Q. For the arrangement shown in figure-$2.168$, the tension in the string to prevent it from sliding down, is :

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A

$6\, N$

B

$6.4\, N$

C

$0.4 \,N$

D

None of these

Solution:

Maximum friction force which can act on block is,
$f_{\max } =\mu N $
$=0.8 \times(1)(10) \cos 37^{\circ} $
$=8 \times \frac{4}{5}=6.4 N$
$m g \sin \theta=10 \times \frac{3}{5}=6 N$
$\Rightarrow m g \sin \theta< f_{\max }$, the block will not slide at also no tension is required in string to hold it at rest.

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