# Forces in Two Dimensions - Mission F2D5 Detailed Help

 An object of mass 'm' is placed upon an inclined plane with an incline angle of 'theta'. The surface is rough and there is a coefficient of friction of 'mu.' The force of friction acting upon the object is equivalent to ____.
 The force of friction depends upon the normal force and the coefficient of friction (see Formula Frenzy section). The normal force is always directed perpendicular to the surface that the object is on - in this case, perpendicular to the inclined plane. For inclined plane problems, the normal force is equal to the perpendicular component of gravity so that there is a balance of forces along an axis perpendicular to the plane (see Math Magic section).
 The force of gravity is neither in the direction of the acceleration nor perpendicular to it. The goal is always to have all forces directed perpendicular or parallel to each other and to the acceleration. So it is the habit on inclined plane problems such as this one to resolve the force of gravity into two components - one being parallel to the inclined plane and the other perpendicular to it. The formulas for resolving the force of gravity into its components are:   Fparallel= m • g • sine(Θ)  Fperpendicular= m • g • cosine(Θ)
 The force of friction (Ffrict) experienced by an object is often calculated using the equation:    Ffrict = mu • Fnorm where mu is the coefficient of friction (dependent predominantly upon the nature of the two surfaces that are in contact) and Fnorm is the normal force. For a static situation (the object is at rest), the static friction force can be less than or equal to mu•Fnorm. Static friction applies an amount of force equal to any accelerating force but eventually reaches its upper limit as more and more accelerating force is applied. When the upper limit is reached, the object budges from its at-rest-state and begins accelerating. It is at this moment that kinetic or sliding friction is the opposing force.

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