# Thread: Force as a vector 2

1. ## Force as a vector 2

A steel wire 40 m long is suspended between two fixed points 20 m apart. A force of 375 pulls the wire down at a point 15 m from one end of the wire. State the tension in each part of the wire.

2. Originally Posted by skeske1234
A steel wire 40 m long is suspended between two fixed points 20 m apart. A force of 375 pulls the wire down at a point 15 m from one end of the wire. State the tension in each part of the wire.
Try drawing a diagram . I got a right angled triangle with horizontal length 20 , vertical length 15 and hypothenus 25 .

The force is pulling downwards from the end of the vertical side .

There will be 2 tensions , T1 and T2 .

T1 will be 375 N .

Resolve T2 vertically T2cos53.13 = 375 , T2=625 N

You can get 53.13 by doing some trigo .

I am not entirely sure with my answers though .

375 N, 0 N...........
and is the 40 m not significant in part of the drawing?

4. Originally Posted by skeske1234

375 N, 0 N...........
and is the 40 m not significant in part of the drawing?
I don see how the tension is 0 N . The wire of 40 m is divided into 2 parts
25m and 15 m ( where the pulling force is ) .

I don see how the tension is 0 N . The wire of 40 m is divided into 2 parts
25m and 15 m ( where the pulling force is ) .
Thanks! I was interpreting this as 15 m measured along the 20 m distance, but then I couldn't find the lengths of the sides!

skeske1234, this gives a triangle with sides of lengths 25, 15, and 20. Since $15^2+ 20^2= 25^2$ this triangle satisfies the Pythagorean theorem and, as mathaddict said before, is a right triangle.