A 1000 kg car is pushing an out-of-gear 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N. Rolling friction can be neglected. (a) What is the magnitude of the force of the car on the truck?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with Newton's 2nd and 3rd laws. So we have a 120 kg motorcycle pulling a 200 kg cart cart is secured using a hook during take off the rear tire exerts a 90 Newton horizontal force against the gravel road. We're asked to determine the magnitude of the force exerted by the motorcycle on the cart, ignore rolling friction. So multiple choice answers are a 56.3 Newtons, B 90 Newtons, C 146 Newtons for D 33.8 Newtons. So the first thing we're going to do is kind of draw a little forced diagram about what is happening here. So we have the motorcycle, we know the motorcycle exerts a force from the tire and that's in the horizontal direction. And then we have the motorcycle attached to the cart. So the motorcycle is going to exert a force on the cart and the cart is going to exert a force on the motorcycle. For Newton's third law, we can recall that those forces are going to be opposite and equal. So from Newton's second law, we can recall that force equals mass times acceleration. So we're going to take the motorcycle and look at the forces acting on the motorcycle. And then we'll take a look at the cart and the forces acting on the cart. So let's look at the motorcycle force first, the sum of the forces on the motorcycle is equal to the mass of the motorcycle times the acceleration of the motorcycle. And when we look at the motorcycle, the forces acting on the motorcycle, it's going to the force of the tire minus force of the cart on the motorcycle. When we look at the cart again, going to be equal to mass of the cart times acceleration of the cart, all we have is the force of the mass on the cart. Okay. So from here from Newton's third law again, we can recall that force of the cart on the motorcycle is equal to the force of the motorcycle on the cart. And knowing that this is all moving together because they are secured the card secured on the motorcycle with a hook. We know that the accelerations are the same A M equals A C. So because the accelerations are the same, we can solve both for the acceleration and then set those accelerations equal to each other. So that looks like F T minus F and we're just gonna call it M C because they're the same motorcycle and car card on motorcycle. It's the same magnitude over mass of the motorcycle is equal to that force over mass of the cart. And we are asked to solve for this F M C. And so we will isolate it, solve for it and then plug in isolated and then plug in what we know to solve for it. So we have F T over mass of the motorcycle F M C for mass of the cart. See over mass motorcycle, we can simplify that further To the FMC and by itself one Over mass in the cart plus one over the mass of the motorcycle. And then from here, we can plug in Our known values. So the force of the tire was given to us as 90 Newtons, The mass of the motorcycle was given to us in the problem as 120 kg Equals FMC Times one over Mass of the cart kg plus one over mass of the motorcycle kg. So we simplify this. It's going to be 0.75 newtons per kilogram and then one over 200 kg plus one, over 120 kg Is 1.3, 3 times 10 To the -2, That Unit one over telegram. So that equals our force. And let me plug that in, simplify it again, we get 56.3 nudes. And that is the answer to this problem. We go back up to our multiple choice solutions. We can see that that is answer choice A so that's all we have for this one. See you in the next video.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Force Interaction

System of Objects