Question

A force of 49N is just sufficient to pull a block of wood weighing 10kg. on a rough horizontal surface calculate the coefficient of friction and angle of friction?​

Answer 1

Answer:

A force of 49N is just sufficient to pull a block of wood weighing 10kg. on a rough horizontal surface.

The coefficient of friction is 0.5

Angle of friction  is $26.34^{0}$

Explanation:

Given :  

Force   (F)  = 49 N

Mass of the block (m) = 10 kg

Acceleration due to gravity (g) = 10m/$s^{2}$

Forces acting on the block are,

• If the  Force of 49N  acting on the block, frictional force (fr) will act in the opposit direction.

• The normal force (N) acts in the upward direction and the weight (W) of the block acts downwards.

When the bock begin to move,

  (fr) = 49 N ...(1)   and

   N  = W    ....(2)

   W =  mg = 10$\times$ 10 = 100N

   N = 100N

we know,

     fr = μ N    where  μ is the coefficient of friction

Using equation  (1)

    μ N = 49N

    μ   =   $\frac{49}{100} = 0.49$  ≈   0.5

     

 The angle of friction is given by,

 cos θ  = $\frac{N}{R}$   where R is the resultant of fr and N.

     R =  ${\sqrt{(\ N )^{2}+(fr)^{2}}}$

$\cos \theta=\frac{\ 100 }{\sqrt{(\ 100 )^{2}+(49)^{2}}} = \frac{2}{\sqrt{5} }$

     θ  =  $cos^{-1} (\frac{2}{\sqrt{5} } ) =26.34^{0}$