It is a general result that if friction on an incline is negligible, then the acceleration down the incline is
Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. Which is the acceleration parallel to the incline when there is 45.0 N of opposing friction. The magnitude of the component of weight parallel to the slope is Then, we can consider the separate problems of forces parallel to the slope and forces perpendicular to the slope. Is drawn as the component of weight perpendicular to the slope. Is drawn parallel to the slope and downslope (it causes the motion of the skier down the slope), and Regarding the forces, friction is drawn in opposition to motion (friction always opposes forward motion) and is always parallel to the slope, This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and the acceleration is downslope. (Motions along mutually perpendicular axes are independent.) We use x and y for the parallel and perpendicular directions, respectively. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. Choose a convenient coordinate system and project the vectors onto its axes, creating two one-dimensional problems to solve. The approach we have used in two-dimensional kinematics also works well here. This is a two-dimensional problem, since not all forces on the skier (the system of interest) are parallel. But it is similar to the sagging of a trampoline when you climb onto it. The table sags quickly and the sag is slight, so we do not notice it. That is the situation when the load is stationary on the table. At this point, the net external force on the load is zero. Thus, when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. The greater the deformation, the greater the restoring force. Unless an object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or a trampoline or diving board).
This would be noticeable if the load were placed on a card table, but even a sturdy oak table deforms when a force is applied to it. But how do inanimate objects like a table support the weight of a mass placed on them, such as shown in (Figure)(b)? When the bag of dog food is placed on the table, the table sags slightly under the load. You must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in (Figure)(a). Weight (also called the force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling.