A uniform wheel of 60 cm diameter weighing 1000 N rests against rectan

1.	A uniform wheel of 60 cm diameter weighing 1000 N rests against rectangular obstacle 15 cm high. Find the least force required which when acting through center of the wheel will just turn the wheel over the corner of the block. Find the angle of force with horizontal.
Answer» Let, P_min = Least force applied as shown in fig. α = Angle of the least force From triangle OBC, BC = BOsin α BC = 30sin α In Triangle BOD, BD = {(BO)² – (OD)² }^1/2 BD = (30² – 15²)^1/2 = 25.98 Taking moment of all forces about point B, We get P_min X BC – W X BD = 0 P_min – W X BD/BC P_min = 1000 X 25.98 /30sin α We get minimum value of P when α is maximum and maximum value of α is at 90º i.e. 1, putting sin α =1 P_min = 866.02N

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

Answer»

Let, P_min = Least force applied as shown in fig.

α = Angle of the least force

From triangle OBC,

BC = BOsin α

BC = 30sin α

In Triangle BOD, BD = {(BO)² – (OD)² }^1/2

BD = (30² – 15²)^1/2 = 25.98

Taking moment of all forces about point B, We get

P_min X BC – W X BD = 0

P_min – W X BD/BC

P_min = 1000 X 25.98 /30sin α

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

P_min = 866.02N