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mybsaccownt
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sorry, couldn't think of a better title
anyways, I'm somewhat stuck on 2 questions i have in my homework assignmentfirst one:Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC= W, WB= kW, determine the angle (theta) for equilibrium.
Given:
k = 0.25
http://www.mydatabus.com/2z/lnubb.pn/mybsaccownt/blocksandcords.bmpalright, for the first one, i thought it would be easy and i thought i had a good plan
Wa = Wc = W Wb = kW and k = 0.25
i want equilibrium so, sum of the forces in x = 0, sum of forces in y = 0
so then i break everything into components to add the x and y forces respectively
in the y the two forces going up should equal the force going down
Wsin(fi) + 0.25Wsin(theta) = - W
W cancels and i end up with
sin(fi) + 0.25sin(theta) = -1
here is where my math skills fail me, instead of trying to write theta or fi in terms of the other one (i get some ugly things), i see what i can do with the other partforces in the x, only two of them, so:
Wcos(fi) = -0.25Wcos(theta)
once again, the W cancels, and i dared to solve for one angle in terms of the other
cos(fi) = -0.25cos(theta)
fi = arccos(-0.25)*theta
fi = 104*theta
theta = fi/104 <---theta will be really really tiny, that's ok, because block b has to fight block c in the x direction and block c has 4 times the mass
anyways, this is where i pretty much got stuck,
have i done something screwy, or is it just a lack of trig knowledge?
andDetermine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s=0 when the cylinders are removed.Given:
d = 0.5m
l1 = 1.5m
l2 = 2m
l3 = 1m
k = 100N/m
g = 9.81m/s^2http://www.mydatabus.com/2z/lnubb.pn/mybsaccownt/springsandmass.bmpfor the 2nd problem
i got stuck pretty early on
i started drawing a FBD at point A, i had t1, t2, and m1
then i started second-guessing my strategy and came to the conclusion that i had no real idea what i was going to do with that
i trigged out some angles, wrote some equations of equilibrium, but...when i saw 5 unknowns, i backed down and decided i'd check if i am on the right path before taking on that beastly chore
i'll just briefly outline where i was headed
Fxa = t1cos48.59 - t2
Fya = t1sin48.59 -m1g
Fxb = t3cos48.59 - t2
Fyb = t3sin48.59 - m2g
i must admit, t2 confuses me in the FBD, i need to have it opposing t3 and t1, which means it opposes itself, hmm
i also tried something like
m1 + m2 = 2kd/g and i get the combined mass of m1 and m2 is 10.2 kg, but surely, it can't be that simplealright...that's where i am :)
any help would be appreciated
anyways, I'm somewhat stuck on 2 questions i have in my homework assignmentfirst one:Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC= W, WB= kW, determine the angle (theta) for equilibrium.
Given:
k = 0.25
http://www.mydatabus.com/2z/lnubb.pn/mybsaccownt/blocksandcords.bmpalright, for the first one, i thought it would be easy and i thought i had a good plan
Wa = Wc = W Wb = kW and k = 0.25
i want equilibrium so, sum of the forces in x = 0, sum of forces in y = 0
so then i break everything into components to add the x and y forces respectively
in the y the two forces going up should equal the force going down
Wsin(fi) + 0.25Wsin(theta) = - W
W cancels and i end up with
sin(fi) + 0.25sin(theta) = -1
here is where my math skills fail me, instead of trying to write theta or fi in terms of the other one (i get some ugly things), i see what i can do with the other partforces in the x, only two of them, so:
Wcos(fi) = -0.25Wcos(theta)
once again, the W cancels, and i dared to solve for one angle in terms of the other
cos(fi) = -0.25cos(theta)
fi = arccos(-0.25)*theta
fi = 104*theta
theta = fi/104 <---theta will be really really tiny, that's ok, because block b has to fight block c in the x direction and block c has 4 times the mass
anyways, this is where i pretty much got stuck,
have i done something screwy, or is it just a lack of trig knowledge?
andDetermine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s=0 when the cylinders are removed.Given:
d = 0.5m
l1 = 1.5m
l2 = 2m
l3 = 1m
k = 100N/m
g = 9.81m/s^2http://www.mydatabus.com/2z/lnubb.pn/mybsaccownt/springsandmass.bmpfor the 2nd problem
i got stuck pretty early on
i started drawing a FBD at point A, i had t1, t2, and m1
then i started second-guessing my strategy and came to the conclusion that i had no real idea what i was going to do with that
i trigged out some angles, wrote some equations of equilibrium, but...when i saw 5 unknowns, i backed down and decided i'd check if i am on the right path before taking on that beastly chore
i'll just briefly outline where i was headed
Fxa = t1cos48.59 - t2
Fya = t1sin48.59 -m1g
Fxb = t3cos48.59 - t2
Fyb = t3sin48.59 - m2g
i must admit, t2 confuses me in the FBD, i need to have it opposing t3 and t1, which means it opposes itself, hmm
i also tried something like
m1 + m2 = 2kd/g and i get the combined mass of m1 and m2 is 10.2 kg, but surely, it can't be that simplealright...that's where i am :)
any help would be appreciated