Laws of Motion 3 Question 24

26. In the figure, the blocks $A, B$ and $C$ have masses $3 kg, 4 kg$ and $8 kg$ respectively. The coefficient of sliding friction between any two surfaces is $0.25 . A$ is held at rest by a massless rigid rod fixed to the wall, while $B$ and $C$ are connected by a light flexible cord passing around a fixed frictionless pulley. Find the force $F$ necessary to drag $C$ along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the figure. i.e. $B$ on $C$ and $A$ on $B$, is maintained throughout.(Take $g=10 m / s^{2}$ ).

(1978)

Integer Answer Type Question

Show Answer

Answer:

Correct Answer: 26. $80 N$

Solution:

  1. Maximum friction between $A$ and $B=\mu m _A g$

or

$$ f _1=0.25(3)(10)=7.5 N $$

Maximum friction between $B$ and $C=\mu\left(m _A+m _B\right) g$

or

$$ \begin{aligned} f _2 & =0.25(3+4)(10) \\ & =17.5 N \end{aligned} $$

Maximum friction between $C$ and ground

$$ \begin{aligned} f _3 & =\mu\left(m _A+m _B+m _C\right) g \\ & =0.25(3+4+8)(10) \\ & =37.5 N \end{aligned} $$

Block $C$ and hence block $B$ are moving in opposite directions with constant velocities and block $A$ is at rest. Hence, net force on all three blocks should be zero. Free body diagrams have been shown below (Only horizontal forces are shown)

For equilibrium of $\boldsymbol{B}, T=f _1+f _2=25 N$

For equilibrium of $C, F=T+f _2+f _3=80 N$



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक