Find out the angle of force:

Q: A uniform wheel having 60 cm diameter weighing 1000 N rests against rectangular obstacle 15 cm high. Find least force required which when acting through the center of wheel will just turn wheel over the corner of the block. Find out the angle of force with horizontal. 

Sol.: Let,

Figure

P_min  = Least force applied as shown in the figure = Angle of least force

From triangle OBC, BC = BOsinθ

BC = 30sinθ

In ?BOD, BD = {(BO)²  - (OD)²}^1/2

BD = (30²  - 15²)^1/2  = 25.98

Taking moment of all the forces about point B, then we get

P_min  X BC - W X BD = 0

P_min  - W X BD/BC

P_min = 1000 X 25.98 /30sinθ

We get minimum value of P whenα is maximum and maximum value of α is at 90º i.e. 1, putting sinα = 1

P_min = 866.02N                                                                            .......ANS