Where $F_s$ is the magnitude of the static friction force. Having said all of this, let me reiterate that kinetic friction always has the magnitude $\mu_k F_N$ regardless of the state of motion of the object. If you have been reading through Lessons 1 and 2, then Newtons first law of motion ought to be thoroughly understood. If you continuously push an object with a force greater than this value, then it will keep accelerating forever. The acceleration of the object is the rate of change of its velocity, so determine the velocity as a function of time you would, in general, have to solve the following differential equation: Lastly, given an applied force $F$, the acceleration of the object will satisfy Newton's second law which says that the net applied force equals the mass of the object multiplied by is acceleration In order for the acceleration to halt, you would need to reduce the applied force so that it equals the force of static friction.Īn object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. This answer is probably long past its sell by date?Īll you have described in your question are ideal cases with the frictional forces and the applied force constant. If those assumptions are made then in Classical Physics the mass will continue to accelerate and not reach a terminal speed. 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.However when one looks at a more realistic situation with the friction force being dependent on the speed of the mass and the limitations of the device applying the force then you might understand why it is that the acceleration might not continue to be constant for all time. You must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in (Figure)(a). 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. 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. 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). The greater the deformation, the greater the restoring force. 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. At this point, the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly and the sag is slight, so we do not notice it. But it is similar to the sagging of a trampoline when you climb onto it. This is a two-dimensional problem, since not all forces on the skier (the system of interest) are parallel. The approach we have used in two-dimensional kinematics also works well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two one-dimensional problems to solve. 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. (Motions along mutually perpendicular axes are independent.) We use x and y for the parallel and perpendicular directions, respectively. 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. Is drawn as the component of weight perpendicular to the slope. 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. Which is the acceleration parallel to the incline when there is 45.0 N of opposing friction. Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. It is a general result that if friction on an incline is negligible, then the acceleration down the incline is
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |