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Dana.A.Al Shammari: Name Qualification: Master physics Solids state: Department بسم الله الرحمن الرحيم

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Find the spring constant of a spiral spring Achieving Hooke’s law

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A spring constant is the measure of the stiffness of a spring

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If a mass m is attached to the lower end of the spring,the spring stretches a distance of d from its is initial position under the influence of the weight W=mg w:weight(N) m:mass(kg) g: Accelration of (gravity(m/s 2 W=mg w:weight(N) m:mass(kg) g: Accelration of (gravity(m/s 2

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2-The upward restoring force F(r)

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In the case of the blance

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m...........X 0=…………cm, X 0 d=∆X=X-X 0 Elongation(m) Distance X(m) Distance X(cm) F(w)=mg (N) Mass (kgm) Mass (gm) ∑ ∆X/n=…… ∑ F/n=……

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∆X(m) F(w) (N) Slope (F(w2), ∆X2) (F(w1), ∆X1) 1- Slope=k= k exp k= K thor

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k exp & k thot

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Dana.A.Al Shammari: Name Qualification: Master physics Solids state: Department بسم الله الرحمن الرحيم

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To find the resultant force by analytically and experimentally (force table)

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A vector is a quantity that possesses both magnitude and direction; examples 1- Velocity (V) 2- Acceleration (a) 3- Force (F)

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In case one vector Each force is resolved into X,Y components:

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In case one vector Find the resultant

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In case mor vectors Find the resultant F X = ∑ F x =F 1 X+F 2 X F Y =∑F Y =F 1 Y+F 2 Y F1 F2 F2 Cosθ

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αβɣF3(N) M1g F2(N) M1g F1(N) M1g M3 (Kg) M2 (Kg) M1 (Kg) No 800.10.071 1000.090.122 1200.08 3 F1,F2=ɣ F2,F3=β F3,F1=α

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F1x= F1 Cosθ1 = 0.686 Cos 0 = 0.686N F1y=F1 Sinθ1 = 0.686 Sin 0 = 0 F2x=F2 Cosθ2 = 0.98 Cos 80= 0.17N F2Y= F2 Sin 80 = 0.98 Sin 80= 0.965N Fx Fy ∑Fx=F1x + F2x = 0.686 + 0.17 = 0.856 N ∑Fy=F1y + F2y = 0 + 0.965 = 0.965 N Magnitude=F3 = R == =1.28 N

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F1 F2 ɣ =80 θ=48.4 β=148.4 α =131.6 Y X F1,F2=ɣ F2,F3=β F3,F1=α -F3 F3

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THEROTICAILEXPERMENTLY F3=1.28 NF3=1.32N ɣ =80 β = 148.4β = 150 α = 131.6 α = 130

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HOOKE’S LAW.

HOOKE’S LAW.

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