Dog Sled Problem

Dog Sled Problem

In this video we’ll be working on a problem involving sled dogs I’ll let you read the question first before we start Let’s picture the situation: we have eight sled dogs pulling the loaded sled forward. Let’s work with symbols first. Before starting the problem we need to define our system. This would be the sleddogs and the loaded sled. They are moving together as one object. Next, we need to define the coordinate system. I’m choosing the left side to be positive which is the direction of motion of the sled. The acceleration and the net force will, therefore, be positive. Any force opposite to acceleration will be negative. Since we have forces, let’s use
Newton’s second law: ∑F=ma. This means the sum of all forces equals the mass of the object and its acceleration. Let’s see what forces we have: we have kinetic friction because the sled is moving; we have the backwards force of the dogs. But these forces are opposite the acceleration, so what force is causing the acceleration forward? Is friction always opposite to the motion? No. How can this be? Doesn’t friction try to stop the motion of an object? Although acceleration is in the forward direction, the tires of the car push against the road. We need a force to push forward and this force is friction So, friction is actually helping the car move forward. It’s always important to draw the free body diagram and analyze the direction of the friction force– it’s not always opposite to the motion! So what does this have to do with our dog-sled problem? Static friction helps the dogs move forward! Let’s see what I mean: When the dog moves, she exerts a backward force against the snow. We’ll call this paw on ground. Now, static friction helps with the forward motion which we’ll call ground on paw. The static friction helps the dog move forward by preventing slippage of the paw and allows muscles to move the dog forward. So while the static friction doesn’t directly help the dog move forward, it provides an anchor point, just like when we walk. Without this static friction, the dog wouldn’t be able to run because she would slip. Now we know that this static friction helps the dog move forward and the dog exerts a backward force. So we have a force pair here– this is Newton’s third law. Ground on Paw and Paw on ground are action reaction forces. The muscle force moving the dog forward is equal in magnitude to the paw on ground force. Let’s go back to the question: what forces do we have? Remember, we consider only forces acting on our system, the dogs and the sled. So ∑F=ma =F (ground on paw) is the propulsion force in the form of static friction that is helping our dogs move forward. The net force would be ground on paw and the kinetic friction by the sled. Let’s take this out of vector notation: Remember that it’s the sum of the forces, so we’re adding forces. For the kinetic friction, we’re adding a negative force. In case you’re wondering why we’re not including the backward force exerted by the dogs, this force is exerted on the ground, so it is not part of the net force acting on our system. Here I rearranged for the acceleration that we’re looking for Notice that I have put an 8 in front of the forward motion of the dogs’ force; this is because we have 8 dogs. Let’s review what we have for this question: we have the force exerted by each dog, all masses and the friction coefficient. And we found the force exerted for each dog which is the force ground on paw. Let’s use some numbers now. So for the acceleration mass, we will use the mass of the 8 dogs + the mass of the loaded sled. So now you can solve the rest of the problem and plug in the numbers. We get 3.62 m/s2 I hope that you found this video helpful!

Antonio Breitenberg

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