Tom enlists the help of his friend John to move his car. They apply forces to he
ID: 1436156 • Letter: T
Question
Tom enlists the help of his friend John to move his car. They apply forces to he car as shown in the diagram. Here = 430 N, 330 N and friction is negligible. Mass of the car = 3.50 into 10^3kg, theta_1 = 12.0 degree , and Theta_2 = 25.0 degree . The diagram below shows the top view of the car which is in the x-z plane horizontal plane). Find the resultant force exerted on the car. (Express your answer in vector notation.) What is the acceleration of the car? (Express your answer in vector notation.)Explanation / Answer
along horizantal
Fx = F1x + F2x = (430*cos12)+(330*cos25) = 719.68 N
along vertical
Fy = F1y + F2y = -(430*sin12)+(330*sin25) = 50.1 N
Fnet = sqrt(Fx^2+Fy62)
F = sqrt(719.68^2+50.1^2)
F = 721.42 N <<<------answer
(b)
acceleration = Fnet/m = 0.206 m/s^2 <<<<<------answer