$${F_{par} = \frac{1}{4}Mg + \frac{1}{2}mg = \frac{1}{4}(2,7\, \mathrm{kg})(9,8\, \mathrm{m/s^2}) + \\ \frac{1}{2}(1,8\, \mathrm{kg})(9,8\, \mathrm{m/s^2}) = 15,44\, \mathrm{N} \approx \underline{\underline{15\, \mathrm{N}}}}$$