Answer:
Explanation:
Given
coefficient of friction [tex]\mu =0.27 [/tex]
It is also stated that block is about to move
suppose W is the weight of block
Force acting on block is Friction force , Gravity Force
Sin component of weight will try to move the block and friction Force will try to stop it
thus [tex]W\sin \theta =f_r[/tex]
and friction force is given by
[tex]f_r=\mu W\cos \theta [/tex]
[tex]W\sin \theta =\mu W\cos \theta [/tex]
[tex]\tan \theta =\mu [/tex]
[tex]\theta =tan^{-1}(0.27)[/tex]
[tex]\theta =15.109^{\circ}[/tex]
Normal Force will be [tex]N=W\cos \theta [/tex]
[tex]N=W\times 0.965[/tex]
[tex]N=0.965 W[/tex]