Determine the amount and direction of the smallest force required to start the wheel as shown in fig over the block what is the reaction of block wt of wheel is 200N

Solution :

Let , the reaction of the block be the R.

The force P is always perpendicular with reaction R.

When the wheel is just on the point of movement up, then it loose contact with inclined plane and reaction at this point becomes zero. 

Consider triangle OPM,

$ \begin {aligned} OM &= 60 cm  \\ OP &= 60–15 = 45 cm  \\ MP &=\sqrt{ {(OM)^2 – (OP)^2}} \\ &= {3600 – 2025}1/2 \\&= 39.68cm \\ tan β &= \frac {MP}{OP} = \frac {39.68}{45}\\ β &= 41.40° \end{aligned} $

Using lamis theorem at point O

$ \begin{aligned} \frac {P}{sin 108.6° } &= \frac{10}{sin 90° }= \frac{R}{sin 161.4°} \\ P &= \frac{(10 × sin 108.6°)}{sin 90°} \\ &=9.4KN \end{aligned} $

Hence smallest force P = 9.4KN.