/wsr/ - Worksafe Requests » Thread #43711

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Anonymous Thu 28 Jan 2016 01:48:37 No.43711 View View Reply Original Report

Quoted By: >>43713 >>43715 >>43717 >>43719 >>43723 >>43724 >>43733 >>43738 >>43743

Hey, /wsr/
Retard here. Can some of you help me with my Physics homework? I have a teacher I can't learn from.

#1: What horizontal force is necessary to accelerate a 215 kg trunk at 1.85 m/s^2 along a flat surface with a coefficient of kinetic friction of 0.32?

#2: A helicopter with a mass of 1450 kg is moving with an acceleration of 1.27 m/s^2 upwards. What is the magnitude of the lift force provided by the propeller?

Anonymous

View Same Google ImgOps iqdb SauceNAO 1441755095072.jpg, 792KiB, 2048x1356

Anonymous Thu 28 Jan 2016 01:50:08 No.43713 Report

Quoted By:

>>43711
Bumping with qt

Anonymous

View Same Google ImgOps iqdb SauceNAO 1420393260003.jpg, 88KiB, 500x552

Anonymous Thu 28 Jan 2016 01:51:15 No.43715 Report

Quoted By:

>>43711
Plz halp.

Anonymous

View Same Google ImgOps iqdb SauceNAO 1443632768918.jpg, 26KiB, 241x274

Anonymous Thu 28 Jan 2016 01:52:53 No.43717 Report

Quoted By:

>>43711

Anonymous

Anonymous Thu 28 Jan 2016 01:54:57 No.43719 Report

Quoted By:

>>43711
Plz /wsr/,
You're my last hope.

Anonymous

Anonymous Thu 28 Jan 2016 01:58:36 No.43723 Report

Quoted By:

>>43711
Plzzzzzzzzzzzzzzzzzzzzzzzz

Anonymous

Anonymous Thu 28 Jan 2016 01:58:57 No.43724 Report

Quoted By: >>43725

>>43711
You want the answer or an explanation?

Anonymous

Anonymous Thu 28 Jan 2016 01:59:37 No.43725 Report

Quoted By: >>43734

>>43724
Both please.

Anonymous

Anonymous Thu 28 Jan 2016 02:00:30 No.43726 Report

Quoted By:

bump

Anonymous

Anonymous Thu 28 Jan 2016 02:05:41 No.43733 Report

Quoted By: >>43735 >>43752

>>43711
1. Ftotal=ma+Ffriction
Ffriction=0.32*2150(the normal force of the trunk)
Therefore F=215*1.85+0.32*2150=1085.75N

2. F=m(a+g) because it has to counteract gravity and provide vertical acceleration, so it's 215*(1.27+10(or 9.8 who cares))=16341.5N

Anonymous

Anonymous Thu 28 Jan 2016 02:06:30 No.43734 Report

Quoted By: >>43737

>>43725
Do you even know how to draw the free-body diagrams?

Anonymous

Anonymous Thu 28 Jan 2016 02:06:36 No.43735 Report

Quoted By:

>>43733
Thank you so much. I love you, anon.

Anonymous

Anonymous Thu 28 Jan 2016 02:08:03 No.43737 Report

Quoted By:

>>43734
Yeah, but past that point I don't get any of this shit. Almost everybody in my class is struggling. I think it might be the teacher.

Anonymous

Anonymous Thu 28 Jan 2016 02:08:09 No.43738 Report

Quoted By:

>>43711
>A helicopter with a mass of 1450 kg is moving with an acceleration of 1.27 m/s^2 upwards. What is the magnitude of the lift force provided by the propeller?
So as we learned, force=mass*acceleration.

Literally all we do is plug the values in, and out comes the answer:

f=m*a
=1.27*1450
=1841.5N <-- remember the units!

EXCEPT!

The question doesn't actually state this (bad Physics question!), but we should probably assume that this helicopter is on Earth.

If it is, the lift force has to hold the helicopter up as well as accelerate it upwards, so we need to calculate the weight of the helicopter and add it to the acceleration force to get the total lift force.

weight = mass * coefficient of gravity
W = m * g

On Earth, g = 9.81, so

W = 1450 * 9.81 = 14224.5N

Total lift = 1841.5 + 14224.5 = 16066N

Anonymous

Anonymous Thu 28 Jan 2016 02:11:34 No.43743 Report

Quoted By: >>43745

>>43711
F_x=F-f=ma_x
F_y=N-mg=0 (it doesn't move in the y-direction)
F=ma_x+f but f=u_k*N and N=mg so
F=m_ax+u_k*N

F_y=F-mg=ma_y
F=ma_y+mg

Anonymous

View Same Google ImgOps iqdb SauceNAO 1424126784437.gif, 2MiB, 200x150

Anonymous Thu 28 Jan 2016 02:12:25 No.43745 Report

Quoted By: >>43754

>>43743
WHAT THE FUCK IS THAT?!

Anonymous

Anonymous Thu 28 Jan 2016 02:15:01 No.43752 Report

Quoted By: >>43757

>>43733
Wait, how is the mass 215 for #2?

Anonymous

Anonymous Thu 28 Jan 2016 02:16:23 No.43754 Report

Quoted By: >>43763 >>43770

>>43745
The equations from the free-body diagrams. The underscores show the axis it is related to so F_x is the sum of the forces in the x-direction and a_x is the acceleration in the x-direction. Also are you under 18? This stuff is sophomore level at best.

Anonymous

Anonymous Thu 28 Jan 2016 02:17:24 No.43757 Report

Quoted By:

>>43752
He made a mistake. Just replace it with the actual mass.

Anonymous

Anonymous Thu 28 Jan 2016 02:20:47 No.43763 Report

Quoted By: >>43768

>>43754
No, I'm just retarded and have a teacher that goes too fast for me.

Anonymous

Anonymous Thu 28 Jan 2016 02:23:21 No.43768 Report

Quoted By: >>43773

>>43763
Did you actually learn anything from this thread?

Anonymous

Anonymous Thu 28 Jan 2016 02:24:11 No.43770 Report

Quoted By: >>43782

>>43754
Maybe that's how you write it in whatever crazy place you come from, but there's no way you wrote that as anything other than a dick-swinging exercise.

You've got someone saying "please explain this physics problem to me, my teacher isn't making sense", and you spout a load of shite that looks like Perl at him.

That's not helping, that's just showing off.

Also you got fristed, so you couldn't even belittle a college kid without fucking it up.

Anonymous

Anonymous Thu 28 Jan 2016 02:24:20 No.43773 Report

Quoted By:

>>43768
Not yet. I'm hoping I can learn from the explanations.

Anonymous

Anonymous Thu 28 Jan 2016 02:32:03 No.43782 Report

Quoted By:

>>43770
Sorry if I somehow offended you.

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