(45%)3Rod AB is pinned to ground at A. There is a prismatic joint between B and
ID: 1719008 • Letter: #
Question
(45%)3Rod 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 as shown below, i.e. L is the length of rod AB. The angle Q shown below is measured counter-clockwise from the global inertial horizontal to the directed line AB- The angle shown below is measured from the global/inertial horizontal clockwiss to the directed line The following values are known/given: 4-4m, R.3m R=9m/s, k 7m/s,9, 3. to the directed line CB. Li REQUESTED: At the instant shown, find the acceleration of point C, in m/s HINTS: () Do not copy the solution for another problem and expect to get a bunch of partial credit. 2) Do not get too bogged down in your calculators. If you problem. credit. (2) Do not get too bogged down in your caleulators. If you have n correct ons in n unknowns that uniquely determine the answer, you have 100% credit for the
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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