TT-MgHomework hints:Free body diagramFa fixes the acceleration of mA+mB. Notice that this is also the acceleration of mAand mBindividually.Now consider what force would be needed to give e.g. mBthat acceleration and what provides that force.Man pulls down with F = -T, but that Fis NOT a force acting on him and the cage.
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FBDθFGmg−fGFsinθFcosθThe block is pulled at constant speed vwith m, Fand θgiven. What is μk ? XxFFcosfma0=θ −==∑fFcos=θBut kfNFcos= μ=θyyFFsinNmgma0=θ +−==∑Along xkFcosNμ=θNeed NAlong yNmgFsin=−θSo,kFcosmgFsinμ=θ−θLast class:N