## Essential University Physics: Volume 1 (3rd Edition)

(a) We know that the horizontal component of force is given as $F_h=\frac{mv^2}{r}$ We plug in the known values to obtain: $F_h=\frac{(45)(6.3)^2}{5.0}=357.21N$ The vertical component of the force is given as $F_v=mg$ We plug in the known values to obtain: $F_v=(45)(9.8)=441N$ (b) The required angle is given as $\theta=tan^{-1}(\frac{F_v}{F_h})$ $\theta=tan^{-1}(\frac{441}{357.21})=51^{\circ}$