|
i need help in physics
|
View this Thread in Original format
| stk |
I have no idea how to solve this
A sphere of mass 3.0x10^-4 is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes an angle of 32degrees. with the vertical at rest.
(a) find the magnitude of that push
(b) find the tension in the chord (what does tension mean?)
I have no idea how to do this question either
A 100 kg crate is pushed at constant speed up the frictionless 30degree ramp.
(A) what magnitude of horizontal force F is reqired?
(B) What force is exerted by the ramp on the crate?
thanks
a |
|
|
| Photo_bot_2k1 |
the answer to both questions is: whothecaresdoyourowninghw
:D |
|
|
| Resnick |
| hahahah dont know what tension is ..HAHAHAHAHAHAHAHAHAHAHAHAHAAHAHHAAHAHAHAHAHAHhA u might as well get drunk off ur ass an at least do something productive :D |
|
|
| TrAnCeAkI |
| i bets all the ppl now are gonna sign up to ta to get their homework done THANKS TO ACID JUNKIE!!!! :P:nervous: |
|
|
| astroboy |
Dude this isn't even a challenging problem. It's basic high-school physics. How many phys periods did you wag anyway?
Answering them isn't interesting since they're basic problems. Explaining the theory to you would just involve rattling off wat you've already got in your textbook. Just read your textbook man, it'll take you an hour max to figure it out. |
|
|
| Mosaic |
review your notes on vector components
Draw a free body diagram to help you
If you need more help, ask, because I can answer all those questions |
|
|
| EriK_V |
1. force diagram
2. sum forces
3. don't forget your sin and cos |
|
|
| raek |
oh my god
i havent done physics in 5 years and i could pull that one out of my ass |
|
|
| getfoul |
| quote: | Originally posted by stk
I have no idea how to do this question either
A 100 kg crate is pushed at constant speed up the frictionless 30degree ramp.
(A) what magnitude of horizontal force F is reqired?
(B) What force is exerted by the ramp on the crate?
thanks
a |
A. the only force ua re fighting is gravity, due to the ramp being frictionless. if u payed attention u should be able to figure it out from there. if not ask the teacher for alil explanation.
b. this one is like a race track in a way, in the corners u have banking and the track pushed the car in a horizontal and vertical direction not just horizontal therefor being harder for it to slide out, so you are fighting horizontal and vertical forces that the crate exerts. |
|
|
| EriK_V |
| leave the person alone if he is asking for help. No one needs people saying "i could do this easily" or "this is easier than jenna jameson". If you aren't going to help, then don't respond/ make the person feel bad. |
|
|
| DrUg_Tit0 |
1) Let's call the force exerted by the breeze Fb, the gravity force Fg, and the total force upon the object Ft. Now, we have a picture that looks like this:
code:
\
\
\
\
\
\
----->O
Fb |\
| \Ft
Fg|32\
| \
V V
Now, after rearranging the force vectors, you get a picture like this one:
code:
-------->
\a Fb |
\ |
\Ft |Fg
\ |
\ |
\ |
\ |
\|
VV
Fg=m*g=2.94e-3N
a=90-32=58 degrees
Ft/sin90=Fg/sina=3.47e-3N
Fb=sqrt((Ft)^2-(Fg)^2)=1.84e-3N
The tension in the chord is equal to the force exerted upon the chord, Ft.
2)Let's call the pulling force Fp and the gravity force Fg.
code:
/|
/ |
/ \/ Fp |
\|/ ---> |
/|Fg |
/ | |
/_30_V_______|
Now, split both of those foces into two components, one being perpendicular to the surface, while the other one being parallel. We need only the parallel components here, which are respectively equal to Fg*cos60 and Fp*cos30. Let's mark them Fg' and Fp'
code:
^
/
Fp' /
/
----->Fp
/|
Fg'/a|Fg
/ |
V V
a=90-30=60 degrees.
Now, for the object to be pushed at a constant speed, Fg' must be equal to Fp'.
Fg*cos60=Fp*cos30
Fp=Fg*cos60/cos30
(Fg=m*g)
Fp=566N
The force exerted upon the surface is the combined sum of the perpendicular forces mentioned earlier but neglected. Now, I guess you can figure out the angles for yourself, so I'll just skip the drawing this time. PM me if somehow you don't manage to do it. Anyway, let's mark those forces as Fp'' and Fg'', while the ramp force is Fr.
Fg''=Fg*cos30
Fp''=Fp*cos60
Fr=sqrt((Fg'')^2+(Fp'')^2)=895N
There, I did it in a hurry and my mechanics are a bit rusty but I hope that helps. |
|
|
| DJ Mikey Mike |
:haha: Seriously, end it all now mate. Reach for those razors ;)
If your having trouble with that you should probably have words with your teacher |
|
|
|
|
