Discussion Forum

A block is moving on an inclined plane making an angle $45^{0}$ with the horizontal and the coefficient of friction $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $N= 10 \mu$ , then N is 

Answer:

Explanation:

$F_{1}=mg \sin \theta+\mu mg \cos \theta$

 $F_{2}= mg \sin \theta-\mu mg \cos \theta$

 Given that , $F_{1}=3F_{2}$

 or  $(\sin 45^{0}+\mu \cos 45^{0})$

 =$3(\sin  45^{0}-\mu \cos 45^{0})$

 On solving  we get , $\mu=0.5$

 $\therefore$  N= $10 \mu =5$

 $\therefore$ Answer is 5