Rod AB is pinned to ground at A. There is a prismatic joint between B and C. The

Question

Rod AB is pinned to ground at A. There is a prismatic joint between B and C. The housing of the prismatic joint is pinned to the rod AB at B. The distance between A and B is L_1 as shown below, i.e., L_1 is length of rod AB. The angle theta_1 shown below is measured counter-clockwise from the global/inertial horizontal to the directed line AB. The angle alpha shown below is measured from the glbal/inertial horizontal clockwise to the directed line CB. The following values are known/given: L_1 = 4m, R = 3m, R = 9m/s^2 , R = 7m/s^2, theta_1 =- 30degree, theta_1 = 5.5rad/s, theta_1 = 3.8rad/s^2, alpha = 40degree, alpha = 7.5rad/s, alpha = 8.2rad/s^2. At the instant shown, find the acceleration of point C, in m/s^2.

Explanation / Answer

For this purpose we need velocity of point C...

cos30=( x2 + L12 - R2) / 2xL1

R2 = -2xL1 cos 30 +x2 +L12

     Taking differential with respect to time

2R . dR / dt = - 2L1 cos 30 dx /dt + 2x dx / dt

2R . dR / dt = (-2L1 cos 30 + 2x) dx / dt

dx / dt = 2R. R(dot) / (-2L1 cos 30 + 2x)

   dx / dt = 2(3)(9) / - 2 (4) cos30 + 2x) ...........1

Using sin law

x / sin ABC = L1 /sin alpha

   x = (sin 110) * 4 / sin 40

   x = 5.84m

   Now putting the value of x in equ1

   dx / dt = 54 / -6.92 + 2(5.84)

dx / dt = 11.34m/s

This is the velocity of point C. Now finding the acceleration of point C ....Now in order to find out the accelaration of point C we will double differentiate

   dx / dt = 2R. R(dot) / (-2L1 cos 30 + 2x)

d2x / dt2 = 2R R( double dot) / -2 L1 cos 30 +( 2 dx / dt)

= 2(3)(7) / -2(4) cos 30 + 2(11.34)

= 42 / -8 cos 30 + 22.68

= 42 / 15.75

= 2.66 m /s2

This is the value of acceleration at point C.

  

     

  

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