![]() ![]() The friction force is not, but the gravitational force has some component in the direction perpendicular to the track !() Where the yellow vector represents the component of gravity in the direction perpendicular to the track. ![]() The normal force is already perpendicular to the track. Anyway, if the acceleration perpendicular to the track is zero, the forces perpendicular to the track must add up to zero (vector sum). ![]() MANY MANY MANY people mess that part up). If his perpendicular velocity stays zero, his (or her) acceleration must be zero perpendicular to the track (notice that the acceleration is zero because the velocity STAYS zero, not because the velocity is zero. This means that his velocity perpendicular to the track is zero and stays at zero. For this case, the line rider stays on the track. The goal is to calculate the ?, so an expression for the normal force is also needed. This gives the following relationship between the magnitude of the normal force and the friction force: !() Where ? is the coefficient of kinetic friction between the two surfaces (in this case the line rider and the track). The harder two surfaces are pushed together, the greater the frictional force. The rider is moving down the incline, and the normal force is perpendicular to the incline, so work would be: !() Now work done by friction: !() Where there is a relationship between the friction force and the normal force (in this model). First, let us look at the work done by the normal force. For this case, I will calculate the work for each individual force. The work can be calculated one of following ways: !() Where the thetas are for the angles between the displacement and each force. !() To calculate the work, all forces will need to be included. Below is a diagram (a free body diagram) representing the forces on the rider as he (or she) is going down the incline. The force the track exerts on the rider can be broken into a component perpendicular to the track (called the normal force) and a component parallel to the track - friction. There is gravitational force and there is the force the track exerts on the rider. For the line rider track, there are only two forces (assuming no or negligible air resistance) acting on the line rider. If you don't like that, you can use the following instead: !() Where F and delta r are now the scalar magnitudes of the vectors and theta is the angle between F and delta r. In this case the dot product (or scalar product) is used. Since these are both vector quantities, you can not simply multiply them. Where work is defined as: !() Where F is the force acting on the object and delta r is displacement. ![]() Basically, the work done on an object changes its energy. Work - Energy Here is a crash course in the work-energy theorem. But, is there friction on non-horizontal portions? To test this, I will use the work-energy principle. Who would want a rider to stop in the middle of the track and be stuck? That wouldn't be fun. Ok, so no friction on the horizontal line. The rider will not stop, but continue on at a constant speed. If you do not believe me (and why should you?) try creating your own line rider track with a long horizontal section. If friction were present, the rider would slow down. This shows the rider traveling at a constant speed of 0.71 m/s. time for the line rider on the first horizontal portion of the track (before he or she goes down the incline). Let me show you something simple before further analysis: !() This is the x-position vs. Friction in Line Rider Is there friction in Line Rider? Does it function as physics would expect? To test this, I set up a simple track: !() Basically, a slope with a flat part to start with and to end with. ![]()
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