Question-39669

Question Number 39669 by rahul 19 last updated on 09/Jul/18

Answered by MrW3 last updated on 09/Jul/18

Before the support is removed: F_k =M_1 g=force in spring F_s =(M_1 +M_2 )g=contact force on support After the remove of the support: M_1 g−F_k =M_1 a_1 M_1 g−M_1 g=M_1 a_1 ⇒a_1 =0 F_k +M_2 g=M_2 a_2 ⇒M_1 g+M_2 g=M_2 a_2 ⇒a_2 =(1+(M_1 /M_2 ))g (↓)

$${Before}\:{the}\:{support}\:{is}\:{removed}: \\ $$$${F}_{{k}} ={M}_{\mathrm{1}} {g}={force}\:{in}\:{spring} \\ $$$${F}_{{s}} =\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right){g}={contact}\:{force}\:{on}\:{support} \\ $$$${After}\:{the}\:{remove}\:{of}\:{the}\:{support}: \\ $$$${M}_{\mathrm{1}} {g}−{F}_{{k}} ={M}_{\mathrm{1}} {a}_{\mathrm{1}} \\ $$$${M}_{\mathrm{1}} {g}−{M}_{\mathrm{1}} {g}={M}_{\mathrm{1}} {a}_{\mathrm{1}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\mathrm{0} \\ $$$$ \\ $$$${F}_{{k}} +{M}_{\mathrm{2}} {g}={M}_{\mathrm{2}} {a}_{\mathrm{2}} \\ $$$$\Rightarrow{M}_{\mathrm{1}} {g}+{M}_{\mathrm{2}} {g}={M}_{\mathrm{2}} {a}_{\mathrm{2}} \\ $$$$\Rightarrow{a}_{\mathrm{2}} =\left(\mathrm{1}+\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{2}} }\right){g}\:\left(\downarrow\right) \\ $$

Commented by rahul 19 last updated on 09/Jul/18