Answer :
Problem 1:
Solomon Grundy applies 1500 N at 20° below, resulting in 1417.3 N horizontally, 509.5 N vertically, and a normal force of 1960 N. Vertical acceleration is 2.548 m/s².
Problem 2:
Workers lift a 140 kg sign. Net horizontal force is 700 N, vertical force is 346.41 N, resulting in horizontal acceleration of 5 m/s², vertical acceleration of 2.474 m/s², and net acceleration of 7 m/s².
Problem 3:
Lifting a 120 kg sign, horizontal force difference is 280.8 N. Horizontal acceleration is 3.744 m/s².
Problem 1: Solomon Grundy Pushing a CrateA
Calculating F_net,x:
F_net,x = F_applied * cos(20°)
F_net,x = 1500 N * cos(20°)
F_net,x ≈ 1417.3 N
Calculating F_net,y:
F_net,y = F_applied * sin(20°)
F_net,y = 1500 N * sin(20°)
F_net,y ≈ 509.5 N
Calculating Normal Force (N):
N = m * g
N = 200 kg * 9.8 m/s^2
N ≈ 1960 N
Calculating a_y:
a_y = F_net,y / m
a_y = (1500 N * sin(20°)) / 200 kg
a_y ≈ 2.548 m/s^2
Problem 2: Construction Workers Lifting a Sign
Calculating F_net,x and F_net,y:
Calculate F_1x, F_1y, F_2x, F_2y using trigonometry.
F_1x = 500 N * 0° = 500 N
F_1y = 500 N * sin(0°) = 0 N
F_2x = 400 N * cos(60°) = 200 N
F_2y = 400 N * sin(60°) = 346.41 N
F_net,x = F_1x + F_2x = 500 N + 200 N = 700 N
F_net,y = F_1y + F_2y = 0 N + 346.41 N = 346.41 N
Calculating a_x and a_y:
a_x = F_net,x / m = 700 N / 140 kg = 5 m/s^2
a_y = F_net,y / m = 346.41 N / 140 kg = 2.474 m/s^2
Calculating Net Acceleration (a_net):
a_net = sqrt(a_x^2 + a_y^2) = sqrt(5^2 + 2.474^2) = sqrt(25 + 6.122) ≈ 7 m/s^2
Problem 3: Construction Workers Moving a Sign on the Ground
Calculating F_net,x:
Calculate F_1x, F_2x using trigonometry.
F_1x = 1500 N * cos(15°) = 1449.43 N
F_2x = 1300 N * cos(25°) = 1168.63 N
F_net,x = F_1x - F_2x = 1449.43 N - 1168.63 N = 280.8 N
Calculating a_x:
a_x = F_net,x / m = 280.8 N / 75 kg = 3.744 m/s^2
