}}] [v=7.23\frac{m}{s}] Problem 13-19 A suitcase of weight W slides from rest a distance d down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C? Given: W = 40 lb θ = 30 deg d' = 201 ft 1. www.docsity.com www.docsity.com h = 4 ft g = 32.2 ft/s² Solution: W sin(θ) = (W/g)a a = g sin(θ) a = 16.1 ft/s² v_B = √(2ad) v_B = 25.377 ft/s t_AB = v_B / a t_AB = 1.576 s Guesses t_BC = 1 s R = 1 ft Given (-g/2)t_BC² - v_B sin(θ)t_BC + h = 0 Find(t_BC, R) t_BC = 0.241 s R = 5.304 ft R = v_B cos(θ)t_BC t_AB + t_BC = 1.818 s Problem 13-39 The conveyor belt delivers each crate of mass M to the ramp at A such that the crate's speed is v directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is μk, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Given: M = 12 kg vA = 2.5 m/s d = 3 m μk = 0.3 θ = 30 deg g = 9.81 m/s² Solution: NC - Mgcos(θ) = 0 NC = Mgcos(θ) Mgsin(θ) - μkNC = Ma a = gsin(θ) - μk(Mgcos(θ)/M) a = 2.356 m/s² vB = √(vA² + 2ad) vB = 4.515 m/s