Problem calculating the force on a tube wall

[adam74269]

Homework Statement

A small ball, considered a material point, moves from point A through a tube whose axis of symmetry is in the plane of the drawing. Find the force exerted by the ball on the wall of the tube at point C. Ignore friction on the curved parts of the trajectory. The ball, after traveling a distance of ℎ0, detaches from the spring.

The following notations are used in the problem:
m (0,3 kg) - mass of the ball;
VA (0 m/s) - initial speed of the ball;
t (0,1) - duration of the ball's movement along segment BD;
f (0,1) - coefficient of friction between the ball and the wall of the tube;
ℎ0 (50 cm) - initial deformation of the spring;
h - maximum compression of the spring;
k (10 N/cm) - spring stiffness coefficient;
H - maximum height the ball reaches;
s - distance travelled by the ball before stopping.
α = 30°, β = 60°, R = 1m

Relevant Equations

mv1-mv0=∫G+F+T dt
G - gravitational force,
F - friction force
T - elastic force

(mv1^2)/2-(mv0^2)/2 = A
A = AG + AT + AF - work that was done by the forces

A-A'
(mv1^2)/2-(mv0^2)/2 = A
A = -AG + AT - AF

AG = mgh = 1.47 J
AT = k/2*h0^2 = 0.0125 J
AF = fmg*cos(60)*h0 = 0.0735 J

A = -AG*cos(30) + AT - AF = -1.334

A-C
(mv1^2)/2-(mv0^2)/2 = A
A = AG

C
NC + Gcos(45) - Φn = 0

 

[Baluncore]

"Find the pressure exerted by the ball on the wall of the tube at point C".
The ball has a specified mass, but is considered a material point, without contact area. The pressure must therefore be infinite.
Check the question, maybe you need to find the force applied by the ball on the wall of the tube at point C.

 

[AR72]

I'm going to try to use your notation for this explanation, hope it makes sense:
The energy you get for the A-A' region should be positive, and this is will give you your total energy which will remain constant until you reach B. Only fluctuating between kinetic and gravitational potential.
A = AT - AG - AF
Also, realize that when integrating hooks law we find the energy imparted by the spring uses x values of initial length from equilibrium and final length from equilibrium, the difference between these is h0. If the spring releases after or before reaching equilibrium your answer will change.
Assuming h>h0:
AT=∫kxdx = (kh^2)/2 -(k(h-h0)^2)/2

For gravity do not use H, the max height of the system. This is where gravitational potential is maximized but this tells us nothing about the region, A-A'.
AG = ∫Gdx = [mg(cosα)](h0) = mg(Δy)

The diagram does not display any dashed lines indicating friction in the region A-A', but if there is friction in this region similar to the friction in region B-D it will have the following work equation.
AF = ∫Fdx = ∫mgf(sinα)dx = mgf(sinα)(h0)

Then use conservation of energy to find the speed of the ball at point C,
A = AG(C) + KE = mg(Δy) + (mv^2)/2

Then use equations for circular motion to find the acceleration of the ball, and subsequently the force via newton's second law.
a = (v^2)/r = (v^2)/(2R)
Newton's third law states that the force on the ball and the force on the wall are the same. Both the Normal force and gravity are acting on the ball at point C.