Chapter 16: Dynamics Problems

University:
Georgia Southern University
Course:
ENGR 2232 | Dynamics of Rigid Bodies
Academic year:

2024
Views:

353

Pages:

2
Author:

Jalen W.

Problem 1: The minimum distance \( b \) between two vehicles shall be as big as the distance which the rear vehicle covers within \( t_s = 2 \) s at its constant velocity. a) Determine the minimum distance \( x_p \) required for passing. b) Determine the minimum time \( t_p \) a car (length \( l_1 = 5 \) m, constant speed \( v_1 = 120 \) km/h) needs to stay on the fast lane for passing a truck (length \( l_2 = 15 \) m, speed \( v_2 = 80 \) km/h) correctly? Disregard the time for changing the lanes. Matlab Code: % Given data l1 = 5; % length of the car in meters v1 = 120 * (1000 / 3600); % speed of the car in m/s l2 = 15; % length of the truck in meters v2 = 80 * (1000 / 3600); % speed of the truck in m/s ts = 2; % time in seconds % a) Determine the minimum distance xp required for passing xp = v2 * ts; % minimum passing distance % b) Determine the minimum time tp needed for passing correctly % The time needed for the car to pass the truck is the time taken to cover the combined lengths of the car and the truck tp = (l1 + l2) / (v1 - v2); % time for passing % Display results fprintf('Minimum passing distance required: %.2f meters\n', xp); fprintf('Minimum time needed for passing correctly: %.2f seconds\n', tp); Problem 2: To simulate absence of gravity, vacuum drop-shafts are used. Given is a shaft with a depth of \( l = 200 \) m. Determine the maximum available test time \( t_1 \) and test distance \( x_1 \) during free fall when the sample, after passing the test distance, is decelerated with \( a_{II} = -50g \) to \( v = 0 \). Matlab Code: % Given data l = 200; % depth of the shaft in meters g = 9.81; % acceleration due to gravity in m/s^2 aII = -50 * g; % deceleration in m/s^2 % To determine maximum available test time and test distance, we'll use equations of motion % During free fall: v^2 = u^2 + 2*a*s % a) Determine the maximum available test time t1 % Initial velocity (u) is zero % Final velocity (v) is also zero at the end of the deceleration % The acceleration (a) is acceleration due to gravity (g) until the deceleration % The deceleration (aII) acts after the test distance (x1) is covered % For the first part of the fall, until deceleration starts: t1_first_part = sqrt(2 * l / g); % For the second part, during deceleration: % v^2 = u^2 + 2*a*s => 0 = v^2 + 2*aII*x1 % So, x1 = -v^2 / (2*aII) x1 = -0^2 / (2 * aII); t1_second_part = abs(x1) / 0; % Since v is 0 % Total time is the sum of time for the first and second parts t1 = t1_first_part + t1_second_part; % b) Determine the test distance x1 % We already calculated it during the second part of the fall % Display results fprintf('Maximum available test time: %.2f seconds\n', t1); fprintf('Test distance during free fall: %.2f meters\n', x1);