Physics Newton Law?

Question

The distance between two telephone poles is 52.0 m. When a 0.900 kg bird lands on the telephone wire midway between the poles, the wire sags 0.210 m. How much tension does the bird produce in the wire? Ignore the weight of the wire

Answer 1

ID the forces.

There are forces where the wire is attached to the poles. There is force where the bird sits on the wire. The sum of these forces f = 0 because nothing is accelerating (or even moving for that matter).

The vertical depression of the wire, where the bird sits, is caused by the weight of the bird w = mg; where m = .9 kg and g = 9.81 m/sec^2. That weight is offset by the vertical pull (fv) of the wire w = fv = 2T sin(theta); where sin(theta) = .210/26 = sag/half the pole distance. T is the tension in the wire on either side of the bird, the answer you are looking for.

Thus, T = w/2sin(theta) = mg/2sin(theta) = (26/.210)(.9*9.81)/2 = 546 Newtons (kg-m/sec^2), which is over 100 pounds. I don't know where your numbers came from, but this is one big bird...maybe from Sesame Street?

Answer 2

assuming no deflection of the poles, but rotation is possible.

let us write an expression that models the displacement between pole1 and pole2, we'll use a binomial to express it.
f(x) = ax^2 + bx + c
f'(x) = 2ax + b
our known displacements are
f(0) = 0
f(26) = -.210
f(52) = 0
we need to solve for the varible a,b, and c

at pole1
f(0) = 0 = a*0^2 + b*0 + c ==> c=0
our equation is in the form
f(x) = ax^2 +bx = x*(ax + b)

at pole2
f(52) = 0 = 52*(a*52 + b)
a = -b/52

midway
f(26) = .210 = a*26^2 + b*26 = 676*-b/52 + b = -.210
solve for b
b = .0175

sub back in a = -b/52
a = -0.000337

displacement function
f(x) = -0.000337x^2 + 0.0175x

f'(0) = +0.0175

Angle of the wire at pole1 is tan^-1( 0.0175) = 1 degree

here is what are system looks like

P1_______ B _______ P2 ( i added ___ for graphics)
---------------- m ----------------
^.................... [] ................... ^

|<------------- L ---------------->|

(locally)
FBD of the pole at P1
Sum of the F's in the y direction =0 = P1 - T*sin(1)
FBD of the pole at P2
Sum of the F's in the y direction =0 = P2 - Tsin(1)
(globally)
Sum of the moments about P1 = 0 = P2*L - m*g*L/2 - T*sin(1)*L
P2 = m*g/2 - Tsin(1)
therefore, P1 = m*g/2 - Tsin(1)
From our FBD 0 = P1 - T*sin(1) = m*g/2 - Tsin(1) - Tsin(1)
2Tsin(1) = m*g/2
T = m*g/(4*sin(1))
T = .9*9.81/(4sin(1))

T = 126 N

----
the above methods assumed a linear displacement of the wire between the pole and the bird, of course that usually is not the case.

Answer 3

cos theta = .008
2T cos theta=mg
by this we get T=557.14 N

Answer 4

The angle (theta) caused by the sag is 89.99

cos(theta) = 0.21/26.0085 = 0.0081

2T*cos(theta) = m*g

T = Unknown
m = 0.9kg
g = 9.8 m/s^2

2T*0.0081= 0.9*9.8
T = (0.9*9.8)/(2*0.0081) = 544.44N

Answer 5

big dobbie?/????????......