Tension of two pieces of string

  • Thread starter whatagun
  • Start date
  • #1
whatagun
I have an easy problem that most of u will answer easily (Yr 11 Physics).

I need to find the tension of two pieces of string that are holding a 1Kg weight.

Hopefully the picture below is easy to understand...there is a 70 degree angle from the horizontal for T1 and a 20 degree angle from the horizontal for T2, and obviously 90 degrees at the bottom. All I need is the tension values T1 and T2 and also an explanation of how to do it (please show all working out). Apparently it has something to with the horizontal and vertical force components?????
___________________
\70) (20/
\ /
\ /
T1 \ / T2
\ /
\90/
|
|
1Kg


Thanks a lot
 
Last edited by a moderator:

Answers and Replies

  • #2
whatagun
well the picture didnt work, but u should be able to understand anyway that theres a 70 degree angle at the left and 20 degree angle at the right, and 90 at the bottom, thanks.
 
  • #3
56
0
Is this the picture?
Code:
    \70)      (20/
     \          /
      \        /
    T1 \      / T2
        \    /
         \90/
          |
          |
         1Kg
 
  • #4
whatagun
yeah thats it, thanks.
so can anybody help plz?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Separate the force vectors into x and y components: drawing a force diagram for the "T1" force, the y (vertical) component is "opposite" the 70 degree angle in a right triangle with T1 as hypotenuse, the y component is T1 sin(70) and the x component is T2 cos(70). Looking at the T2 force, where the angle is 20 degrees, the y component is
T2 sin(20) and the x component is T2 cos(20).

It is the y components that support the weight so they must add to 9.8 Newtons (1 kg times 9.8 m/s2):
T1 cos(70)+ T2 cos(20)= 9.8.

Since the the mass is not moving horizontally, the x components must balance (clearly one is to the left, the other to the right):
T1 sin(70)= T2 sin(20)

Solve those two equations to find T1 and T2.

(Notice that 20 degrees is a pretty small angle compared to 70 degrees. T2 should be holding most of the load!)
 
  • #6
whatagun
Thanks for that, but I had pretty much worked it out up to there...the main thing i had trouble with was how to work out the x and y-components...im not asking for the answer here I am just a bit stumped on the working out. Have I just completely overlooked the obvious here???? I would have thought you would need to figure out how much of the 9.81N that pulls on the 2 wires is delivered vertically to each, but I can't seem to figure out how to do that.
 
  • #7
HallsofIvy
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I'm a little puzzled. Did you write your last post before or after reading my last post?
Tx
------- This is a diagram of left wire.
\70 | sin(70)= Ty/T1 so Ty= T1 sin(70)
\ | cos(70)= Tx/T1 so Tx= T1 cos(70)
T1\ | Ty
\ |
\ |
\ |
\|
The point that I made before was that the two y- components are what support the weight:
T1 sin(70)+ T2 sin(20)= 9.8
while the two x components (horizontal) must be equal in order to offset (so there is no motion
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,833
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I'm a little puzzled. Did you write your last post before or after reading my last post?
Tx
------- This is a diagram of left wire.
\70 | sin(70)= Ty/T1 so Ty= T1 sin(70)
\ | cos(70)= Tx/T1 so Tx= T1 cos(70)
T1\ | Ty
\ |
\ |
\ |
\|
The point that I made before was that the two y- components are what support the weight:
T1 sin(70)+ T2 sin(20)= 9.8
while the two x components (horizontal) must be equal in order to offset (so there is no motion horizontally):
T1 cos(70)= T2 cos(20).

If your problem is how to solve these two equations- that easy:
From the second equation, T1 cos(70)= T2 cos(20) we get
T2= T1 (cos(70)/cos(20)).

Putting that into the first equation
T1 sin(70)+ T2 sin(20)= T1 sin(70)+ (T1(cos(70)/cos(20)))sin(20)= 9.8
T1( sin(70)+ cos(70)sin(20)/cos(20))= 9.8
According to my calculator sin(70)+cos(70)sin(20)/cos(20)= 1.06 so (1.06)T1= 9.8 or T1= 9.21 N. while T2= 9.21(cos(70)/cos(20))= 3.35 N.

(I don't know where my head was when I said "T2 should be holding most of the load". It's the opposite. I must have been thinking of the angles at the bottom of the triangle of wires.)
 

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