Substitute |N| by M g cos(35°) - |T| sin (25°) in eq 1 to get We now need to solve the system of two equations with two unknowns |T| and |N|. T = (Tx, Ty) = (|T| cos (25°), |T| sin (25°) )į s = (- |F s|, 0) = ( - μ s |N|, 0), where μ s is the coefficient of friction between the box and the inclined plane. T tension of string, W weight of the box, N force normal to and exerted by the inclined plane on the box, F s is the force of friction b)įorces and their components on the x-y system of axis. The coefficient of friction between the box and the inclined plane is 0.3.Ī) Draw a Free Body Diagram including all forces acting on the particle with their labels.ī) Find the magnitude of the tension T in the string.Ĭ) Find the magnitude of the force of friction acting on the particle. The string makes an angle of 25 ° with the inclined plane. A string is used to keep the box in equilibrium. |N| = 50 cos (30) = 25 √3 ≈ 43.3 N Problem 3Ī box of mass M = 10 Kg rests on a 35° inclined plane with the horizontal. The F a + W + N + F s = 0 in components form: The components form of all forces (vectors) acting on the box are: The box is at rest, hence its acceleration is equal to 0, therefore the sum of all forces acting on the box is equal to its mass times its acceleration which is zero. In the diagram below, W is the weight of the box, N the normal force exerted by the inclined plane on the box, F a is the force applied to have the box in equilibrium and F s the force of friction opposite F a. A force F a of magnitude 30 N acts on the particle in the direction parallel and up the inclined plane.Ī) Draw a Free Body Diagram including the particle, the inclined plane and all forces acting on the particle with their labels.ī) Find the force of friction acting on the particle.Ĭ) Find the normal force exerted by the inclined on the particle. |N| = |W| cos (27°) = 2 × 10 cos (27°) ≈ 17.8 N Problem 2 A particle of mass 5 Kg rests on a 30° inclined plane with the horizontal. Y-components are equal : - |W| cos (27°) + |N|= 0 X-components are equal : |W| sin (27°) + 0 = M |a| (|W| sin (27°), - |W| cos (27°)) + (0, |N|) = M (|a|, 0)įor two vectors to be equal, their components must be equal. In components form, the above equation becomes Use Newton's second law to write that the sum of all forces on the box is equal to the mass times the acceleration (vector equation) W = (Wx, Wy) = (|W| cos (27°), - |W| sin (27°))Ī = (a x, a y) = (|a|, 0), box moving down the inclined plane in the direction of positive x hence a y = 0. Use system of axes x-y as shown to write all forces in their Two forces act on the box: the weight W of the box and N the force normal to and exerted by the inclined plane on the box (blue point) View Notes.How to solve the house problem of the box on the inclined plane and label all forces acting on the box.ī) Determine the acceleration a of the box down the plane.Ĭ) Determine the magnitude of the force exerted by the inclined plane on the box. Technical information, teaching suggestions, and related resources that complement this Concept Builder are provided on the Notes page. Learners and Instructors may also be interested in viewing the accompanying Notes page. However, the 8 different groups of questions can be printed. There is no need for an activity sheet for this Concept Builder. Users are encouraged to open the Concept Builder and explore. Using the Concept Builder with our Task Tracker system allows for the storage of progress records. Numerical information in the situation is generated using a random number generator and will thus vary from person to person and from session to session. Question-specific help is provided for each of the 6 situations. There are two different situations to analyze in each level. There are three levels of difficulty with each level adding an additional complexity. The analysis involves the use of vector resolution and Newton's second law. The Solve It! - Inclined Planes Concept Builder challenges a learner to analyze a physical situation involving the acceleration of an object along an inclined planes. Concept Builders » Forces in 2-Dimensions » Solve It! - Inclined Planes
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |