For the given figure,if the block remains in | vedclass

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(B) Given: Mass of the block $m = 5 \ kg$,Applied force $F = 1000 \ N$,Coefficient of friction $\mu = 0.1$,Acceleration due to gravity $g = 10 \ m/s^2

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) Given: Mass of the block $m = 5 \ kg$,Applied force $F = 1000 \ N$,Coefficient of friction $\mu = 0.1$,Acceleration due to gravity $g = 10 \ m/s^2$.$1$. The block is pressed against the wall by a horizontal force $F = 1000 \ N$. This force acts as the normal force $N$ exerted by the wall on the block,so $N = F = 1000 \ N$.$2$. The maximum static frictional force $f_{max}$ that can act is given by $f_{max} = \mu N = 0.1 	imes 1000 = 100 \ N$.$3$. The gravitational force acting downwards on the block is $W = mg = 5 	imes 10 = 50 \ N$.$4$. Since the block is in equilibrium,the upward frictional force $f$ must balance the downward gravitational force $W$. Therefore,$f = W = 50 \ N$.$5$. Since $f = 50 \ N$ is less than $f_{max} = 100 \ N$,the block will remain in equilibrium. Thus,the frictional force is $50 \ N$.

For the given figure,if the block remains in an equilibrium position,then find the frictional force between the block and the wall in $N$.

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