Two Dogs Pulling a SledDate: 10/15/2001 at 17:32:43 From: mike Subject: Son's homework question Two dogs are pulling a sled across an icy road. We assume that the friction between the sled and the road is negligible. A person keeps the sled on the proper path by pulling on it with a rope attached at the point where the ropes from the dogs are attached. A noise startles the dogs and they move toward opposite sides of the road, one at an angle of 60 degrees with the centerline and the other at an angle of 30 degrees with the centerline, each with a force of 200 N. What are the magnitude and direction of the force that the person will have to exert on the sled to keep it moving along the centerline of the road at constant velocity? Thank you for your help, Mike Date: 10/17/2001 at 10:45:19 From: Doctor Code Subject: Re: Son's homework question Hi Mike, First let's draw a picture of the situation: Y D ^ / | / | / | / 60 deg -------> X P<-------S---------------------road \ 30 deg \ \ \ D S is the sled, P is the person, the D's are the dogs, and the horizontal line is the center line of the road. The angles on the picture look the same because I can't draw them correctly on the keyboard, but they should be 60 degrees and 30 degrees from the center line. I have also drawn the positive X and Y axes. You can put these axes in any orientation, but I have chosen this one so that the X axis coincides with the direction of the road. Notice that if both dogs were pulling with equal force at the same angle (both at 60 degrees or both at 30 degrees), the sled would travel along the center line. But since the angles aren't equal, the sled is being pulled off course. The first thing to do is to find the total force the dogs are exerting on the sled. The forces are vectors. We can't just add the vectors together, because they are pointing in different directions. In order to add them, we must first resolve the vectors into X and Y components. For example, the 60 degree dog's vector is: D / / / / S If we resolve it into X and Y components, it will look like this: ^ | | | | S----> The above picture shows one vector along the X direction and one along the Y direction. If we do the same thing with the other dog, we can add the two X vectors together, and the two Y vectors. In order to turn the vectors into component vectors, we have to use some trigonometry: D /| / | / | /60 | S----> The angle from the X axis to line SD is 60 degrees. We know that SD is 200 N. We need to find the bottom leg of the triangle: X component = 200 * cos 60 The Y component is found in the same way, except you use sin instead of cos. Now we can add the components together: X: Fx = 200 * cos 60 + 200 * cos 30 = 273.2 N Y: Fy = 200 * sin 60 - 200 * sin 30 = 73.2 N Note that the second force in the y direction is negative, because the 30 degree dog is on the other side of the road. The total value of this new force is: F = sqrt((Fx)^2 + (Fy)^2) = sqrt(273.2^2 + 73.2^2) = 282.8 N And the direction of this force can be found with the tangent: tan(angle) = 73.2/273.2 angle = arctan(73.2/273.2) = 15 degrees This means that the two dogs working together produce the same force as a big dog pulling at 15 degrees above the road line with a force of 282.8 N. Since we want the sled to travel with constant velocity along the road, the force that the person exerts must exactly cancel the combined force of the dogs. This will produce zero acceleration and constant velocity. So the person must pull with 282.8 N on the other side of the sled, but 15 degrees below the center road line: D / / / / 60 deg road-----S---------------------road 15 / \ 30 deg / \ / \ / \ P D Write back if you need more help. - Doctor Code, The Math Forum http://mathforum.org/dr.math/ |
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