Physics Help - Static Equilibrium

  • #1
God, I took twenty minutes to prepare this, and then it logged me out when I tried to post it, and now I have to retype the whole damn thing. So here goes:

Here is the diagram: http://photobucket.com/albums/v517/shirozu/?action=view&current=sharkproblem.jpg

Since its almost past ten, and I'm losing my coherency :eek: , I will make this really easy.

That diagram and problem requires me to find the sum of the torques. The torque that the cord performs on the rod, and the torque that the weight of the shark creates on the rod. First the shark:

I simplified the force by using a triangle. Through thinking, I eventually found that the angle inside of this triangle is 60 degrees. I am positive this is the right angle. So, I did the equation to find the torque:

cos(60)=x/10000
10000cos(60)=x
x=5000

the I came to the part where I actually find the torque, but in the equation it says to neglect the weight of the rod... do I just multiply 5000 by the lenght of the rod, which is given? If not, what do I do?

Also, finding the torque of the cord requires finding the tension. What exactly do I do here? I made the simplifying triangle, and eventually got to:

Tcos(10)=x

But what do I do? I have two variables, and the tension I don't think should be used if the length of the pole in the first torque equation is used. Its inconsistent, if you catch my drift.

Please help, I have a physics test on this tomorrow, and I have no one to ask except this forum. I would appreciate any help at all.

Thanks in advance.

Daniel
 

Answers and Replies

  • #2
231
1
Why, you should get the solution by resolving horizontal and verticle components of forces. You need not calculate moments.
 
  • #3
thats essentially what I was doing.. I think...
 
  • #4
231
1
What is the problem, then? Put the equations for verification.
 
  • #5
My problem is as stated... I don't know if I should multiply the perpindicular weights by the object lengths... like the rod is 4 meters or so, do I multiply that by 5000?
 
  • #6
OlderDan
Science Advisor
Homework Helper
3,021
2
Inv.e[ln(t*r)]=t said:
God, I took twenty minutes to prepare this, and then it logged me out when I tried to post it, and now I have to retype the whole damn thing. So here goes:

Here is the diagram: http://photobucket.com/albums/v517/shirozu/?action=view¤t=sharkproblem.jpg

Since its almost past ten, and I'm losing my coherency :eek: , I will make this really easy.

That diagram and problem requires me to find the sum of the torques. The torque that the cord performs on the rod, and the torque that the weight of the shark creates on the rod. First the shark:

I simplified the force by using a triangle. Through thinking, I eventually found that the angle inside of this triangle is 60 degrees. I am positive this is the right angle. So, I did the equation to find the torque:

cos(60)=x/10000
10000cos(60)=x
x=5000

the I came to the part where I actually find the torque, but in the equation it says to neglect the weight of the rod... do I just multiply 5000 by the lenght of the rod, which is given? If not, what do I do?

Also, finding the torque of the cord requires finding the tension. What exactly do I do here? I made the simplifying triangle, and eventually got to:

Tcos(10)=x

But what do I do? I have two variables, and the tension I don't think should be used if the length of the pole in the first torque equation is used. Its inconsistent, if you catch my drift.

Please help, I have a physics test on this tomorrow, and I have no one to ask except this forum. I would appreciate any help at all.

Thanks in advance.

Daniel
It will be convenient to call the length of the rod L, which is given in the problem as 4.00 m. The torques can be calculated about the base of the rod. For the shark the torque is clockwise and has magnitude

[tex]\tau = 10,000N L cos 60[/tex]

The tension in the rope can be broken into horizontal and vertical components. Each component will produce a counterclockwise torque. The magnitudes are

[tex]\tau = T_v L cos 60[/tex]

[tex]\tau = T_h L sin 60[/tex]

You can find the ratio of the vertical to horizontal components of tension because you are given the direction of the rope. That information is enough for you to find the tension.

At the base of the rod there will be a force that can be broken into horizontal and vertical components. The vertical component of this force, plus the vertical component of the tension must equal the weight of the shark. The horizontal component has to offset the horizontal component of the tension. I'll leave it to you to translate this verbal description of the force at the base into the equations you need to combine with those above to solve the problem.
 
  • #7
231
1
You never have to multiply the weights(or forces) with lengths in this problem. First resolve the forces horizontally. You will get one equation for tension in the wire and compression in the rod. Secondly resolve the forces vertically. This gives you another equation. So, two equations and two unknowns.
 
  • #8
FredGarvin
Science Advisor
5,066
9
Like Quark said, you have to simply resolve the 3 main forces into the X and Y components.

The three main forces are the weight of the shark, the tension and the reaction force in the pole. The shark's weight obviously acts in the negative Y direction. All you have to do is resolve the tension and the reaction into their X and Y components. Then when you apply the conditions of static equilibrium, i.e. [tex]\Sigma F_x = 0[/tex] and [tex]\Sigma F_y = 0[/tex] you will end up with 2 equations and two unknowns.
 
  • #9
OlderDan
Science Advisor
Homework Helper
3,021
2
quark said:
You never have to multiply the weights(or forces) with lengths in this problem. First resolve the forces horizontally. You will get one equation for tension in the wire and compression in the rod. Secondly resolve the forces vertically. This gives you another equation. So, two equations and two unknowns.
You are right of course. It is also true that you don't have to find the force acting on the rod to find the tension in the rope. It may not have been essential to approach the problem using torques, but there is nothing wrong with doing it either. Since that was the approach taken by the OP, who seemed to be having trouble figuring out how to calculate the torques, I thought it worthwhile to follow that line of thought.

I should have taken the time to acknowledge that your earlier comment was correct that the torque calculations are not necessary. I apologize for that oversight.
 

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