# questions in physics help

• Dec 10th 2010, 05:36 AM
r-soy
questions in physics help
Attachment 20050

1 ) A traffic light weighing 100 N hangs from a vertical cable to two other cables that are fastened to a support.
The upper cabels make angle of 37 and 53 with the horizontal . Find the tension in each of the three cables

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Here I now first must find x component and y component

How I find the
x component and y component in every Q I do mistake

and why you say -T cos 37 why ( -) here ??

plese Help me
http://www.mathhelpforum.com/math-he...isc/pencil.png
• Dec 10th 2010, 05:58 AM
here is my solution

for the equilibrium of the traffic light,

$T_3 = 100N$

considering the equilibrium of the joining point of cables,

$T_2cos37 = T_1cos53$

$T_1 = \frac{T_2cos37}{cos53}$

on Y direction,

$T_1sin37+T_2sin53=T_3$

you have three equations

I cannot find where you go a minus sign
• Dec 10th 2010, 06:11 AM
r-soy
How you get the point of cables ?
• Dec 10th 2010, 06:34 AM
since all the points on the system is at a static equillibrium, you can apply $\sum F = 0$

so applying that to the point where three cables meet you can get an equation with three tensions.
• Dec 10th 2010, 06:41 AM
r-soy
ok the system at equillibrium

then

F = 0

Fx = T2 cos 53 + T1 cos 37 = 0
Fy = T2 sin 35+ T1 sin 37 = 0

After that what I do ?
• Dec 10th 2010, 06:43 AM
e^(i*pi)
Since you have two equations and two unknowns solve simultaneously
• Dec 10th 2010, 06:47 AM
Fy is incorrect reffer above post
• Dec 10th 2010, 06:59 AM
r-soy
Fx = T2 cos 53 + T1 cos 37 = 0
Fy = T2 sin 53+ T1 sin 37 - 100 = 0

sorry but now what will I do
• Dec 10th 2010, 07:13 AM
infact Fx=> T2cos53 -T1cos 37 =0

take T2 (or T1) from 1st equation and substitute it into the second equation.
• Dec 10th 2010, 07:28 AM
r-soy
Ok Now I take T2

T2 = T1cos37/cos53 = 1.327T1

Then I substitute this value ( 1.327 T1 ) second equation
T1 sin37 +(1.327T1)sin53 - 100 = 0
T1[sin37 +(1.327)sin53 ] = 100
T1 = 100/sin37 +(1.327)sin53 = 167
• Dec 10th 2010, 07:39 AM
correct...
• Dec 23rd 2010, 02:42 AM
r-soy
...
but the answer is wrong not as ny book
• Dec 23rd 2010, 02:56 AM
CaptainBlack
Quote:

Originally Posted by r-soy
Ok Now I take T2

T2 = T1cos37/cos53 = 1.327T1

Then I substitute this value ( 1.327 T1 ) second equation
T1 sin37 +(1.327T1)sin53 - 100 = 0
T1[sin37 +(1.327)sin53 ] = 100
T1 = 100/sin37 +(1.327)sin53 = 167

T1 = 100 / [sin(37) +(1.327) sin(53)]

Redo your arithmetic, and make sure the calculator is in the correct mode and you put the brackets in the right place..

CB
• Dec 23rd 2010, 03:03 AM
r-soy
Yes now ok