The discussion focuses on calculating the magnitude and direction of the normal force acting on the second tier of a three-tiered birthday cake, with weights of 16 N, 9 N, and 5 N for each tier, respectively. The normal force (FN) is determined using the equation FN = FWcos(theta), although the primary consideration is the vertical forces since the system is static. The total weight on the second tier is the sum of the weights of the tiers above it, which is 5 N (top tier) + 9 N (second tier) = 14 N acting downward. Therefore, the normal force acting on the second tier is 14 N acting upward.
PREREQUISITESThis discussion is beneficial for physics students, educators, and anyone studying mechanics, particularly those interested in understanding forces acting on static objects.
Where did you get this equation? Resist the temptation to memorize equations from other example problems, that will only serve to confuse you. Keep it basic. Everything is in the vertical direction, and nothing is moving, so sum forces in the vertical direction = 0 , when looking at Free Body Diagrams. Please get familiar with them, as they are essential in the study of mechanics. In a Free Body Diagram, you isolate the system , or part of the system, and examine the forces acting on it, both contact forces (like normal forces) and non-contact forces (like gravity), and apply Newton's laws. Give it a try.becsantos said:Homework Statement
A three-tiered birthday cake rests on a table. From bottom to top, the cake tiers weigh 16 N, 9 N, and 5 N, respectively. What is the magnitude and direction of the normal force ating on the second-tier?
Homework Equations
FN=FWcos(theta)