Question

A block of mass 5kg is placed on a horizontal table, when a force of 20N is applied at an angle of 30 degrees with the horizontal, the block just begins to slide. Find the coeffecient of static friction

Answer 1

Answer:

Coefficient of static friction is √3/5.

Explanation:

Horizontal component of force is equal to the frictional force.

20cos30 = k x 5 x g

10√3 = 50k

Therefore, k = √3/5.

Answer 2

Coeffecient of static friction is 0.35 .

Given :

Mass of object , m = 5 kg.

Force applied , F = 20 N.

Angle of force from horizontal surface, $\theta=30^o.$

Now,

Horizontal component of force is the reason of its movement.

Therefore, horizontal component of force must be equal to maximum friction force.

So, $F\times cos \ 30^o=\mu \times N$

Here, N is normal reaction and it is equivalent to mg  , $g=9.8\ m/s^2.$

$F\times cos \ 30^o=\mu \times m\times g$

Putting all values we get ,

$\mu=0.35$

Hence, this is the required solution.

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