Monday, 22 October 2007 - 8:51 PM EDT

Does anyone know how to do the "Torque on a out-streched Arm" prolem? 

Tuesday, 23 October 2007 - 12:32 AM EDT

Torque on an Outstretched Arm:

a) add the distance from the elbow to the wrist + the center of weight and multipy by the weight given to you in the problem and your answer should be in m*N

 b) add the distance between the shoulder and elbow + the elbow to the wrist + the cw and multiply that by the weight and your answer again is in m*N

 Does anyone know how to do the crane problem?

Tuesday, 23 October 2007 - 10:19 AM EDT

Does anyone have the extra problems that Dr. Lind gave out on Monday?

Tuesday, 23 October 2007 - 11:13 AM EDT

does anyone know how to do the arm strength problem?

Tuesday, 23 October 2007 - 12:40 PM EDT

Arm Strength:

Fm = Dmg/ dsin(theta)

Anyone know how to do the Crane and Hanging Mass problems?

 I was told that this is how you do parts a and b, but it didn't work for me, and the person that gave me the equations kept getting the same answer that I did, but it worked for them.

Crane part a:
you need to solve for T(tension)
so do T*sinB - mg*1/2*cos(theta) - m1g*cos(theta)=0 and solve for T
B= theta-alpha

Crane part b:
do cos(alpha)=Tx/your answer for part a
you need to solve for Tx (which is the x component of the tension in the cable)

Tuesday, 23 October 2007 - 1:18 PM EDT

The hanging mass problem is a bitch.  Here goes...

First find the tension (T1) between the knot and m1.  This comes from solving the opposing force (F friction) which comes from multiplying FN by your coefficient of static friction.  Once you get this, tension on the x-componet equals Tx=cos*Theta*T1.  Then it follows that the tension on your y-componet is equal to Ty=sin*theta*m2*g.  Here comes the weird part.... I dont know why, but Ty=Tx, something about balancing out.  set the 2 equations equal to each other and you eventually get tan*theta=(m2*g)/(T1)....remember that t1 is the tension between the knot and m1.   Then finally solve for m2 and you have it.  I left my answer in newtons and it was 95.95 N if you want a reference.  Hope this helps, this problem took me forever. 

Tuesday, 23 October 2007 - 2:22 PM EDT

Does anyone know how to do the balance on a seesaw and diving board ones? Thanks!

Tuesday, 23 October 2007 - 3:08 PM EDT

Does anyone know how to do the crane part c? I got the first two but cannot figure out the third part! Thanks!

Tuesday, 23 October 2007 - 3:10 PM EDT

Diving Board:

F=ma (mass of the diver x 9.8N/kg)

T=Fr so r=T/F ... so you plug in T that was given in the problem and divide that by the answer you got for the force above.

Does anyone know how to do part c of the crane problem? 

Tuesday, 23 October 2007 - 3:16 PM EDT

On Diving Board
The weight is given in kg so multiply by 9.8 to get Newtons....then divide by the number of actual newtons they give you and that is your answer in meters

Tuesday, 23 October 2007 - 3:18 PM EDT

Anybody have the horizontal bar problem?

Tuesday, 23 October 2007 - 3:20 PM EDT

Figured out part c to the crane problem

Fy= Tsin(alpha) +mg +M1g

T being the answer you got for part (a)

Tuesday, 23 October 2007 - 3:20 PM EDT

Hanging mass final formual  is tan theta=m2/t1

there shouldnt be g (gravity)  otherwise that is great thanks!

Tuesday, 23 October 2007 - 3:27 PM EDT

Does anyone have the horizontal bar or balance on seesaw?

Tuesday, 23 October 2007 - 4:52 PM EDT

I have everything but horizontal bar part a... i can help with anything else... does anybody have part a?

Tuesday, 23 October 2007 - 5:21 PM EDT

Horizontal bar:
A: torque(A) = -xMg- (L/2)mg + T(max)Lsin(theta) = 0
rearrange: x = [ TLsin(theta) - (L/2)mg ] / Mg
you dont actually multiply by g because the givens are already in Newtons.

