OpenStax College Physics for AP® Courses, Chapter 4, Problem 15 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We're going to assume that the force exerted on the artillery shell by the boat is the net force on the artillery shell which isn't strictly true which is why I've drawn a squiggly sort of approximately equal sign here because there is also the earth exerting a force on the artillery shell. So this force exerted by the boat isn't the entire force, it's also the earth's force. But this force due to gravity is small in comparison to the force exerted by the boat because the force exerted by the boat is, you know, with this assumption we can calculate it to be 2.6 times ten to seven newtons. But the weight of the artillery shell being it's mass times gravitational field strength of the earth is 1100 kilograms and let's say times ten approximately and that's about eleven thousand newtons which is 1.1 times ten to the four newtons which is three orders of magnitude smaller than this force. So this weight would affect some decimal way over here somewhere after the six and -- but this number is precise only to two significant figures anyway, so this would have no effect that would make any difference on our calculation. So this assumption is okay. So there is a force 2.6 times ten to the seven newtons applied by the boat on to the artillery shell. The force applied on the boat is equal and opposite to the force applied on the artillery shell. So let's write that a bit better here. So the force on the boat due to the artillery shell is in the opposite direction to the force on the artillery shell due to the boat. These are Newton's third law pairs because they are the force that one body exerts on another. Well that body, that other thing will exert the same force on the original thing but in the opposite direction. All right. So, 2.6 times ten to the seven newtons in the opposite direction to the shell's acceleration.