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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Dec 2007 23:18:55 IST
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Q. A rectangular plate ABCD of mass m and dimensions a x b is supported in a vertical plane by two hinges at A & B as shown in the figure. Find the instantaneous reaction of the hinge at A immediately after B is removed.
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11 Dec 2007 23:26:59 IST
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assuming AB = a and BC = b,
im getting inst acc about A as alpha = 3ag/(a^2 + b^2)^3/2
but im not able to proceed further.
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11 Dec 2007 23:32:40 IST
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WERES D PICTURE YAAAAAAAAAAAAAR
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11 Dec 2007 23:35:45 IST
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HEY PANNAGUMA IM UNABLE 2 understand how u got d acceleration yaar plz explain
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12 Dec 2007 15:06:26 IST
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hey that alpha i posted is wrong.
alpha = 3ag/4*[a^2 + b^2]^1/2
and instantaneous reaction comes out
Mg/4 * [ 25 - {9b^2 / a^2 + b^2} ]^1/2
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12 Dec 2007 15:13:05 IST
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The correct answer is: Vertical Component (a2+4b2)mg ----------------- 4(a2+b2) Horizontal Component 3abmg ------------- 4(a2+b2) If somebody is able to get the answer,please give me the solution.
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12 Dec 2007 15:35:25 IST
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man that is what i have got, i just took the resultant.
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12 Dec 2007 15:35:43 IST
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rate me.
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12 Dec 2007 15:42:58 IST
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my vertical component is Mg + [3Mga^2 / 4(a^2 + b^2)]
horizontal is same.
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14 Dec 2007 00:14:13 IST
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Can u please give me the solution
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14 Dec 2007 00:24:58 IST
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ok in several parts ..............
MOI thru com for plate = M(a^2 + b^2)/12
MOI abt A = M(a^2 + b^2)/3
now find @.
T = I@
T = FL = Mg*a/2
substituting gives
@ = 3g*a/2(a^2 + b^2)
thus Acom = @R = 3ag/4*[a^2 + b^2]^1/2
since R = [(a^2 + b^2)^1/2] /2
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14 Dec 2007 00:26:43 IST
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taking theta with vertical,
costheta = b/(a^2 + b^2)^1/2
sintheta = a/(a^2 + b^2)^1/2
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14 Dec 2007 00:30:03 IST
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AcomX = Acom*costheta
and AcomY=Acom*sintheta
now write force eqns,
Nvetical - Mg = M*AcomY
Nhorizontal = M*AcomX
substitute and solve.
Plz rate.
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