Question Number 49093 by peter frank last updated on 02/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
$$\left.\mathrm{12}\right){Normal}\:{reaction}\:{N}=\mathrm{5}×\mathrm{10}=\mathrm{50}{N} \\ $$$${limiting}\:{friction}\:{f}_{{L}} =\mathrm{0}.\mathrm{42}×\mathrm{50}=\mathrm{21}{N} \\ $$$${kinetic}\:{friction}\:{f}_{{k}} =\mathrm{0}.\mathrm{15}×\mathrm{50}=\mathrm{7}.\mathrm{5}\:{N} \\ $$$$\left.{a}\right){applied}\:{force}=\mathrm{15}{N}<\left({f}_{{L}} =\mathrm{21}{N}\right) \\ $$$${so}\:{frictional}\:{force}\:{also}\:\mathrm{15}{N} \\ $$$${net}\:{force}=\mathrm{15}{N}−\mathrm{15}{N}=\mathrm{0} \\ $$$${hence}\:{accelaration}\:{is}\:{zero} \\ $$$$\left.{b}\right){applied}\:{force}\:\mathrm{25}{N}>\left({f}_{{L}} =\mathrm{21}{N}\right) \\ $$$${net}\:{force}=\mathrm{25}{N}−\mathrm{7}.\mathrm{5}{N} \\ $$$${acc}=\frac{\mathrm{17}.\mathrm{5}}{\mathrm{5}}=\mathrm{3}.\mathrm{5}{meter}/{sec}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
$$\left.\mathrm{13}\right){normal}\:{reaction}=\mathrm{42}×\mathrm{10}=\mathrm{420}{N} \\ $$$${limiting}\:{friction}\:{f}_{{L}} =\mathrm{420}×\mathrm{0}.\mathrm{44}=\mathrm{184}.\mathrm{8} \\ $$$${let}\:{accelation}\:{is}\:{a}\:{so}\:{pseudo}\:{force}\:{on}\:{refrigerator} \\ $$$${is}\:\:\mathrm{42}{a} \\ $$$${when}\:\mathrm{42}{a}\leqslant\mathrm{184}.\mathrm{8}\:{the}\:{refrigerator}\:{will}\:{not}\:{slide} \\ $$$${so}\:{a}\leqslant\frac{\mathrm{184}.\mathrm{8}}{\mathrm{42}} \\ $$$${a}\leqslant\mathrm{4}.\mathrm{4}{meter}/{sec}^{\mathrm{2}} \:{is}\:{pickup}\:{accelaration} \\ $$$${pls}\:{check}… \\ $$