A graph of force (F) versus elongation (x) shown in figure below. Find the spring constant!
Solution
Hooke's law formula :
k = F / x
F = force (Newton)
k = spring constant (Newton/meter)
x = the change in length (meter)
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o sa sa s y es gn gn m a ons s ne necessary to determine the effect of the 2-kN tension in the cable on the shear, tension, . purpose replace this force by its equivalent of two forces at A, Ft parallel . n Dete Deterrmine mine Ft and Fn.
= t F t = 1.28 kN F n = 2sin50 = . n
Determ ne t e magn tu e Fs o e ens e spr ng orce n or er s
ma nitude R of this vertical resultant force.
cos
=
F s = 60
sin60 =
s
120
120 R = 103.92 lb
In t e es gn o a contro mec an sm, s e erm ne a
the x and P.
tanθ =
scalar com onents of
5 -1
⎛5⎞ ⎝ 12 ⎠
= . P = −260cos 22.62 P x = −240 N P y = −260sin 22.62 − y
For t e mec an sm s own, e erm ne e sca ar t
n
, ,
.
5 tan α = 12
α = 22.62 β = 30 − α = 30 − 22.62
β = 7.38 Pn = 260cos 7.38 Pn = 257.84 N
Pt = 260sin ( 7.38 . t
)
I t e equa tens ons T n t e pu ey ca e are , express
. ma nitude of R.
R x = ∑ F x R x = 400 + 400 cos 60
= R y = F y R y = 400sin60 R y = 346.41 N x
R = 600i + 346.41 j R =
=
( 600 ) + ( 346.41) 2
.
2
, exerts a 180 N force P as shown Determine the components of P which are parallel and perpendicular to the incline.
Pt = 180cos25
Pt = 163.13 N P = −180sin25
Pn = −76.07 N
The normal reaction force N and the tangential friction force F act on the tire of a front wheel drive car as shown. forces in terms of the unit vectors a i and j along the x-y axes and (b) et and en along the n-t axes shown.
R x = 400cos15 + 900cos105 i lb . ˆ x R = 400sin15 + 900sin105 ˆj lb
(a)
R = 972.86 ˆj lb
R = 153.43iˆ + 972.86 ˆj lb
(b)
R = 400eˆt + 900eˆn lb
e erm ne e resu an o e wo orces app e to t e rac et. r te n terms of unit vectors along the x-and y-axes.
R x = ∑ F x R x = 200cos35 + 150cos120
x
=
R y =
. F y
R y = 200sin 35 + 150sin120
R y = 244.61 N
R = 88.83i + 244.61 j N
drag force D for the simple airfoil is L/D=10. If the lift force on a short sect on o t e a r o s 50 , resultant force R and the an le θ which it makes with the horizontal.
L
= 10, L=50 lb 50 10 2
2
R =
50 2 + 5 2 = 50.25 lb −1 ⎛ L ⎞ −1 ⎛ 50 ⎞ = tan θ = tan D 5 = .
