A force "A" is addes two force B =(i-2j+4k) and C=(2i+3j-6k) and given a unit vector force along the x axis as their resultant. Find the acceleration of the body 5 kg body , if A acts on the body
A→=B→+C→=(i→−2j→+4k→)+(2i→+3j→−6k→)=\overrightarrow{A}=\overrightarrow{B}+\overrightarrow{C}=(\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k})+(2\overrightarrow{i}+3\overrightarrow{j}-6\overrightarrow{k})=A=B+C=(i−2j+4k)+(2i+3j−6k)=
=3i→+j→−2k→=3\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}=3i+j−2k
a→=A→m=3i→+j→−2k→5=0.6i→+0.2j→−0.4k→\overrightarrow{a}=\frac{\overrightarrow{A}}{m}=\frac{3\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}}{5}=0.6\overrightarrow{i}+0.2\overrightarrow{j}-0.4\overrightarrow{k}a=mA=53i+j−2k=0.6i+0.2j−0.4k
∣a→∣=0.62+0.22+0.42=0.75 (m/s2)|\overrightarrow{a}|=\sqrt{0.6^2+0.2^2+0.4^2}=0.75\ (m/s^2)∣a∣=0.62+0.22+0.42=0.75 (m/s2)
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