Blocks A (mass 2m),B (mass 4m),and C (mass m) | vedclass

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Question

(A) Let $a_A$,$a_B$,and $a_C$ be the accelerations of blocks $A$,$B$,and $C$ respectively. Let $T$ be the tension in the string.From the free body dia

Vedclass · Accepted Answer

$a_B = \frac{3F}{21m} \, m/s^2$

Vedclass · Answer

(A) Let $a_A$,$a_B$,and $a_C$ be the accelerations of blocks $A$,$B$,and $C$ respectively. Let $T$ be the tension in the string.From the free body diagrams:For block $A$ (mass $2m$): $F - 2T = (2m)a_A$For block $B$ (mass $4m$): $3T = (4m)a_B$For block $C$ (mass $m$): $T = (m)a_C$From the constraint relation of the pulleys,the acceleration of the string segments gives: $2a_A = 3a_B + a_C$Substituting $a_A = \frac{F-2T}{2m}$,$a_B = \frac{3T}{4m}$,and $a_C = \frac{T}{m}$ into the constraint equation:$2(\frac{F-2T}{2m}) = 3(\frac{3T}{4m}) + \frac{T}{m}$$\frac{F-2T}{m} = \frac{9T}{4m} + \frac{4T}{4m} = \frac{13T}{4m}$$4F - 8T = 13T \implies 21T = 4F \implies T = \frac{4F}{21}$Now,$a_B = \frac{3T}{4m} = \frac{3}{4m} 	imes \frac{4F}{21} = \frac{3F}{21m} \, m/s^2$.

Blocks $A$ (mass $2m$),$B$ (mass $4m$),and $C$ (mass $m$) are arranged as shown in the figure. $A$ force $F$ is applied to block $A$. Find the acceleration of block $B$.

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