Question

A heavy wheel of radius 20cm and weight 10kg
is to be dragged over a step of height 10cm
by a horizontal force F applied at the centre of
the wheel. The minimum value of F is​

Answer 1

Question:

A heavy wheel of radius 20cm and weight 10kg

is to be dragged over a step of height 10cm

by a horizontal force F applied at the centre of

the wheel. Find the minimum value of F .

Answer:

F = 10√3 kg.wt.

or F = 10√3 kg.f.

or F = 98√3 N

Diagram:

Please refer to the attachment.

Solution:

Here,

We need to drag the wheel over a step. If we drag the wheel over the step , it will rotate about the point P.

Also,

When the value of horizontal force F is minimum then the clockwise torque and the anticlockwise will be balanced for a while at the point P.

Thus,

Balancing the torque about the point P , we have;

=> Clockwise torque = Anticlockwise torque

=> mg × RP = F × QP

{ mg = 10 kg.wt. , QP = 10 cm , RP = 10√3 cm }

=> 10 kg.wt. × 10√3 cm = F × 10 cm

=> F = (10 kg.wt. × 10√3 cm)/10 cm

=> F = 10√3 kg.wt.

OR F = 10√3 kg.f.

OR F = 10√3 × 9.8 N { 1 kg.f. = 9.8 N }

F = 98√3 N

Hence,

The required value of F is 98√3 N .

Answer 2

Answer:

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