Question-69183

Question Number 69183 by Tony Lin last updated on 21/Sep/19
Commented by Tony Lin last updated on 21/Sep/19
$$\mathrm{49}.\:{m}_{\mathrm{1}} {g}=\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){a} \\$$$$\Rightarrow{a}=\frac{{m}_{\mathrm{1}} {g}}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} } \\$$$$\Rightarrow{F}=\left({M}+{m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){a}=\frac{{m}_{\mathrm{1}} {g}\left({M}+{m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} } \\$$$${I}'{m}\:{not}\:{sure}\:{of}\:{the}\:{direction}\:{of}\:{a}_{{m}_{\mathrm{1}} } \\$$$${is}\:\rightarrow\:{or}\:\downarrow? \\$$$${Or}\:{my}\:{method}\:{has}\:{mistakes}? \\$$
Commented by Tony Lin last updated on 21/Sep/19
$$\mathrm{53}.\:{Does}\:{this}\:{example}\:{satisfies}\:{the}\: \\$$$${conservation}\:{of}\:{horizontal}\:{momentum}? \\$$$${If}\:{Yes},{how}\:{to}\:{solve}\:{it}? \\$$
Answered by mr W last updated on 21/Sep/19
Commented by mr W last updated on 21/Sep/19
$$\left({note}:\:{in}\:{my}\:{solution}\:{m}_{\mathrm{1}} \:{corresponds}\right. \\$$$${with}\:{m}_{\mathrm{2}} \:{in}\:{the}\:{question}\:{and}\:{m}_{\mathrm{2}} \:{with} \\$$$$\left.{m}_{\mathrm{1}} \right) \\$$$$\\$$$${all}\:{blocks}\:{must}\:{have}\:{the}\:{same}\:{horizontal} \\$$$${motion}\:{and}\:{no}\:{vertical}\:{motion}. \\$$$$\\$$$${T}={m}_{\mathrm{2}} {g} \\$$$${N}_{\mathrm{2}} ={m}_{\mathrm{2}} {a} \\$$$${N}_{\mathrm{1}} ={m}_{\mathrm{1}} {g} \\$$$${T}={m}_{\mathrm{1}} {a}={m}_{\mathrm{2}} {g} \\$$$$\Rightarrow{a}=\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }{g} \\$$$${F}−{N}_{\mathrm{2}} −{T}={Ma} \\$$$$\Rightarrow{F}={Ma}+{N}_{\mathrm{2}} +{T}={Ma}+{m}_{\mathrm{2}} {a}+{m}_{\mathrm{1}} {a} \\$$$$\Rightarrow{F}=\left({M}+{m}_{\mathrm{2}} +{m}_{\mathrm{1}} \right){a}=\frac{\left({M}+{m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){m}_{\mathrm{2}} {g}}{{m}_{\mathrm{1}} } \\$$
Commented by Tony Lin last updated on 21/Sep/19
$${Thanks},{I}\:{got}\:{it}. \\$$