A mass M of 4.80E-1 kg slides inside a hoop of radius R=1.20 m with negligible friction. When M is at the top, it has a speed of 5.67 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 27.0°.

this is what I did
F = mg cos theta + mv^2/r
= (0.480)(9.81)cos27 + [(0.480)(5.67)^2]/1.20
= 17.050 N

12.859+ 4.191= 17.050N

I got a wring answer :( what's wrong?

Remember that the speed of 5.67 m/s is when the mass is at the top. When it gets to 27 degrees it has gained some kinetic energy (equal to the decrease in potential energy from having slid downhill), so at that point it's going faster than 5.67 m/s.

Also, be careful with the signs of the forces. Depending on how the 27 degrees is measured, it could be that the normal force due to gravity is toward the inside of the loop, not the outside. Since you didn't define where the 27 degrees is measured from, it's impossible for me to tell whether that should be a positive or a negative number.

I did not get it..
Well I tried another way and I got wrong answer too

Thanks I'm done ;)
I reread your comment again, and got it