+Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or...

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Transcript of +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or...

Page 1: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of
Page 2: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of
Page 3: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of
Page 4: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of
Page 5: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of
Page 6: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of
Page 7: +Q2 +2pQ (P+Q)2 R=P+Q when e-1800 ; cos 180--1 ; Sin 180 = o = p2 +Q2 _2PQ = (p-Q)2 R=P-Q a = or 1800 a) Define moment of force about a point. State the condition of equilibrium of

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A small semantics quiz. 2 Guess the determiner... 1. P Q x(P(x)&Q(x)) 2. P Q x(Plural(x)&P(x)&Q(x)) 3. P Q x(P(x) Q(x)) 4. P Q x(P(x)& y(P(y)

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1 Methods of Proof. 2 Consider (p (p→q)) → q pqp→q p (p→q)) (p (p→q)) → q TTTTT TFFFT FTTFT FFTFT.

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