Moments Question

Original post by joyoustele

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Reply 2

Original post by RDKGames

Attachment not found.

It works now. But how can i make it so the image stays on the page, instead of just providing a link?

Reply 3

Original post by joyoustele

It works now. But how can i make it so the image stays on the page, instead of just providing a link?

Haven't encountered this type of problem before, but just to check before I guide you to the answer, what's the answer to the question?

Reply 4

Original post by RDKGames

Haven't encountered this type of problem before, but just to check before I guide you to the answer, what's the answer to the question?

The Answer: Tension =7.28N

:frown:

I've spent so long on this

Reply 5

Original post by joyoustele

The Answer: Tension =7.28N

:frown:

I've spent so long on this

Oh fair enough, OK then.

So, on that diagram, the CoM of the hoop in the diagram is dead in the middle of it. Mark that point and draw a force of 20N going down (for obvious reasons). Then, you have

F(t)

going horizontally from the bottom of the hoop in the diagram. (I chose

F(t)

since F is a gradually increasing force as a formality, but since you are looking at an instant of time, just say that force is

F

). Then you want to take moments about the pin, and considering those two forces, proceed to solve for F. Call the radius

r

, you'll quick notice that it cancels out so it isn't really a big thing here as to what it is.

Reply 6

Original post by RDKGames

Oh fair enough, OK then.

So, on that diagram, the CoM of the hoop in the diagram is dead in the middle of it. Mark that point and draw a force of 20N going down (for obvious reasons). Then, you have

F(t)

going horizontally from the bottom of the hoop in the diagram. (I chose

F(t)

since F is a gradually increasing force as a formality, but since you are looking at an instant of time, just say that force is

F

). Then you want to take moments about the pin, and considering those two forces, proceed to solve for F. Call the radius

r

, you'll quick notice that it cancels out so it isn't really a big thing here as to what it is.

where did i go wrong?

F(2r)=20r ?

Reply 7

Original post by joyoustele

where did i go wrong?

F(2r)=20r ?

Well I dunno how you ended up with that equation so I dunno where you went wrong.

The perp distance to F from the pin is not 2r since the pin isn't all the way at the top of the circle when at the angle.

Reply 8

Original post by RDKGames

Oops.....Rookie mistake...

Reply 9

Original post by RDKGames

I cant find the Distance. Thats the main bit i have struggled with

Reply 10

Original post by joyoustele

I cant find the Distance. Thats the main bit i have struggled with

Perp distance to F is the black line + the red dashed line. What's the length of the red dashed line...?

[NOTE: This is a skill you'll need to use if you'll study vertical circular motion - to find this sort of length in a circle]

(edited 6 years ago)

Reply 11

Original post by RDKGames

Perp distance to F is the black line + the red dashed line. What's the length of the red dashed line...?

Note: This is a skill you'll need to use if you'll study vertical circular motion.

rcos40?

Reply 12

Original post by joyoustele

rcos40?

Should be good to finish it off now then.

Reply 13

Original post by RDKGames

Should be good to finish it off now then.

F(r+rcos(40)=20 \sqrt (r^2-r^2cos40)

F=\dfrac{ 20\sqrt (r^2-r^2cos40) }{r+rcos40}

.....have i gone wrong? ..

(edited 6 years ago)

Reply 14

Original post by joyoustele

F(r+rcos(40)=20g(\sqrt (r^2-r^2cos40)

F=\dfrac{ 20g\sqrt (r^2-r^2cos40) }{r+rcos40}

.....have i gone wrong? ..

Yeah.

First line:
1. LHS is fine though missing a bracket at the end
2. Where did

g

come from...?
3. The perp. distance to the weight is just, yknow,

r\sin(40)

maybe...??

Reply 15