Question

the block a in the given figure weight 100 Newton the coefficient of friction between the block and the surface on which it is rest is 0.30 the weight is 20 Newton and the system is in equilibrium find the frictional force exerted on block a

Answer 1

Block a=100 N
100*20=2000

Answer 2

Friction force exerted on block A is 20N.

Given

Given a equilibrium system

The coefficient of friction between the block and the surface on which it is rest is 0.30

The weight is 20 Newton.

To Find

We need to find the frictional force acting on the block.

Solution

The Tension acting upon the wire T attached to the hook can be determined as T sin(θ )w.

T = w / ( sin Ф)

The force exerted due to the  static friction on block A is equal to the horizontal component of the tension experienced.

The expression for frictional force is given by the formula,

$F_{f}$ = T cos  Ф

Substitute the equation for T in this equation for frictional force

Now we get,

$F_{f}$ = ( w × cos   Ф) / sin   Ф

    =  w cot   Ф

Substituting the values

 w = 20 N            Ф = 45°

$F_{f}$  = 20 cot  45°

    = 20 N

Therefore friction force exerted on block A is 20N.

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