Laws Of Motion Question 411
Question: A perfectly straight portion of a uniform rope has mass M and length L. At end A of the segment, the tension in the rope is $ T _{A} $ and at end B it is $ T _{B}(T _{B}>T _{A}) $ . Neglect effect of gravity and no contact force acts on the rope in between points A and B. The tension in the rope at a distance L/5 from end A is
Options:
A) $ T _{B}-T _{A} $
B) $ (T _{A}+T _{B})/5 $
C) $ (4T _{A}+T _{B})/5 $
D) $ (T _{A}-T _{B})/5 $
Show Answer
Answer:
Correct Answer: C
Solution:
The F.B.D. of section of rope between A and B having acceleration a towards left is
Applying Newton’s second law on section AB and section AC of rope we get
$ T _{B}-T _{A}=Ma $ and $ T _{C}-T _{A}=\frac{M}{5}a $
Solving $ T _{C}=\frac{T _{B}+4T _{A}}{5} $
BETA
