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 = (30^{2} - 15^{2})^{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