Statics: System in equilibrium - determine height (h)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
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: [itex]\rightarrow[/itex])
Td sin(θ) + Tcsin(α) - W= 0 (along y-axis: [itex]\uparrow[/itex])

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:
[itex]\frac{sin180-α-θ}{a}[/itex] = [itex]\frac{sin(α)}{r}[/itex]

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).