Dana .A.Al Shammari: NameQualification: Master physicsSolids state: Department

بسم الله الرحمن الرحيم

Notifications

The Spring

Name of experiment

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=mgw:weight(N)m:mass(kg)

g: Accelration of (gravity(m/s2

1-The downward force F(W)

m:Mass(kg)g=9.8 m/s2

Forcees that affect the spring

F(w)= m g {1}

2-The upward restoring force F(r)

According to Hooke's law

K: the spring constant (N/m)(-): The F(r) is in opposite

direction to its elongationd:Elongation(m)

F(r)= - k d {2}

Forcees that affect the spring

F(r) α d

Forcees that affect the spring

F(w)= m g

F(r)= - k d

F(w)= m g

F(r)= - k d∑F=0F(w)-F(r)=oF(w)=F(r)

Mg=kdK=mg/d(N/m)

In the case of the

blance

DataTable

m...........X0=…………cm , X0

d=∆X=X-X0Elongation(m)

DistanceX(m)

DistanceX(cm)

F(w)=mg(N)

Mass(kgm)

Mass(gm)

∑∆X/n……=

∑F/n……=

Graphing

∆X(m)

F(w)(N)

Slope

(F(w2), ∆X2)

(F(w1), ∆X1)

1 -Slope=k= k exp

k= K thor

Error

Conclusion

k exp & k thot

In the end

Dana .A.Al Shammari: NameQualification: Master physicsSolids state: Department

بسم الله الرحمن الرحيم

Force table The

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 vectorsFind the resultant

FX=∑Fx=F1X+F2XFY=∑FY=F1Y+F2Y

F1

F2

F2 Cosθ

Expermently

α β ɣ F3(N)M1g

F2(N)M1g

F1(N)M1g

M3(Kg)

M2(Kg)

M1(Kg)

No

80 0.1 0.07 1

100 0.09 0.12 2

120 0.08 0.08 3

F1,F2=ɣF2,F3=βF3,F1=α

TheorticalF1x= 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

Theortical

Theortical

F1

F2

ɣ=80

θ=48.4

θ=48.4

β=148.4

α =131.6

Y

X

F1,F2=ɣF2,F3=βF3,F1=α

-F3

F3

Theortical

THEROTICAIL EXPERMENTLY

F3=1.28 N F3=1.32N

ɣ =80 ɣ = 80β = 148.4 β = 150

α = 131.6 α = 130