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بسم الله الرحمن الرحيم. Dana . A.Al Shammari : Name Qualification : Master physics Solids state : Department. Notifications. Name of experiment. The Spring. objective. Find the spring constant of a spiral spring. Achieving Hooke’s law. Theory.
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بسم الله الرحمن الرحيم Dana .A.AlShammari:NameQualification: Master physics Solids state:Department
Name of experiment The Spring
objective Find the spring constant of a spiral spring Achieving Hooke’s law
Theory A spring constant is the measure of the stiffness of a spring
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/s2
Forcees that affect the spring • 1-The downward force F(W) • m:Mass(kg) • g=9.8 m/s2 F(w)= m g {1} • 2-The upward restoring force F(r)
Forcees that affect the spring F(r) α d • According to Hooke's law F(r)= - k d {2} K: the spring constant (N/m) (-): The F(r) is in opposite direction to its elongation d:Elongation(m)
Forcees that affect the spring F(r)= - k d F(w)= m g
In the case of the blance F(r)= - k d • ∑F=0 • F(w)-F(r)=o • F(w)=F(r) • Mg=kd • K=mg/d(N/m) F(w)= m g
Graphing 1- Slope=k= k exp F(w) (N) K thor k= (F(w2), ∆X2) (F(w1), ∆X1) Slope ∆X(m)
Conclusion k exp & k thot
بسم الله الرحمن الرحيم Dana .A.AlShammari:NameQualification: Master physics Solids state:Department
objective To find the resultant force by analytically and experimentally (force table)
Theory A vector is a quantity that possesses both magnitude and direction; examples 1- Velocity (V) 2- Acceleration (a) 3- Force (F)
Theory In case one vector Each force is resolved into X,Y components:
Theory In case one vector Find the resultant
Theory In case mor vectors Find the resultant FX=∑Fx=F1X+F2X FY=∑FY=F1Y+F2Y • F2 F1 • F2 Cosθ
Expermently F1,F2=ɣ F2,F3=β F3,F1=α
= = Theortical = 0.686 Cos 0 = 0.686N F1x= F1 Cosθ1 = 0.686 Sin 0 = 0 F1y=F1 Sinθ1 Fx =0.98 Cos 80= 0.17N F2x=F2 Cosθ2 Fy F2Y= F2 Sin 80 = 0.98 Sin 80= 0.965N ∑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
Theortical • Y • F2 F3 β=148.4 ɣ=80 θ=48.4 • X • θ=48.4 F1 -F3 F1,F2=ɣ F2,F3=β F3,F1=α • α =131.6