/sci/ - Science & Math

[Anonymous No.3778617]

The horizontal force F holds the ball steady. Given F = 8.0 N, and theta = 10o.

>>Part One
Calculate the mass of the ball.

>>PART TWO
What is the tension of the strings?

[Homework Helper No.3778630]

1) E = mc^2
Since you don't know how high off the ground the ball is, you can't figure out its energy, meaning there isn't enough info in the problem, OP. Tell your teacher.

2. ~1.37 N

[Anonymous No.3778634]

trigonometry and x/y force components

[Anonymous No.3778643]

>> incentive... 1/??

[Anonymous No.3778648]

2/??

[Anonymous No.3778644]

USE FEMA
FeMA is the formular u warnt

[Anonymous No.3778664]

,,, 3/??

[Anonymous No.3778672]

You'll get banned for posting those, OP.

[Anonymous No.3778681]

1
fema you fuckface
f=ma
forcemassacaelreration
2
>>3778634
>>3778634
>>3778634

[Anonymous No.3778693]

>>3778681
>>3778681

OP HERE...

F = M A

F/A = M

But A = 0... so WTF???

[Anonymous No.3778701]

Damn, those are some sexy bitches.


>> Part 1
1) Write Newton's Second Law in the <span class="math">x[/spoiler] direction for the ball.
(Remember the total force is <span class="math">0[/spoiler] since the ball is stationary (<span class="math">a = 0[/spoiler])).

2) Write NSL in the <span class="math">y[/spoiler] direction for the ball.
Don't forget gravity pointing down.

Divide the two equations (to cancel the <span class="math">T[/spoiler]s, and recall that <span class="math">\frac{\sin \theta}{\cos \theta} = \tan \theta[/spoiler]. Solve for <span class="math">m[/spoiler].

>> Part 2
Referencing either of your 2 original equations, solve for T.

[Anonymous No.3778717]

>Doing kids math the board.

[Anonymous No.3778742]

>>3778701
>>3778701

Thats great! Thanks... But I'm still confused on how the A = 0 comes into play. Any suggestions?

[& !!tSqR6hj9NpI No.3778741]

As much as the tits are appreciated, I do believe you just broke two rules:

1) No homework threads
2) Work safe (As in, no 18+ content) board

The secomd being worse (punishment wise... Well, that depends really) then the first. I would highly reccomend you delete those pictures before some fag reports you, or a janitor/moderator comes across this thread.

[Anonymous No.3778752]

>>3778701
>>3778701
>>3778701
>>3778701
>>3778701

Thats great! Thanks... But I'm still confused on how the A = 0 comes into play. Any suggestions?

I'd post more tits, but they really seem frowned upon here...

[Anonymous No.3778853]

(Thisfag again >>3778701)
I hope you're still here.

>>3778752
Newton's Second Law should be (can be) interpreted in the following way:

<span class="math">\sum \vec F = m \vec a[/spoiler], or in 1 dimension:

<span class="math">\sum F_x = m a_x[/spoiler].

Where we sum over all the forces in that direction.
The interpretation is that on the left side, you write the forces that are acting on X (whatever body you are considering). And on the right side, you write the accelerations that X is experiencing.

Thus, if there are 2 forces <span class="math">\vec F_1[/spoiler] and <span class="math">\vec F_2[/spoiler] acting upon X in opposite directions, AND we know that X is not experiencing any accelerations, then:

<span class="math">F_1 - F_2 = 0[/spoiler].

Thus we can conclude that <span class="math">F_1[/spoiler] and <span class="math">F_2[/spoiler] are equal in magnitude (<span class="math">F_1 = F_2[/spoiler]), which is by all means logical.

So even when <span class="math">a = 0[/spoiler], you never divide by <span class="math">a[/spoiler], so it doesn't really matter.
Do all your problems like this, and you will never have a problem.

[Anonymous No.3778887]

(Thisfag again >>3778853)

Just wanted to mention one more thing for clarity.

I see when you were writing
>F = MA
>F/A = M
>...

that you were having a problem getting the mass of the ball from here.
Don't worry, always leave it as <span class="math">ma = 0[/spoiler], and manipulate the forces on the other side of the equation.
The solution for the mass of the ball will not come from the <span class="math">m[/spoiler] in <span class="math">F = ma[/spoiler], it will come from one of the forces on the left.
(that is what I was trying to remind you of when I wrote:
>Don't forget gravity pointing down.)

