When the 5kg box reaches point A it has a speed of VA 10ms D

When the 5-kg box reaches point A, it has a speed of V_A =-10m/s. Determine the normal force the box exerts on the surface when it reaches point B. Neglect friction and size of the box.

Solution

Data Given

Mass of the box = 5 kg

Linear velocity = 10 m/s

At point B, x and y are equal and it has to satisfy the equation x^0.5 + y^0.5 = 3

Since x = y at point B, equation becomes square root * x + square root * x = 3

ie; 2 * square root x = 3

squaring both sides

4 * x = 3

x = y = 3/4

Radius of the profile at B = 9m

Angular velocity at B = linear velocity/ radius at B

= 10/9 = 1.11 radians/ second

The force(F) acting on the profile at point B due to angular acceleration 1.11 radians/ second

F = Mass * angular velocity^2 * Radius of the profile

= 10 * 1.11 * 9

= 99.9 Newtons