Statics Problem, two disks in equilibrium

In summary, the two disks each have a mass of 18kg and are attached at their centers by an elastic cord that has a stiffness of k = 1.6kN/m. The cord has a stretch of 10mm when the system is in equilibrium and an angle of θ of 30.96 degrees. The horizontal force of the left disk is equal to .6*18kg = 12.8kg and the vertical force of the right disk is equal to .6*16kg = 11.6kg. The tension in the cord is 12.8*kN and the extension is .6*(k+10mm) = 7.92mm.
  • #1
leomclaughlin
9
0

Homework Statement



http://imgur.com/yVOXkAT

The two disks each have a mass of 18kg and are attached at their centers by an elastic cord that has a stiffness of k = 1.6kN/m . Determine the stretch of the cord when the system is in equilibrium and the angle θ of the cord. (Figure 1)

Homework Equations



F_x=0
F_y=0

The Attempt at a Solution



So far I've determined that the forces in the X-direction equals zero so cos(theta)lengthk=294.34(horizontal force of the left disk)
and in the y-direction,
sin(theta)lengthk=176.58(the y force of the right disk)
I believe this to be all I need, but I'm having trouble solving two equations with a sin and cos, and even then I'm not entirely sure if my calculations are correct. Any help is appreciated.
 
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  • #2
"length" should get its own variable instead of a word (the sketch uses l).
You can solve for l*k in one equation and use this in the other. Alternatively (and quicker), divide one equation by the other.
 
  • #3
According to me finding tension in the cord would work good.Once you get the tension you can easily find the extension.
 
  • #4
well i was wrong, if i divide the equations by each other to get tan(theta)=.6, solving for theta gives me 30.96, which ironically is the lower left angle of the triangle, and its incorrect. Now I'm not sure where to go from here
 
  • #5
i was attempting to find the tension in the cord by splitting it into its x and y components, from there I could solve for length and theta, or take the x and y components and use Pythagoreans theorem to get the hypotenuse?
 
  • #6
mfb said:
"length" should get its own variable instead of a word (the sketch uses l).
You can solve for l*k in one equation and use this in the other. Alternatively (and quicker), divide one equation by the other.
The length, x, to use in the T = kx equation is the extension of the cord, which will be rather less than l. The tension needs to be calculated from the FBDs of the discs, and the extension deduced from that.
 
  • #7
Ah right, l is the rest length. Okay, that makes more sense.
 
  • #8
mfb said:
Ah right, l is the rest length. Okay, that makes more sense.
No, from the diagram I think l is the extended length, but how that's made up of rest length and extension is unknown.
 
  • #9
So is it impossible to answer because I was not given a rest length of the cord? Everyone is mentioning variable without a whole lot of focus on the problem at hand. The only horizontal force is that which the left disk provides, so the cosine of the string theta multiplied by k and stretched length l should equal the horizontal force provided by the left disk, but that is one equation and two unknowns. The other equation would be the sin of the string theta * l*k equals the vertical component of both disks I believe
 
  • #10
leomclaughlin said:
So is it impossible to answer because I was not given a rest length of the cord? Everyone is mentioning variable without a whole lot of focus on the problem at hand. The only horizontal force is that which the left disk provides, so the cosine of the string theta multiplied by k and stretched length l should equal the horizontal force provided by the left disk, but that is one equation and two unknowns. The other equation would be the sin of the string theta * l*k equals the vertical component of both disks I believe
Don't worry about the stretching of the cord for the moment. Just treat it as a string of known length under unknown tension T. There is also its unknown angle and two unknown normal forces. You can write down four equations concerning the horizontal and vertical forces at the discs. Solve to find T.
 

1. What is the definition of equilibrium in a statics problem?

Equilibrium in a statics problem refers to a state in which the net force and net torque on an object are equal to zero. This means that the object is not accelerating or rotating.

2. How do you determine the forces acting on each disk in a statics problem?

In order to determine the forces acting on each disk, you must first draw a free-body diagram for each disk. This diagram should include all the external forces acting on the disk, such as weight, friction, and normal forces. Then, using Newton's second law, you can calculate the net force and net torque on each disk.

3. What is the difference between a statics problem and a dynamics problem?

A statics problem deals with objects that are not moving or are in a state of equilibrium, while a dynamics problem deals with objects that are moving and experiencing acceleration. In a statics problem, the sum of the forces and torques is equal to zero, while in a dynamics problem, the sum of the forces and torques is equal to the mass times the acceleration.

4. How do you know if a system of two disks is in equilibrium?

A system of two disks is in equilibrium if the net force and net torque on each disk is equal to zero. This means that the forces and torques acting on each disk are balanced, and the disks will not accelerate or rotate.

5. What are the key steps to solving a statics problem with two disks?

The key steps to solving a statics problem with two disks are:
1. Identify all the external forces and torques acting on each disk.
2. Draw a free-body diagram for each disk.
3. Write out the equations for the sum of the forces and sum of the torques for each disk.
4. Solve the equations simultaneously to find the unknown forces and torques.
5. Check if the solution satisfies the condition of equilibrium (net force and net torque equal to zero).

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