Understanding Point B: Solving for Normal Force at Point B

[Miliman13]

I need help understanding point B
Point A i get.
Normal force at B = m x v2/r - m x g

= 1.68 x 12 x 12 /5 - 1.68 x 9.8

= 31.92 N

I can't make sense of why the forces are subtracting if A radial and MG are both positive

 

[Miliman13]

http://www.cmt.ua.ac.be/golib/PHYS101/exams/HW_ch06_Lec02.pdf

 

[haruspex]

View attachment 212372

I need help understanding point B
Point A i get.
Normal force at B = m x v2/r - m x g

= 1.68 x 12 x 12 /5 - 1.68 x 9.8

= 31.92 N

I can't make sense of why the forces are subtracting if A radial and MG are both positive

The centripetal force is not an applied force, it is a resultant force, so you do not add it to other applied forces. Instead, sum of applied forces = resultant.

 

[Miliman13]

The centripetal force is not an applied force, it is a resultant force, so you do not add it to other applied forces. Instead, sum of applied forces = resultant.

my textbook says the same thing Sum of forces and Radial Acceleration always point in.

But i still don't know how to draw a diagram that allows me to visually see how the algebra equates the solution,

I keep wasting paper trying to draw it but nothing matches the math

 

[haruspex]

how to draw a diagram

The diagram at A would show normal force up, mg down, resultant up: N-mg=mv²/r.
At B, normal force down, mg down, resultant down. How you write that as an equation depends on your sign convention. If you choose up as positive for all forces, N-mg=-mv²/r, N being negative now. Or you can choose down as positive for N and write N+mg=mv²/r.