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Help needed - Newtons first law - particles in equilibrium

  • Thread starter Hemmelig
  • Start date
  • #1
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For some odd reason, i'm having no problems with the rest of the exercises, though this one is really annoying as i just can't seem to get it right.

Here's a picture of the exercise

h t t p://img2.freeimagehosting.net/image.php?4bc2e12442.jpg
http://img2.freeimagehosting.net/image.php?4bc2e12442.jpg [Broken]
Could someone please give me a detailed description on how to solve it ?

I've used m=w/g to find the weight of the person (882.9 N )
But when i try to use sin and cos with the angle, i don't get the right result
 
Last edited by a moderator:

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
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55
Why don't you show us what you have done?

What angles did you use in the sin and/or cos functions?

Please provide us with more information about what you have done.
 
  • #3
22
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I tried dividing it up in two sets, where the rope to the left was A and the rope to the right was B

Since it's in equilibrium, the sum of forces in the x direction (Fx) is 0, Fy is also 0

Since Fx=0 : B * sin 10 + (-A)
Since Fy=0 : B * cos 10 + (-w)

I then used the second one to find the tension from part B, but i'm not allowed to sum them together (?), so i don't really know what to do.

Basicly, i haven't got a clue right now
 
  • #4
alphysicist
Homework Helper
2,238
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Hi Hemmelig,

Newton's law in the x-direction, for the case of no acceleration in the x-direction is [itex]\sum F_x =0[/itex]. Writing this out for the case of your three forces, A, B, and w gives:

[tex]
A_x+ B_x + w_x =0
[/tex]

Your got that w_x=0 (so the weight force does not appear in the x direction equation); however, it appears that you did not get the right x-components for A and B.

In the y direction, the same thing is happening. Your y-component for the weight is okay ([itex]w_y=-w[/itex]) but the y-components for A and B are incorrect.
 
  • #5
22
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hmm, what are the right components for x and y then ?
 
  • #6
22
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I got the right result (well, in a way)

If i call the rope C

C * sin 10 + (-w)

w/sin10 = C

But for some reason, the answer i get is exactly twise the correct result

Am i doing it right and i'm just supposed to divide the answer with two ?
 
  • #7
alphysicist
Homework Helper
2,238
1
For the part of the rope pulling to the right (force B), it is pulling both upwards and to the right. It's x-component is the horizontal component of the force and if you draw a triangle for the components of B you can see that the horizontal component is adjacent to the angle.

So, for example, [itex]B_x = +B \cos\theta[/itex]. If you find [itex]B_y, A_x, A_y[/itex] you can then find the tension.
 
Last edited:
  • #8
alphysicist
Homework Helper
2,238
1
I got the right result (well, in a way)

If i call the rope C

C * sin 10 + (-w)

w/sin10 = C

But for some reason, the answer i get is exactly twise the correct result

Am i doing it right and i'm just supposed to divide the answer with two ?
If you solve the equation for forces in the x-direction and plug it into the force equation for the y direction, you'll get the above formula except with a factor of 2.

Solving the x-direction equation will also tell you something important about the tension along ropes in these problems that will show you why the above equation is so close.
 

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