A smooth track in the form of a quarter circle of radius $$6m$$ lies in the vertical plane. A particle moves from $$P1$$ to $$P2$$ under the action of forces F→$$1$$, F→$$2$$ and F→$$3$$. Force $$F1$$ is always toward $$P2$$ and is always $$20$$ N in magnitude. Force F→$$2$$ always acts horizontally and is always $$30$$ N in magnitude. Force F→$$3$$ always acts tangentially to the track and is of magnitude $$15$$ N. Select the correct $$alternative(s)$$

Question

A smooth track in the form of a quarter circle of radius $$6m$$ lies in the vertical plane. A particle moves from $$P1$$ to $$P2$$ under the action of forces F→$$1$$, F→$$2$$ and F→$$3$$. Force $$F1$$ is always toward $$P2$$ and is always $$20$$ N in magnitude. Force F→$$2$$ always acts horizontally and is always $$30$$ N in magnitude. Force F→$$3$$ always acts tangentially to the track and is of magnitude $$15$$ N. Select the correct $$alternative(s)$$

Infinity Learn · Accepted Answer

Work done by F→$$1$$ is $$W1=$$∫$$P1P2$$ $$F1cos$$⁡θdsWhere, $$ds=(6)d($$−$$2$$θ$$)=$$−$$12d$$θ and $$F1=20N$$⇒$$W1=$$−$$240$$∫π$$/40$$ $$cos$$⁡θdθ⇒$$W1=240sin$$⁡π$$4=1202JF$$→$$1$$ is conservative because it is always directed towards a fixed point $$P2$$. Therefore, $$W1$$ can be directly calculated $$asW1=F1P1P2=(20)(62)=1202JSimilarly$$, $$W2=F2OP2=(30)(6)=180$$ Jand $$W3=$$∫$$06($$π$$/2)$$ $$F3ds=$$∫$$03$$π $$15ds$$       $$W3=15s03$$π$$=45$$π J

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