Three blocks of masses 2m,4m,and 6m are place | vedclass

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Question

(A) The total mass of the system is $M = 2m + 4m + 6m = 12m$.The driving force along the incline for the blocks on the right side is $F_1 = (4m + 6m)g

Vedclass · Accepted Answer

$a=\frac{17}{30} g$

Vedclass · Answer

(A) The total mass of the system is $M = 2m + 4m + 6m = 12m$.The driving force along the incline for the blocks on the right side is $F_1 = (4m + 6m)g \sin 53^{\circ} = 10mg \sin 53^{\circ}$.The opposing force along the incline for the block on the left side is $F_2 = 2mg \sin 37^{\circ}$.Using Newton's second law for the system,$F_{net} = Ma$,where $F_{net} = F_1 - F_2$:$10mg \sin 53^{\circ} - 2mg \sin 37^{\circ} = 12ma$Substituting the given values $\sin 53^{\circ} = \frac{4}{5}$ and $\sin 37^{\circ} = \frac{3}{5}$:$10mg \left(\frac{4}{5}\right) - 2mg \left(\frac{3}{5}\right) = 12ma$$8mg - 1.2mg = 12ma$$6.8mg = 12ma$$a = \frac{6.8}{12} g = \frac{68}{120} g = \frac{17}{30} g$.

Three blocks of masses $2m$,$4m$,and $6m$ are placed as shown in the figure. If $\sin 37^{\circ}=\frac{3}{5}$ and $\sin 53^{\circ}=\frac{4}{5}$,the acceleration of the system is:

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