Question

$M_A=3$kg, $M_B=4$kg and $M_C=8$kg. $\mu$ between any two surfaces is $0.25$. Pulley is frictionless and string is massless Block. A is connected to the wall through a massless rigid rod as shown in figure. The value of F to keep C moving with constant speed is $10X$ N. Find value of $X(g=10m/s^2)$.

Answer 1

$N_{AB}=M_{A}g$ $f_{BC}=(M_{A}g+M_{B}g)\mu$

$T+f_{AB}=0$ $f_{AB}=M_{A}g\mu$

$T=0$ $T^{\prime }-f_{AB}-f_{BC}=M_{B}a$

$N_{BC}=M_{A}g+M_{B}g$

$F-f_{BC}-T^{\prime }-\mu N=0$ $N=M_{C}g+M_{B}g+M_{A}g$

$F=2(M_{A}g+M_{B}g)4+M_{A}g4+4(M_{A}g+M_{B}g+M_{C}g)$

$F=\left[2\right(30+40)+30+(30+40+80\left)\right]0.25$

$=\left[2\right(70)+30+150]0.25$

$=320\ast \frac{1}{4}$

$=80N$

$\therefore x=80/10=8$