Question Number 214395 by mr W last updated on 07/Dec/24
Commented by mr W last updated on 07/Dec/24
$${find}\:{acceleration}\:{of}\:{wedge}\:{A}=? \\ $$
Answered by mr W last updated on 07/Dec/24
$${A}={acceleration}\:{of}\:{M}\:\left(\leftarrow\right) \\ $$$${a}_{{x}} ={acceleration}\:{of}\:{m}\:\left(\rightarrow\right) \\ $$$${a}_{{y}} ={acceleration}\:{of}\:{m}\:\left(\downarrow\right) \\ $$$${N}={contact}\:{force}\:{between}\:{M}\:{and}\:{m} \\ $$$${N}\:\mathrm{sin}\:\alpha={MA}={ma}_{{x}} \\ $$$$\Rightarrow{a}_{{x}} =\frac{{MA}}{{m}} \\ $$$$\Rightarrow{N}=\frac{{MA}}{\mathrm{sin}\:\alpha} \\ $$$${mg}−{N}\:\mathrm{cos}\:\alpha={ma}_{{y}} \\ $$$${a}_{{y}} ={g}−\frac{{MA}\:\mathrm{cos}\:\alpha}{{m}\:\mathrm{sin}\:\alpha} \\ $$$$\mathrm{tan}\:\alpha=\frac{{a}_{{y}} }{{a}_{{x}} +{A}}=\frac{{g}−\frac{{MA}\:\mathrm{cos}\:\alpha}{{m}\:\mathrm{sin}\:\alpha}}{\frac{{MA}}{{m}}+{A}} \\ $$$$\Rightarrow{A}=\frac{{g}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\frac{{M}}{{m}}+\mathrm{sin}^{\mathrm{2}} \:\alpha}=\frac{{g}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\left(\frac{{M}}{{m}}+\mathrm{1}\right)−\mathrm{cos}^{\mathrm{2}} \:\alpha} \\ $$
Commented by ajfour last updated on 07/Dec/24
$${yes}\:{right},\:{i}\:{remember}\:{the}\:{result}\:{too}. \\ $$
Commented by mr W last updated on 07/Dec/24
$${in}\:{case}\:{of}\:{a}\:{rolling}\:{object},\:{the} \\ $$$${acceleration}\:{of}\:{wedge}\:{should}\:{be} \\ $$$${smaller}\:{than}\:{this}.\:{so}\:{the}\:{result}\:{in} \\ $$$${your}\:{video} \\ $$$${A}=\frac{{g}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{M}}{{m}}+\mathrm{1}\right)−\mathrm{cos}^{\mathrm{2}} \:\alpha}\:{with}\:{rolling} \\ $$$${cyclinder}\:{is}\:{by}\:{trend}\:{right}. \\ $$