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# phy (force + turning moment)

A worker tries to roll a roller of weight 300 N and radius 25 cm up a step of height 10

cm by exerting a force on the axle of the roller. Assume the weight of the roller acts

through the centre.

(a) What is the minimum horizontal force required?

(b) What is the minimum force required?

(c) If the force can be applied to other part of the roller, what is then the minimum

force required?

picture : http://img269.imageshack.us/img269/2844/rw333z.png

( I need step )

### 1 Answer

- 六呎將軍Lv 71 decade agoFavorite Answer
(a) From the given, the axle is 15 cm vertically above the edge of the step, so taking moment about the edge of the step:

Moment of the weight of the roller = 300 x 0.2 = 60 Nm (By Pyth. thm, the horizontal distance of the axle from the edge is 20 cm)

Moment of the horizontal force = 0.15F

Thus:

0.15F = 60

F = 400 N

(b) In order to maximize the perpendicular distance between the line of force and the edge, the force should be acted perp. to the radius joining the edge and the axle with magnitude:

F x 0.25 = 60

F = 240 N

(c) For the max. perp. distance = 0.5 m, the force should be acted tangentially to the roller at the point just opposite to the edge. Hence:

F x 0.5 = 60

F = 120 N

Source(s): Myself