Final answer:
To find the tension in the chain pulling a 200kg weight on a horizontal surface at a 30° angle with an applied force of 3000N, we need to break down the applied force into its horizontal and vertical components. The tension in the chain is found by adding the vertical component of the applied force and the force due to the weight. The tension in the chain is approximately 3460N.
Step-by-step explanation:
To find the tension in the chain, we need to break down the applied force into its horizontal and vertical components. The horizontal component is given by F_h = F * cos(theta), where F is the applied force and theta is the angle. In this case, F_h = 3000N * cos(30°). The vertical component is given by F_v = F * sin(theta), where F is the applied force and theta is the angle. In this case, F_v = 3000N * sin(30°).
Since the weight is on a horizontal surface, the vertical force exerted by the weight is equal to the tension in the chain, T. Therefore, T = F_v + mg, where m is the mass and g is the acceleration due to gravity. In this case, T = F_v + mg = 200kg * 9.8m/s^2 + 3000N * sin(30°).
Calculating the values, we get T = 1960N + 1500N = 3460N. Therefore, the tension in the chain is approximately 3460N.