Statics: System in equilibrium - determine height (h)

Click For Summary
SUMMARY

The forum discussion revolves around solving a statics problem involving a system in equilibrium, specifically determining the height (h) in a given setup. The problem provides parameters such as r = 0.575 m, a = 1.05 m, W = 4.905 N, and Wc = 4.905 N. The user initially derived two equations based on equilibrium conditions but struggled to find a third equation to solve for three unknowns. After guidance, the user corrected their approach using a TI Voyage 200 calculator, leading to the correct values of Td = 5.06 N, α = 27.99 degrees, and θ = 31.00 degrees.

PREREQUISITES
  • Understanding of statics and equilibrium principles
  • Familiarity with free body diagrams
  • Knowledge of trigonometric functions and the sine law
  • Experience using a TI Voyage 200 calculator for solving equations
NEXT STEPS
  • Review the principles of static equilibrium in mechanical systems
  • Practice solving problems involving free body diagrams
  • Learn advanced trigonometric applications in physics
  • Explore the functionalities of the TI Voyage 200 calculator for complex equations
USEFUL FOR

Students studying physics, particularly those focusing on statics and equilibrium, as well as educators looking for practical examples of problem-solving techniques in mechanics.

gate13
Messages
3
Reaction score
0

Homework Statement



Good day to all,
I have been given a problem in statics, ore specifically a system that is in equilibrium, but a system I am having some trouble with. I have attached (in pdf) the problem's context and the free body diagram I have drawn.

The data we are give are as follows:
r = 0.575 m, a=1.05 m, W=4.905 N, Wc = 4.905 N

Homework Equations



We are asked to find h.

The Attempt at a Solution



What I have so far is as follows:

Using the fact the the hoop is in equilibrium, I obtained the two equations:

Td cos(θ) - Tccos(α) = 0 (along x-axis: \rightarrow)
Td sin(θ) + Tcsin(α) - W= 0 (along y-axis: \uparrow)

We know W and Tc and so we have two equations with three unknowns so far. Then I thought of using the problem's geometry to obtain a third equation.

Ans so using the sine law, I obtained:
\frac{sin180-α-θ}{a} = \frac{sin(α)}{r}

But when I try solving the system with the values, I get:
α=90, θ=88.17 and Td = -1* 10-81

which I am quite certain are wrong. I can't seem to pinoint the third equation to help solve the problem. Any help would be greatly appreciated.

Thank you.
 

Attachments

Physics news on Phys.org
hi gate13! :wink:

gate13 said:
… But when I try solving the system with the values, I get:
α=90, θ=88.17 and Td = -1* 10-81

your method looks ok :confused:

show us how you got those figures :smile:
 
Hello tiny-tim,

First of all, I wish to thank you for taking time to answer my question. The values I listed above were obtained using a TI Voyage 200 calculator. What I entered (in the calclulator) was the following:
solve(Td*cos(theta)-Tc*cos(alpha)=0 and Td*sin(theta)+Tc*sin(alpha)-W = 0 and r*(180-alpha-theta) = a*sin(apha), {alpha, theta, Td})
I just realized I should be entering:
solve(Td*cos(theta)-Tc*cos(alpha)=0 and Td*sin(theta)+Tc*sin(alpha)-W = 0 and r*sin(180-alpha-theta) = a*sin(apha), {alpha, theta, Td}).
Upon correcting this, I obtained:
Td = 5.06 N, alpha = 27.99 degrees, theta = 31.00 degrees.

Once more I wish to thank you for your help ( and I feel a bit embarassed with respect to my mistake).
 

Similar threads

Replies
7
Views
3K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
3
Views
3K