B: sum of Fx = Fax - Tcos(theta)

C: sum of Fy = Fay - mg - Mg + Tsin(theta)

Tuesday, 23 October 2007 - 5:26 PM EDT

balance on a see-saw:

Tpivot = x1m1g- x2m2g - ym3g = 0

 x1 and x2 are each half of the L

Tuesday, 23 October 2007 - 5:28 PM EDT

Lisa can you please explain how to do part A better? I am so lost!

Tuesday, 23 October 2007 - 5:30 PM EDT

Nevermind.....I figured it out! Thanks

Tuesday, 23 October 2007 - 6:54 PM EDT

Lisa can you please explain how to do part B and C of the horizontal bar better, I am confused on what Fay and Fax are!

 Also for the seesaw, what is the y value or T pivot?

Tuesday, 23 October 2007 - 7:22 PM EDT

Yeah I am really lost on part B and C of the horizontal bar...any help would be awesome  

Tuesday, 23 October 2007 - 7:28 PM EDT

Does anyone know how to do Leg Cast and Arm Strength??? I have the others =)

Tuesday, 23 October 2007 - 7:39 PM EDT

Horizontal bar:

Fax and Fay are what you are trying to solve for. (I should have included earlier that the sum of Fx is zero, and same for Fy.) so the equations rearrange to become Fax = Tcos(theta) and Fay = mg + Mg - Tsin(theta).  Remember that g is included in the given values of m, so you don't need to multiply by it.

Tuesday, 23 October 2007 - 7:42 PM EDT

seesaw:

Tpivot is the midpoint on the seesaw, where the wood is balanced on the rock. this is the point where the measurements are taken from. Y is the unknown, the distance the 3rd boy needs to be from Tpivot.

I hope this helps! 

Tuesday, 23 October 2007 - 7:53 PM EDT

leg cast:

F2*R2 - F1*R1 = 0  . F2 is your unknown (and remember F = mg). R2 is the distance from hip to sling, and R1 is hip to CG. the mass is given for F1, so just remember to multiply by g. This problem is the same as 9-3 in the book, and it was an assigned problem so the solution is online. if my explanation doesnt make sense, it might help to look at that.

arm strength:

d*Fmsin(alpha) - D*mg .  you are solving for Fm so it rearranges to: Fm = (mg/sin(alpha))(D/d) 

Tuesday, 23 October 2007 - 8:07 PM EDT

I know it's a really easy one but I must be doing something wrong b/c I can't figure out how to do the Leg Cast-Torque problem.

Does anyone know how to do this one?

Tuesday, 23 October 2007 - 8:12 PM EDT

could someone further explain the seasaw, i'm having difficulty. what are x1 and x2?

Tuesday, 23 October 2007 - 8:18 PM EDT

hey what are the units for the leg cast???

Tuesday, 23 October 2007 - 8:19 PM EDT

Nevermind I just saw the comment right above mine. Sorry I'm dumb.

Tuesday, 23 October 2007 - 8:45 PM EDT

the girl in the green shirt smells

Tuesday, 23 October 2007 - 8:53 PM EDT

Does anyone know the problems that are supposed to be on the test or might help with the test?

Tuesday, 23 October 2007 - 9:05 PM EDT

Leg cast units are in kg

 seesaw: x1 and x2 are the distances of the boys from the center. both are half the total distance L.

Tuesday, 23 October 2007 - 9:11 PM EDT

Disregard the last post I sent.

Tuesday, 23 October 2007 - 9:18 PM EDT

Hey can anyone help me with the Crane problem please?

Tuesday, 23 October 2007 - 10:54 PM EDT

For Crane part C:

FyH= Mg+Mg+T sin alpha

Tuesday, 23 October 2007 - 11:39 PM EDT

can anyone help with crane part a

Wednesday, 24 October 2007 - 12:59 AM EDT

could someone please tell me what dr. lind said to study?

Friday, 26 October 2007 - 11:36 AM EDT

Hey I've started some of capa 9 already.. i have the conceptual done, part of the scuba diving and i have equations for the tire pressure but am not quite getting it

Saturday, 27 October 2007 - 12:09 AM EDT

Hey, I have the first four and part a of tire pressure.... could you help me with the scuba?nlet me know