I can't solve this basic physics problem and I just don't see how you solve the force F.
A guy is pulling a fridge (m=50kg) at constant velocity so there's no acceleration. This means the friction force is as great as the x-component of pulling force F.
But I only know the mass of the fridge, the angle (30 degrees) and the friction factor of 0,40.
How do you even solve it with just this information?
Here's how far I've got.
don't bother, physics is complete bullshit
my physics teacher claimed that every force has an opposite counter force of the same strength pushing against it. I said if that was the case then there would need to be a counterforce to the counterforce, and a counterforce to the counterforce of the counterforce, which would essentially mean you had unlimited energy. I got out my deck and demonstrated that while I can play mirror force and you can play solemn judgement and we keep chaining traps against each other, eventually we will run out of cards because those are a resource. he was completely speechless. I just walked out of class and never went back
I am a literal mathematical retard and can't speak this Swedish-meatball gobbledygook so I can't help you and have no reason to be posting in this thread, goodbye.
Find the Y component of the force he is pulling with, It'll give you how much of the weight of the fridge he is nullifying. Then you can just use that weight to work out the friction and stuff.
U trollin bro?
The first force F_a has a counter force F_b that has a counter force of F_a and there are only those two forces in the system, there's no implication that an infinite amount of counter forces are generated.
>don't bother, physics is complete bullshit
Physics is the science of applying mathematics to reality. If you want something more exact than physics you need to dig deeper and go with math.
>Physics is bullshit!!!!
>Uses technology that the advancement of physics made it possible to type these sentences
Gas yourself brainlet
OP here, I thought of that but I can't find F_y any discrete value.
It's just F*sin30 or mg - N
The problem is, I usually have the total force F but this time I don't, so I'm confused.
Here's where I'm at now, just running in circles.
Just noticed the math in the end makes no sense
wowzas and yikes
>my teacher came up with this wild idea of counter force
>proved him wrong with Yugioh card games
Newton would be proud.
I think your second equation is W-Fsin30 = N, for N is the resultant force of y component. Substite Friction = 0.4N = Fcos30 to your second equation and I think you'll get your answer
>unable to solve 9th grade physics problem
Underage fag detected
I bet you can't finish it
Heh i am genius
F_g can't be equal to F_n because pulling on the rope takes away some of the normal force.
That's the trick number one in this problem and most people don't see it.
F is closer to 260 N but I'm not sure how to get there.
Fx = Fcos30, Fx = (sqrt(3)/2)*F
Also Fx = uN, Fx = 0.4*N, (sqrt(3)/2)*F = 0.4*N
Fy = Fsin30, Fy = 0.5*F, mg = 50*9.8
N = 490 -0.5*F, substitute into previous equation
(sqrt(3)/2)*F = 0.4*(490-0.5*F)
I dont have a calculator on me, solve for F.
It's not that hard, you have 2 unknowns F and N, and 2 equations (sqrt(3)/2)*F = 0.4*N and N = 490 -0.5*F. Next time try substituting in the values you know to narrow it down.
F*(1+cos30/0,4sin30)=mg/0,4sin30
-->F= (mg/0,4sin30)*1/(1+.....)
There you go! Is this school work?
OP is close to finishing it, along the x axis, the eqn is
>Fcos30 =mu*N
Then along y is
>N = mg - Fsin30
substitute for N into the first eqn to get
>Fcos30=mu(mg-Fsin30)
>>Fcos30=mu(mg-Fsin30)
Well that's exactly the part that I can't solve. My equation solving skills aren't that great.
I got F on both sides meaning they cancel each other out..??
At this point OP has to be trolling.
I'm not trolling, here's where I'm at.
I think I solved it finally
nah they dont cancel, next step is to distribute the mu through and take out the F once you get it on the same side of the eqn
>Fcos30=mu(mg-Fsin30)
>Fcos30=mu*mg-mu*Fsin30
>Fcos30+mu*Fsin30=mu*mg
>F(cos30+mu*sin30)=mu*mg
>F=(mu*mg)/(cos39+mu*sin30)
Fcos30=mu(mg-Fsin30)
Fcos30/mu=(mg-Fsin30)
Fcos30/mu + Fsin30 = mg
F(cos30/mu + sin30) = mg
F = mg/(cos30/mu + sin30)
The joke
Is
Over
Your head