II Figure 3 illustrates link 2 rotating at a constant angula

II. Figure 3 illustrates link 2 rotating at a constant angular velocity of 10 rad/s ccw while the sliding block 3 slides toward point A on the link 2 at the constant rate of 5 in./s. At the instant indicated, -4.0 in. Find the absolute acceleration of point P of block 3. (you must use i,_j, k unit vectors in your procedures and all acceleration components have to be clearly calculated). (15 points) 3\' 36,87 Figure 3 The dimensions are =3.0 in. and 6.0 in.

Solution

solution:

1) for given link 3 sliding on link2 has position such that x coordinate is 3 and y coordinate is 4,hence length of link o2p=Rp=5 in

2) let its positon is given for angle a in vector form as

R=Rpcosai+Rpsinaj

3) on differentiating withrespect to time we get

R\'=(Rp*w*cosa+sina*Vs)i+(-Rp*w*sina+cosa*Vs)j

on again differentiating we get accelaration of point

ap=R\'\'=[w(-Rp*w*cosa-sina*vs)+-sina*Vs+f*cosa]i+[w(-Rp*w*sina+cosa*Vs)+Vs*w*cosa+sina*f]j

for constant vs,f=0

angle of link 3 wrt link 2 is

a1=tan^-1(4/3)=53.13 degree

final angle

a=a1+link 2 angle=53.13+36.87=90 degree

for a=90 above equation reduce to

ap=-vs*wi-Rp*w^2j

ap=-corolis accelaration-centrifugal accelaration

here due to constant w,tangential accelaration is zero

on putting value we get

ap=-2*5*10i-5*10^2j

ap=-100i-500j

hence absolute magnitude is

ap=-509.90in/s^2