Equilibrium Homework: Solving Simple Force Problem with Multiple Solutions

[ToeNugget]

Homework Statement

Hi everyone, I've been stuck on this relatively very simple problem for quite a while now and even after solving I'm still getting two answers while the book only lists one.

Question: If block D weighs 300 lb and block B weighs 275 lb, determine the required weight of block C and the angle (theta) for equilibrium.

So I did the usual since it's in equilibrium then F_x = 0 and F_y = 0.

So, F_c(cos30) - 275(cos$\theta$) = 0
and F_y(sin30) + 275(sin$\theta$) - 300 = 0

Then after finding cos$\theta$ in terms of F_c and substituting, you get this equation: 158.7cos$\theta$ + 275sin$\theta$ = 300

The answer listed is: $\theta$ = 40.9 and F_c = 240lb

Can anyone please tell me how to solve the equation above? I tried using the trig identity (square of cos plus square of sin = 1) but I got two answers, one was similar to the book ($\theta$ = 40.9, and one was $\theta$ 79). Thing is, when I use the second value I got (79) in the equations above, it works for all.

I'd appreciate some clarification on this problem! Thanks.

 

[ehild]

Both angles are correct.

ehild

 

[Rayquesto]

first realize that (275lbs)/tantheta= C/tan30 since the weights are all in the y direction right?

where C= 275lbs(tan30)/tantheta and theta=inverse tan ((275lbs)(tan30)/C)

and so, you must find c first.

c would be found using forces in the y direction such that sum of all forces is zero since equalibrium occurs.

so, 300lbs + 275lbs + D=0N

D=575N

what do you say to this?

 

[Rayquesto]

sorry C=575N

 

[ToeNugget]

Rayquesto, I'm sorry I don't understand exactly what you're trying to do or how you're doing and why we need to find C first? Also, wouldn't the force in the y direction need to take into account the angle the string is making with the horizontal?

ehild, so then there are also two possible weights as well, correct?

 

[ehild]

ehild, so then there are also two possible weights as well, correct?

Yes, of course. There is less weight for the greater angle.

ehild

 

[grzz]

Yes the two answers for the angle are correct and the less weight mentioned by ehild is 59.9lb.

 

[Rayquesto]

yes, but ok, well I am confused too then, because wouldn't the weights already be acting in the y direction? I mean, to me, your calculuation would be correct if the forces were affected such that it can be treated as a hypotenuse of the force, but It looks like the weights can only be thought of as a Y leg.

 

[grzz]

I cannot follow what Rayquesto is saying. Can he be more specific?
The weights are acting in the Y-direction (i.e. vertically).But here we are examining the equilibrium at point A and the equations in the first post are correct.