Question

A hydraulic cylinder causes the distance between points A and O to decrease at a constant rate of 3 inches per second. a) Determine the speed of the slider when the support bar [OB] orientation angle, theta , is 55 degrees. b) Determine the velocity vector of the center of the bar [ABC] when , theta = 50° .

Answer 1

Answer:

a) The speed of the slider is 4.28 in/s

b) The velocity vector is 2.33 in/s

Explanation:

a) According to the diagram 1 in the attached image:

$r_{C/A} =12*cos55i-12*sin55j\\r_{C/A}=6.883i-9.829j$

Also:

$v_{C} =v_{A}+w_{AC}*r_{C/A}\\v_{Ci}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC} \\6.883&-9.829&0\end{array}\right]\\v_{Ci}=-3j+(0+9.829w_{AC} i-(0-6.883w_{AC})j\\v_{Ci}=9.829w_{AC}i+(-3+6.883w_{AC})j$

If we comparing both sides of the expression:

$-3+6.883w_{AC}=0\\w_{AC}=0.435rad/s$

$v_{C}=9.829*0.435=4.28in/s$

b) According to the diagram 2 in the attached image:

$r_{C/A}=12cos50i-12sin50j=7.713i-9.192j\\r_{B/C}=-3.856i+4.596j$

$v_{C}=v_{A}+w_{AC}r_{C/A}\\v_{C}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC}\\7.713&-9.192&0\end{array}\right] \\v_{Ci}=-3j+(9.192w_{AC})i+7.713w_{AC}j\\v_{Ci}=9.192w_{AC}i+(7.713w_{AC}-3)j$

Comparing both sides of the expression:

$7.713w_{AC}-3=0\\w_{AC}=0.388rad/s\\v_{C}=3.575i$

$v_{B}=v_{C}+w_{AC}r_{B/C}\\v_{B}=3.57i+\left[\begin{array}{ccc}i&j&k\\0&0&0.388\\-3.856&4.59&0\end{array}\right] \\v_{B}=3.57i+(0-1.78)i-(0+1.499)j\\v_{B}=1.787i-1.499j\\|v_{B}|=\sqrt{1.787^{2}+1.499^{2} } =2.33in/s$