A horizontal (2.0-mathrm{kg}) rod is (2.0 mathrm{~m}) long. An (8.0-mathrm{kg}) block is suspended from its left end,
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A horizontal \(2.0-\mathrm{kg}\) rod is \(2.0 \mathrm{~m}\) long. An \(8.0-\mathrm{kg}\) block is suspended from its left end, and a \(4.0-\mathrm{kg}\) block is suspended from its right end.
(a) Determine the magnitude and direction of the single extra force necessary to keep the rod in mechanical equilibrium.
(b) At what distance from the left end of the rod must the point of application of this force be?
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To keep the rod in mechanical equilibrium the net torque and net force acting on it must be zero Lets consider the torque first a The torque about the ...View the full answer
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Godswill Okorie
M.Sc chemistry specialization in organic chemistry, B.ed .I am having industry experience of seven years working with Ranbaxy nd Shimadzu analytical India by working as an application chemistry.I am having good practical experience on chromatography techniques,which later helped me in my teaching.I worked as PGT chemistry teacher with KV and APS.
As a teacher I was able to achieve good results with my students.I used to take 11th and 12th chemistry and science to classes 7th ,8th and ninth. While teaching I used to guide students for various carrier opportunities.
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