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Forces and Vectors

  • #1
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I've had a bit of a problem with the following question. I have the answer I did it properly except for one part i can't comprehend.

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

Ok I know the answer but I don't understand why one of them is 0N i mean can someone explain it very simply and clearly............Please
 

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  • #2
Tom Mattson
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thomasrules said:
Ok I know the answer but I don't understand why one of them is 0N
The tension is not 0 N in either part of the wire. It can't be, because then the components of tension in the x-direction cannot balance out.
 
  • #3
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mr.mattson trust me one part of the wire is 375 and the other is 0
 
  • #4
Tom Mattson
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I interpret your verbal description roughly as follows.

Code:
I*              *I
I    *         * I
I        *    *  I
I            *   I
I            |   I
I            V   I
I                I
The I's denote the supports that are 20 m apart, the *'s denote the 40 m cable, and the downward arrow at the kink in the cable denotes the downward 375 N force.

Have I got that right?
 
  • #5
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yes you have but if you actualy do the question and try finding the angles you find that one is 90 degrees :D. So now you have a different diagram right?
 
  • #6
Tom Mattson
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My diagram can still be used for the purpose of obtaining the equations of static equilbrium. So let the left end of the rope have tension [itex]T_1[/itex] and make an angle [itex]\theta_1[/itex] with the negative horizontal axis, and let the right end of the rope have tension [itex]T_2[/itex] and make an angle [itex]\theta_2[/itex] with the positive horizontal axis.

Summing the forces on the point of application of the force gives:

[tex]-T_1\cos(\theta_1)+T_2\cos(\theta_2)=0[/tex]

[tex]T_1\sin(\theta_1)+T_2\sin(\theta_2)=375[/tex]

Now, you say that one of these angles is 90 degrees, right? (I'm taking your word for it) Let's say for definiteness that the 90 degree angle is [itex]\theta_1[/itex].

The first equation then gives:

[itex]-T_1\cos(90)+T_2\cos(\theta_2)=0[/itex].

The first term goes to zero, which leaves:

[itex]T_2\cos(\theta_2)=0[/itex].

Now it's clear that [itex]\theta_2\neq90[/itex] (or else you would have two segments of rope along their respective supports with a flat horizontal segment in between, something you can't accomplish with just one force). So since [itex]\cos(\theta_2)\neq0[/itex], it therefore follows that [itex]T_2=0[/itex].
 
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  • #7
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sorry but I don't get how you got t1cos90+t2cosA=0
 
  • #8
Tom Mattson
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Draw a free body diagram at the point of application of the force. Sum the forces in the x-direction to get my first equation, and sum the forces in the y-direction to get my second equation.
 
  • #9
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Tom Mattson said:
Draw a free body diagram at the point of application of the force. Sum the forces in the x-direction to get my first equation, and sum the forces in the y-direction to get my second equation.
ok first of all lol i don't know what "point of application" is exactly. And I don't know how to draw the diagram your talking about. I don't know what i'm adding I'm confused lol......In my answer I had a right triangle with the force 375 going in the same way as the tension force.
 
  • #10
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thomasrules said:
ok first of all lol i don't know what "point of application" is exactly.
"point of application of the force"="point along the cable at which the force is applied"

And I don't know how to draw the diagram your talking about.
*cough* *sputter*

Your teacher hasn't taught you how to draw a free body diagram??? :surprised:

How have you been analyzing 2D force problems?

See this website:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/Forces/FreeBodyDiagram.html [Broken]
 
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  • #11
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lol i know how to draw free-body diagrams but then I don't know what you want me to add up......how you get fsin whatever......whats ur adding there not sure.....btw this is grade 12
 
  • #12
Tom Mattson
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You are supposed to add up the forces where the 375 N force is applied. Since that point is not moving, the vector sum of all the forces acting on it must be zero.

The vector equation [itex]\Sigma\vec{F}=\vec{0}[/itex] reduces to the following two scalar equations:

[itex]\Sigma F_x=0[/itex]

[itex]\Sigma F_y=0[/itex].

That is, the sum of the x-components of all the forces must equal zero, and the sum of the y-components of all the forces must equal zero. That is Newton's second law, and I would be very surprised if it were not presented in a Grade 12 Physics course. I recall that it was in my Grade 11 course.

So, that's where those equations come from. I resolved the tensions into their x- and y-components, and summed them as per Newton's 2nd law.
 
  • #13
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-t1cosA+t2cosB=0

You don't know that it's equal zero when your doing the question though.....so why did you assume its zero? ......:D I'm sorry but I'm having a hard time comprehending this shizzle.......Um like I have question all over the place, why - in front? Why are you adding stuff lol i'm confused with the adding part...I'm sorry but you will have to go even simpler , sorry for the trouble
 
  • #14
Tom Mattson
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thomasrules said:
You don't know that it's equal zero when your doing the question though.....so why did you assume its zero? ......:D
Yes, you do know it's zero. It's not moving, remember?

why - in front?
Because the x-component of that tension vector points to the left, and I'm taking the left as the negative direction.
 
  • #15
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well we were learning it with the paralellogram law of addition...none of the fsintheta stuff.........don't get why and how your using the angles in the fsintheta+fsinalpha=0 equations.....
 
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  • #16
Tom Mattson
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thomasrules said:
well we were learning it with the paralellogram law of addition...
OK, so you've got 3 forces to add up. One you know (375 N), and two you don't. The known force points straight down. You know that one of the unknown forces points straight down, and that the other one doesn't.

If the resultant is supposed to be zero (static equilibrium), than it should be obvious that the second unknown force (the one that is neither straight up nor straight down) is zero. If it weren't then there would be a net force in the x-direction.

none of the fsintheta stuff.........don't get why and how your using the angles in the fsintheta+fsinalpha=0 equations.....
It just comes from right triangle trigonometry, which presumably you've taken already. Make a triangle out of each tension vector and find the lengths of the legs.
 

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