Good luck.

[Anonymous No.3778893]

>>3778853
>>3778853
>>3778853

Yeah I'm still here. I'm not sleeping until I get this. Thanks for taking the time to write that all out. I'm trying to apply it now... I swear I'm going to office hours after I get this done and in....

[Anonymous No.3778896]

>>3778693
You are an idiot.
F/M = A is the proper form. More specifically, <span class="math">\frac{\sum{\sigma}F}{M} = A.[/spoiler]
Because the net force on the ball is zero, the acceleration is zero.

Stop being a fucking moron.

[Anonymous No.3778931]

>>3778893
Yeah, office hours are a great idea, they're invaluable.

I also want to stress the importance and power of the free-body diagram.
If you're good at drawing FBDs and writing equations from there, you'll just fly with this stuff. It will be so easy, trust me.

But I've met a lot of kids that just hate drawing them for whatever reason, and they seems to make a lot of trouble for themselves that way.

So I just want to say that if you are one of those kids, you really ought to learn how to draw good FBDs, they can save your live in this field.

[Anonymous No.3778943]

The force downward D, satisfies

D/F = tan(10o)

when you consider that F counters the vector combination of the downward force and the force perpendicular to the string that points to the left 10o below the horizontal.

This makes D=1.4 N, and the mass 0.14 kg.

The tension in the string is 1.4*cos(10o) N = 1.4 N to 2 sig figs.

[Anonymous No.3778960]

And if i need 3 sig figs... it would be 0.144 kg??

I've done that, and typed that in....

[Anonymous No.3778973]

>>3778943
>>3778960
>>3778943
>>3778931

Is the mass 0.143940... kg
---> 3 sig figs = 0.144 kg

?????

Is it the damn software??? >>3778943
>>3778943
>>3778931
>>3778931
>>3778931
>>3778931

[Anonymous No.3779003]

>>3778973
I'm getting <span class="math">m = 4.6249... ~\mathrm {kg}[/spoiler]

[Anonymous No.3779033]

>>3779003
>>3779003

Ugh, that's right!! WTF?

Well, >>3778943, was getting the same answer as me, so there must be some sort of common flaw...

[Anonymous No.3779041]

>>3779003

should my calculator be in radians or degree??

[Anonymous No.3779050]

I've drawn a pretty little triangle for you. x is the force of gravity acting on the ball.

So you can see tan 10 = 8/x
x = 8/tan 10 = 12.3388084 kg

F = ma
12.3388084 kg = m10 (we are taking 10 m/s as the acceleration due to gravity)

m = 1.23388084

Everybody else in this thread is a massive faggot

[Anonymous No.3779055]

>>3779050
I meant
x = 8/tan 10 = 12.3388084 N

[Anonymous No.3779066]

>>3779041
Yeah definitely degrees dude.
If the angle is given to you in degrees, use degrees.

[Anonymous No.3779077]

>>3779050
You labeled your angle wrong.
You fucking moron.

[Anonymous No.3779079]

>>3779050
>>3779055


Mass is 4.63 kg dude.

[Anonymous No.3779088]

>>3779077
I don't understand. What exactly is wrong?

>>3779079
Is that answer from the textbook?

[Anonymous No.3779100]

>>3779088
>>3779088

We submit our HW's online. (CAPA)

>>3779003 said 4.62... I submitted it. and it said yes, it is within the Stand Dev of 4.63.

[Anonymous No.3779106]

>>3779088
He was in radians, not degrees.

<span class="math">m = 4.63 \mathrm{kg}[/spoiler] is correct.
How cares if it's in the text book? If it's not, that book seriously fucked up.

[Anonymous No.3779138]

>>3779106
Oh, okay.
I was confused because I couldn't find where I miscalculated.

[Anonymous No.3779621]

bumpp

[Anonymous No.3779692]

triangle of forces:

F/(m*g)=tan(10°)
m=F/(g*tan(10°))
m=4.62kg

F/T=sin(10°)
T=F/sin(10°)
T=46N

or

(m*g)/T=cos(10°)
T=(m*g)/cos(10°)
T=46N

or

T^2=F^2+(m*g)^2
T=sqrt(F^2+(m*g)^2)
T=46N