Figure illustrates the driving mechanism in a saber saw At t

Figure illustrates the driving mechanism in a saber saw. At the instant shown, the crank is rotating at a constant rate of 300 rpm clockwise. Graphically determine the linear acceleration of the saw blade.

Solution

solution:

saber screw is similar to slider crank mechanism which consist of crank r,connecting rod l,and slider at distance x

1)as we dont know the length of of connecting rod,slider position hence from geometry we draw mechanism as follows

1)draw line of crank oa of length 25 mm in -45 direction,then connecting rod is 90degree to crank at this position hence draw linw perpendicular to oa in leftward direction and draw perpendicular line from point o in downward direction we get our mechanism,

length of slider is=x=35 mm

l=25 mm

2)here velocity of crank is

vr=r*w=r*2*pi*300/60=78.53 cm/s

3)as r and l are same and both have same angular velocity and angular ccelaration to be zero

Vl=Vr=78.53cm/s

4)here instead of drawing velocity polygon we directly draw accelaration polygon

centrifugal accelaration of crank and connecting rod is samea and due to constant velocity tangential accelartion is zero,

Fcr=Fcl=V^2/r=78.53^2/2.5=2466.78 cm/s2

4)for accelaration polygon draw centrifugal accelaration parallel to crank r to get point a,then draw centrifugal accelaration paralllel to connecting rod as line of unknown length,

through origin o draw line along slider as mechanism and wherever they meet is point b

hence length origin o to point b is accelartion of slider and it is tangential accelaration and its centrifugal accelaration is zero

OB=3.5 cm

where

OA=2.5 cm

factor=2466.78/2.5=986.71 cm/s2

Ftob=3.5*986.71=3453.49 cm/s2=linear accelaration of slider or screw blade