How Can I Share a Scanned Document for Help with Torque Physics Problems?

[laker_gurl3]

Physics help please :( Torque Questions...

umm okay i have 4 physics questions that i was hoping someone could help me through it..but it's on a sheet of paper and i scanned it through the computer..i don't know if there's anyway i can put the question on here? or if someone is willing to help me, it deals with torque..if you want, can you please private message me with ur email address and i'll send you the file personally..lemmi know if you private msged me also, so i can check :P
thanks a lot you guys..
it's part of a lab and I am loost 'cuz i missed a couple days :(

 

[quasar987]

there's an option when you post a message that let's you upload a file.

See "Additional option" --> "Attach Files" Right under the emoticons and "submit reply" etc buttons

 

[laker_gurl3]

ahh ijust tried to upload it that way..it says my file is too big though..my file is 238kb...

 

[laker_gurl3]

okaay so i made it smaller! so can anyone help guide me thru it...

 

[quasar987]

Note that they are currently "pending approval". I'm not sure what this means, but other people won't have acess to it until they get approved. (By who?)

 

[laker_gurl3]

hmm...so you can't check it out yet? because i can..lol maybe it's because i put it up though...ahhh!

 

[laker_gurl3]

can you guys see the pics yet? if not..can someone msg me and i'll email it..cuz i need it by monday.. :( thanks you guys

 

[whozum]

Its still pending. Itll be up by tomorrow since youve apparently gone to bed already.

 

[laker_gurl3]

hmm so how is it now?

 

[whozum]

For problem 1 (first picture),

you have a bar balanced at the center, and a torque applied on the right side of force 12 N at a distance of 80cm. Calculate the torque about the axis ofrotation. Since the bar would rotate about the triangle in the center, your radius is 80cm.

Now the question is asking, what torque when applied at 30cm to the left of the axis of rotation will keep the bar balanced?

remember the bar will balance if both torques are equal:

$$\tau_r = \tau_l$$

 

[whozum]

No. 2 is the pretty much the same thing

3 is a bit tougher, you want all the torques to sum up to zero, so equilibrium can be achieved. I would add up all the torque vectors starting fom the left, and usingthe left end (0cm) as your axis of rotation. Add allthe torques together, and then find the torque needed on the 'x' vector to make it 0.

 

[laker_gurl3]

okay..so for 1a.) asking for the magnitude of the force,i did

.3m x 12N = 3.6 N/M
3.6N/M=.2m(X)
X= 18...im not sure what's the unit for this though? N?

 

[laker_gurl3]

I'm still having problems with 1b though..

 

[whozum]

okay..so for 1a.) asking for the magnitude of the force,i did

.3m x 12N = 3.6 N/M
3.6N/M=.2m(X)
X= 18...im not sure what's the unit for this though? N?

The units for torque are Newton meters, (Nm).

You are applying a force of 18N at .2m away, so the units would be N.

 

[whozum]

1b is just asking the same question if the force was applied at 10cm point. Also I just noticed, in your solution to 1b, you want the distance from the fulcrum (the triangle), not the distance from the edge. The '30cm' labelled is the distance from the edge, and the fulcrum is at 50cm, so the distance is 20cm. This will give you a different answer for 1a. Always pay attention to things like this.

 

[laker_gurl3]

okay..so for 1a, in what i did, i put in .2m where i put .3m before..and i got an answer of 12N, is that correct?

And for 1b, is the answer 6N?
soo sorry, i just have so much trouble understanding this.. :(

 

[whozum]

The torque on the right side is 0.3m * 12N = 3.6N

The torque on the left side is 0.2m * x N

For equilibrium they will be equal:

3.6Nm = 0.2x Nm

Divide both sides by 0.2m (equal to multiplying by 5)

18N = x N

Sorry you did the first one correctly. I don't know how I thought otherwise.

1b)

Torque on the right = 3.6Nm
Torque on the left = .4m*xN

3.6Nm = .4x Nm

Divide by .4m

9N = x N

 

[laker_gurl3]

okay..ahh thanks so much for that! I am going to try number 2 now..ahh! lol

 

[laker_gurl3]

for number 2, i got 1m...is that right?

Torque on the right:
2.5m * 20N = 50Nm
to be equilibrium...
X50N = 50Nm
divide by 50 N and i get 1m...

 

[whozum]

You got it.

 

[laker_gurl3]

okay thanks a bunch! I am working on number 3..i added up the torques together and left out the 40cm at the end..(not sure if that's what I am supposed to do?) well i got a totall of 11.5Nm
Where do i go from there?

 

[laker_gurl3]

i did some calculations, and i got an answer of 28.75N for 3a.)
after adding up all the torques i got 11.5Nm = .4mN
divided it by .4 and got my answer...

 

[whozum]

For static equilibrium you want all the forces and torques to equal 0.

You have a total force of 40N downwards, and 10N upwards, so you know F needs to be pointing upwards and have magnitude 30N, that way the net force is 0.

Notice however the torques arent even, so you need to find a distance 'x' to where the torques will be even, given that F = 30N.

 

[whozum]

The torque of the known forces is:

.1m*10N + .1m*25N = 3.5N in the clockwise direction.
Find the required torque to make this 0. Notice that since the 15N is applied at x = 0, it exerts NO torque.

 

[laker_gurl3]

hmmm okay..i now understand in order to make it equilibrium the force needs to point up and have a magnitude of 30N...

so should i find the distance x first and then find the force of F? meaning should i do part b first?

 

[whozum]

Nope, you can't do (b) without doing (a). We solved (a) to be 30N, so work (b) with F = 30N, I don't think there's another way.

 

[laker_gurl3]

okay so i got an answer of .28m..?

 

[whozum]

3.5Nm in the clockwise direction

so we need 3.5N in the counterclockwise direction.

3.5Nm = 30N* x m

Divide by 30N.

Important: Always make sure to include a direction with torques, it is possible in some instances that you want x to be negative, if you need a torque in the opposite direction.

0.116m = x m

 

[laker_gurl3]

ahh,, thanks soo much, you've helped alot! i have one more question, but by the time i upload and its approved, i don't think i have enough time, so I'm going to try it and pray it's right..haha, thanks soooooo much for this