The wheel shown in the diagram (Fig.1.18) has a fixed axle passing through O. The wheel is kept stationary under the action of (i) a horizontal force F_1 at A and (ii) a vertical force F_2 at B. (a) Show the direction of force F_2 in the diagram. (ii) Which of the force F_1 or F_2 is greater (c) Find the ratio between the forces F_1 and F_2 Given: AO =2.5 cm BO'=1.5 cm and O'O=2.0cm

The wheel shown in the diagram (Fig.1.18) has a fixed axle passing thr

1.	The wheel shown in the diagram (Fig.1.18) has a fixed axle passing through O. The wheel is kept stationary under the action of (i) a horizontal force F_1 at A and (ii) a vertical force F_2 at B. (a) Show the direction of force F_2 in the diagram. (ii) Which of the force F_1 or F_2 is greater (c) Find the ratio between the forces F_1 and F_2 Given: AO =2.5 cm BO'=1.5 cm and O'O=2.0cm
Answer» Solution :The force `F_1` applied at A produces a CLOCKWISE moment on the wheel. It can be balanced by applying the force `F_2` at B in a direction such that it produces an anticlockwise moment. Therefore the VERTICAL force at B should be applied in the downward direction as SHOWN in FIg.1.19 In equilibrium `F_1 times OA=F_2 times OO.` (b) SInce the perpendicular forces `F_1` from O is greater than the perpendicular distanceOO. of point of APPLICATION of force `F_2` from O, so in magnitude, force `F_2` is greater than force `F_1` (c) Moment of force `F_1` about`O= F_1 times OA` (clockwise ) Moment of force `F_2` about `O=F_2 times OO.` (anticlockwise) When the wheel is in equilibrium position, Clockwise moment= Anticlockwise moment i.e., `F_1 times OA=F_2 times OO.` `therefore F_2/F_1= (OA)/(OO.)`........(i) Given OA=2.5 CM and OO.=2.0 cm Substituting the values of OA and OO. in eqn. (i), the ratio of forces `F_2/F_1=2.5/2.0 or F_2:F_1=5:4`