Question :-
A block of mass m is pulled by a uniform chain of mass m tied to it by applying a Force F at the other end of the chain . The tension at a point which is at a distance of quarter of length of the chain from the free end , will be
a) [tex]\texbf{\textsf{\dfrac{4F}{5}}}[/tex]
b) [tex]\textbf{\textsf{\dfrac{7F}{8}}}[/tex]
c) [tex]\textbf{\textsf{\dfrac{4F}{7}}}[/tex]
d) [tex]\textbf{\textsf\dfrac{6F}{5}}}[/tex]
Diagram :-
[tex]\setlength{\unitlength}{1mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(3,0){3cm}}\put(1,1){\boxed{\Huge\bf M}}}\put(7.2,2.2){\boxed{\qquad\qquad\;\;\;}}\put(26.6,2.48){\vector(3,0){0.5cm}}\put(33,2){\bf F}\put(22,4.8){\vector(3,0){4mm}}\put(22,4.8){\vector(-3,0){4mm}}\put(21,6.7){\sf ^l\!/_4$ }\put(0,0){\line(1,-2){2mm}}\put(4,0){\line(1,-2){2mm}}\put(8,0){\line(1,-2){2mm}}\put(12,0){\line(1,-2){2mm}}\put(16,0){\line(1,-2){2mm}}\put(20,0){\line(1,-2){2mm}}\put(24,0){\line(1,-2){2mm}}\end{picture}[/tex]
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Answers & Comments
Verified answer
Explanation:
Mass of the block = m
Mass of the chain = m
Total mass = 2m
Force applied = F
Let the point at which tension is to be calculated is A.
Then,tension at point A will be due to mass of the block and the mass of the of the chain.
As it is a uniform chain of mass " m " ,
∴ Mass of the length of the chain will be
Hence,the mass of the system on which the tension acts at point A will be
Total mass =
∴The tension at point A = total mass at point A × acceleration at point A
Tension =
∴ The tension at a point which is at a distance of quarter of length of the chain from the free end , will be