Menu Close

Question-64478




Question Number 64478 by necx1 last updated on 18/Jul/19
Commented by necx1 last updated on 18/Jul/19
please help me analyse this free body  diagram. I′ll be grateful sirs.
$${please}\:{help}\:{me}\:{analyse}\:{this}\:{free}\:{body} \\ $$$${diagram}.\:{I}'{ll}\:{be}\:{grateful}\:{sirs}. \\ $$
Commented by necx1 last updated on 18/Jul/19
please find R_l  and R_r
$${please}\:{find}\:{R}_{{l}} \:{and}\:{R}_{{r}} \\ $$
Answered by behi83417@gmail.com last updated on 18/Jul/19
R_1 ×8+R_2 ×4=10×?+10×5  R_1 +R_2 =10+10+5  ⇒ { ((2R_1 +R_2 =2.5×?+12.5)),((R_1 +R_2 =25)) :}  R_1 =2.5×?−12.5,R_2 =37.5−2.5×?  ?=distance from: left 10^N  to 5^N  force.(not given)
$$\mathrm{R}_{\mathrm{1}} ×\mathrm{8}+\mathrm{R}_{\mathrm{2}} ×\mathrm{4}=\mathrm{10}×?+\mathrm{10}×\mathrm{5} \\ $$$$\mathrm{R}_{\mathrm{1}} +\mathrm{R}_{\mathrm{2}} =\mathrm{10}+\mathrm{10}+\mathrm{5} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2R}_{\mathrm{1}} +\mathrm{R}_{\mathrm{2}} =\mathrm{2}.\mathrm{5}×?+\mathrm{12}.\mathrm{5}}\\{\mathrm{R}_{\mathrm{1}} +\mathrm{R}_{\mathrm{2}} =\mathrm{25}}\end{cases} \\ $$$$\mathrm{R}_{\mathrm{1}} =\mathrm{2}.\mathrm{5}×?−\mathrm{12}.\mathrm{5},\mathrm{R}_{\mathrm{2}} =\mathrm{37}.\mathrm{5}−\mathrm{2}.\mathrm{5}×? \\ $$$$?=\mathrm{distance}\:\mathrm{from}:\:\mathrm{left}\:\mathrm{10}^{\mathrm{N}} \:\mathrm{to}\:\mathrm{5}^{\mathrm{N}} \:\mathrm{force}.\left(\mathrm{not}\:\mathrm{given}\right) \\ $$
Commented by necx1 last updated on 18/Jul/19
yes....That′s where i also do not  understand.Though a friend got 10N  and 15N but the process he applied i  dont get.He was using something  +ve and −ve in relation to a freely  supported and cantilever beem. I also  think the question is kinda not complete  in terms of given parameters.
$${yes}….{That}'{s}\:{where}\:{i}\:{also}\:{do}\:{not} \\ $$$${understand}.{Though}\:{a}\:{friend}\:{got}\:\mathrm{10}{N} \\ $$$${and}\:\mathrm{15}{N}\:{but}\:{the}\:{process}\:{he}\:{applied}\:{i} \\ $$$${dont}\:{get}.{He}\:{was}\:{using}\:{something} \\ $$$$+{ve}\:{and}\:−{ve}\:{in}\:{relation}\:{to}\:{a}\:{freely} \\ $$$${supported}\:{and}\:{cantilever}\:{beem}.\:{I}\:{also} \\ $$$${think}\:{the}\:{question}\:{is}\:{kinda}\:{not}\:{complete} \\ $$$${in}\:{terms}\:{of}\:{given}\:{parameters}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *