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Question-208931




Question Number 208931 by byaw last updated on 27/Jun/24
Answered by mr W last updated on 27/Jun/24
(a)  R=m(g+a)=0.5×(10+2)=6 N    (b)(i)  Mg×0.8+100×0.3+80×1.3−90×2.4=0  ⇒M=10.25 kg  (b)(ii)  R_B =102.5+100+80−90=192.5 N
$$\left({a}\right) \\ $$$${R}={m}\left({g}+{a}\right)=\mathrm{0}.\mathrm{5}×\left(\mathrm{10}+\mathrm{2}\right)=\mathrm{6}\:{N} \\ $$$$ \\ $$$$\left({b}\right)\left({i}\right) \\ $$$${Mg}×\mathrm{0}.\mathrm{8}+\mathrm{100}×\mathrm{0}.\mathrm{3}+\mathrm{80}×\mathrm{1}.\mathrm{3}−\mathrm{90}×\mathrm{2}.\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{M}=\mathrm{10}.\mathrm{25}\:{kg} \\ $$$$\left({b}\right)\left({ii}\right) \\ $$$${R}_{{B}} =\mathrm{102}.\mathrm{5}+\mathrm{100}+\mathrm{80}−\mathrm{90}=\mathrm{192}.\mathrm{5}\:{N} \\ $$

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