# Forces, blocks, cords, and springs

1. Sep 18, 2006

### mybsaccownt

sorry, couldn't think of a better title

anyways, i'm somewhat stuck on 2 questions i have in my homework assignment

first 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.bmp

alright, 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 part

forces 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?

and

Determine 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^2

http://www.mydatabus.com/2z/lnubb.pn/mybsaccownt/springsandmass.bmp

for 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 simple

alright...that's where i am :)

any help would be appreciated

2. Sep 19, 2006

### nazzard

Hello mybsaccownt,

be careful whenever you try to compare forces. As vectors they have both: magnitude and direction. Your approach to look at the components of the acting forces is promising. By comparing the magnitudes, the 2 equations you'll get in your first problem will be:

$$\sin{\phi}+\frac{1}{4}\sin{\theta}=1 \,\,\,(1)\\$$ and
$$\cos{\phi}=\frac{1}{4}\cos{\theta} \,\,\,(2)$$

The following step in your attempt was wrong:

$$\arccos(x*y)\neq\arccos(x)*\arccos(y)$$

To solve the problem I'd use the following equation:

$$\sin^2{x}+\cos^2{x}=1 \,\,\,(3)$$

Starting with rearranging and squaring (1):

$$\sin^2{\phi}=(1-\frac{1}{4}\sin{\theta})^2 \,\,\,(1')$$

squaring (2) and using (3) to get:

$$1-\sin^2{\phi}=\frac{1}{16}\cos^2{\theta} \,\,\,(2')$$

Using (1') in (2'),...

Regards,

nazzard

Last edited: Sep 19, 2006
3. Sep 20, 2006

### andrevdh

For the first one:

Using Newton's second law in component form we have that

$$\Sigma F_Y = 0$$

Wsin(fi) + 0.25Wsin(theta) = - W

Should actually be

Wsin(fi) + 0.25Wsin(theta) - W = 0

But I would rather use the fact that the forces form a closed triangle. Two of the legs are the same length and one is a quarter of the length of the other. So one get an isosceles triangle with the two legs of length 1 and the base of length a quarter. You just need to figure out where the angles should go in such a triangle.

4. Sep 20, 2006

### mybsaccownt

thanks a lot nazzard! :-)

i got it, theta = arcsin(1/8)

my trig skills need some work, hehe

andrevdh, blocks b and c will lift block a

that's why their vertical components will be equal to a, but opposite in direction, in order to achieve equilibrium

so, equal but opposite, hence vertical components of b, c = -Wa

and Wa is just W

so

sum of vertical components of b and c = - W

also, i'm not sure i understand what you mean by the forces forming a closed triangle

i'm just breaking the forces into components and applying the equations of equilibrium, it's a good method and it's simple, i'm not saying your method wouldn't work, but it seems like i should just stick to what i know

i think the only trouble i had was with my trig skills (i didn't remember the identity that was needed to solve for theta)

the TA mentioned it today in tutorial, but i still would have been stuck for a while without some help

i appreciate your input, thanks for the help, guys

Last edited: Sep 20, 2006
5. Sep 21, 2006

### andrevdh

Yes their magnitudes need to be the same in order to cancel. In Newton's equation direction is taken into account. That is from

$$\Sigma F_y = 0$$

which is the same condition as your remark that the forces in the y-direction need to balance - their sum need to be zero. So from this we have that

$$W_{BY} + W_{CY} - W_{AY} = 0$$

this is the scalar component form of N2 = 0 (taking up as positive). Which then comes to

$$W_{BY} + W_{CY} = W_{AY}$$

6. Sep 21, 2006

### mybsaccownt

oh...I see what you mean

alright, thanks i got it