A ball of weight $1\,\text{N}$ is placed on a slope $25°$ to the horizontal. Find the normal force experienced by the ball. I know this seems like a really simple question but I would normally take $R$ as $W\cos(25^{\circ})$ as it is perpendicular to the slope but the answer key states that $R\cos(25^{\circ}) = W$. Is there something that I am missing? The question is part of the bigger problem but to solve for $R;$ the answer key only states this : $R\cos(25^{\circ})= 0.1 g$
Original Problem : A hollow container consists of a smooth circular cylinder of radius 0.5 m, and a smooth hollow cone of semi-vertical angle 65◦and radius 0.5 m. The container is fixed with its axis vertical and with the cone below the cylinder. A steel ball of weight 1 N moves with constant speed 2.5 m s−1in a horizontal circle inside the container. The ball is in contact with both the cylinder and the cone (see Fig. 1). Fig. 2shows the forces acting on the ball, i.e. its weight and the forces of magnitudes RN and SN exerted by the container at the points of contact. Given that the radius of the ball is negligible compared with the radius of the cylinder, find R and S
As we are looking at the picture in the problem, we see that the ball is on the left side of the hollow container. Define $x$ positive to the right, and $y$ positive up. The forces on the ball are as follows: \begin{align*} mg &\quad \text{in } -y\;\text{direction: weight}\\ \frac{mv^2}{r} &\quad \text{in } -x\;\text{direction: centripetal force}\\ S_N&\quad \text{in } +x\;\text{direction: normal exerted by cylinder}\\ R_N\sin(25^{\circ})&\quad \text{in } +x\;\text{direction: } x \; \text{component of } R_N\\ R_N\cos(25^{\circ})&\quad \text{in } +y\;\text{direction: } y \; \text{component of } R_N. \end{align*} Now, viewed from the $xy$ axis, the system is in equilibrium. So we sum the forces and solve: \begin{align*} S_N+R_N\sin(25^{\circ})-\frac{mv^2}{r}&=0\quad x \text{ direction}\\ R_N\cos(25^{\circ})-mg&=0\quad y \text{ direction.} \end{align*} Then immediately we see that \begin{align*} R_N&=\frac{mg}{\cos(25^{\circ})}\\ S_N+\frac{mg}{\cos(25^{\circ})}\,\sin(25^{\circ})-\frac{mv^2}{r}&=0\\ S_N+mg\tan(25^{\circ})-\frac{mv^2}{r}&=0\\ S_N&=\frac{mv^2}{r}-mg\tan(25^{\circ}). \end{align*} So you were missing two things: