Question

Q1 (b) A truck A pulls a crate B up a slope that is inclined at an angle of 150 to the horizontal, as shown in Figure Q1(b). The rope passes over a pulley of negligible size at C. At the instant shown, r 30 m, 0-45o and the truck has a speed of vA 20 km/h and an acceleration ax-0.2 m/s i) Express the velocity components of the crate B along the rope (4 marks) BC and transverse to it, in terms of the polar coordinates, r, θ , and their time derivatives. Write your answer in the form Ug = ( ), + ( )ee-and show the unit vectors, êr and êe, on a diagram. Hence calculate the speed, vB, of the crate B, and the rate of change of the angle (6 marks) (ii) 0 at the instant shown (3 marks) (iii) Calculate the acceleration, aB, of the crate B. Solution: VA, aA VB. aB 15°

Answer 1

Concept: velocity and acceleration along the rope will be same because for different velocity along the rope, length of rope will change and that is not possible.

Concept: velocity and acceleration along the rope will be same because for different velocity along the rope, length of rope will change and that is not possible. Velocity and acceleration of cart B along rope will be same because length of rope is constant V_B cos(theta - 45 degree) = V_A V_B cos(45 - 15 degree) = V_A V_B squareroot 3/2 = 20 km/hr V_B = 23 middot perpendicular km/hr V_B along r^^= V_B cos(45 - 15 degree) = 20 km/hr V_e theta = V_B sin(45 - 15 degree) = 11.55 km/hr V^vector_B = 20 e^^_r + 11.55 e^^_theta \r\n (ii) speed of rote B = V_B = 23 middot perpendicular km/hr Now, we know - rote of change of angle theta = > d theta/dt = w w = angular velocity = V_e theta/r d theta/dt = (11.55 km/hr/30m) = (11.55 times 5/18 m/sec/30m) d theta/dt = 0.107 rad/sec^2 acceleration a_B cos(theta - 15 degree) = a_A a_B = a_A /cos 30 degree = 0.2