Inclined Plane without Friction

A very important application in mechanics is the inclined plane. The following incline plane is assume to be frictionless. Inclined planes are defined based on the angle with the horizontal, angle q in the figure below. Given the angle q = 350 and the mass, m = 220 Kg, of the box. How should this problem be solved when calculating for the acceleration and the normal force?

Find the acceleration of the box:
 N a) 5.62 m/s2 b) 9.80 m/s2 c) 4.90 m/s2 d) 2.15 m/s2 e) None of the above.

and the normal force

 a) 956 N b) 1116 N N c) 1766 N d) 2156 N e) None of the above.

 In addition to follow the recommendations stated in Application I, this problem should be initially solved by drawing the actual vectors present in the problem . These vectors are either forces or accelerations. In the absence of friction, the vectors present are: Forces The weight of the box. This force is always vertical directed to the center of the Earth. And, The normal force on the box due to the reaction of the surface of the inclined. This force is always perpendicular to the reacting surface. Acceleration Since there is not any friction, nothing prevent the box for sliding down the inclined. Therefore, there is an acceleration along the direction of motion along the inclined surface. Based on the vectors of the figure, a different coordinate system should be selected. The standard horizontal x-axis and vertical y-axis are not the most convenient one for this situation. Notice that there are two vectors that are already perpendicular to each other, the acceleration and the normal force. Thus, the coordinate system should be selected along these directions . In this coordinate system, the only vector that need to be broken down into components is the weight of the box. A very important step in the solution of this problem is to find out where between the vectors and the coordinate system is the angle q replicated  . Notice that the extension of the y-axis to the horizontal, the horizontal, and the inclined plane itself form a right angle triangle with known angles, 900 , 900 -  q, and q . A second triangle is formed by the same extension of the y-axis to the horizontal, the horizontal and the vertical extended from the vector representing the weight of the object . The x-component of the weight correspond to the opposite side with respect to the angle q . After finding the y-component of the weight (adjacent side with respect to the angle q), the diagram associated with this problem is completed . Finally, the FBD of the problem can be obtained . Now, from the diagram, the equations involving the unknowns, a and FN, can be obtained. This equations are obtained by analyzing each one of the components. x-Component: From the diagram (Last figure of the previous sequence), the net force in the x direction is   Applying Newton's second law to this component, The first equation is obtained equating the two previous expressions for the normal force, From where the algebraic form of the acceleration is obtained y-Component: And the net force in the y direction is Applying Newton's second law to the y-component, Notices that there is not motion along the y-direction. Therefore, the acceleration along this direction is zero. Equating the two previous relations, the second equation is obtained, Thus, The numerical solution is Thus, the acceleration is ;and, the normal force is .

 by Luis F. Sáez, Ph. D. Comments and Suggestions: LSaez@dallaswinwin.com