36 Chapter 2 • Statics Solution:


(a) MA= load = 490


= 3,5


effort 140


load = 50 × 9,8 = 490 N; D = 0,75 m; effort = 140 N; d = 0,15m (b) VR = s


s


E L


= π


\VR = D ,


π d


D d


= 075 = 5


015 ,


Example 2.10


The length of the crank of a simple winch is 400mmand is connected directly to the drum, which has a diameter of 159 mm. A mass of 75 kg is raised by applying an effort of 200 N. Calculate for thismachine and at this load; (a) (b)


the mechanical advantage, and the velocity ratio.


Solution: D = 2ℓ load = 75 × 9,8 = 735 N = 2 × 0,4 = 0,8m


(a)


MA= load = 735


effort 200


= 3,675


effort = 200 N d = 0,159m


(b) VR = s = π = D


s


π d


,


E L


D d


= 08 = 5,03


2.8 Moments


Fig. 2.12 shows a spanner being used to tighten a nut. Tis is probably the most common example of the moment of a force. Force P, applied to the spanner, will obviously turn or tend to turn the nut in a clockwise direction.


Perpendicular distance (Radius “r”)


P Effort


0159 ,


Fig. 2.12


Fig. 2.12


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