Physics 4.7 Friction & Forces at Angles (1 of 8) Horizontal Surface: 1
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- čas přidán 23. 11. 2016
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In this video I will find maximum force=? pushed at 30 degrees on to a 5kg box on a horizontal surface with a static friction of 0.4 before the box starts moving.
(Bloopers at the first :30 seconds)
Next video in this series can be found at:
• Physics 4.7 Friction...
6 years later and this video is still simply amazing!
Glad you found it and that it is helpful.
Haha, is the beginning a blooper reel? Even professors have fun sometimes!
@Derek Walrond you're right no one gives a damn
this guy is singlehandedly saving my AP physics grade
In the same boat man.
This is the first video I have watched. I've been tearing my hair out on a question for the last 4 hours. And you just helped me solve it in 6 minutes. I am now subscribed.
Thank you.
Welcome to the channel! 🙂
That intro was great. (I mean, honestly, all your videos are great in general...!)
Beautifully done. Thank you!
The coefficient of friction between ground and ground is 0. Is 1. 15 kg is the value of the force f if the offender receives 3ms ^ -2 armature when the f key is applied to the offender.(fcos 30, f sin 30)
Thank you so much! Your explanations are amazing and easy to understand :)
You're very welcome!
Graduated with a degree in Computational Physics and still come here for the basics when I need help. Thank you sm :')
You are welcome. Glad you like our videos.
I was going to say "this guy is a god".. i look in the comments and somone already said it lol
Thank you soo much, only video that helped me with my homework and it's 2023!!
Glad you found our videos! 🙂
This concept is very helpful thank sir 👍
What if velocity is unknown and you have mass force angle and distance?
What a legend! Great work Sir
Just wondering what grade you guys are in ?
hi boss, the friction force also depends on contact area between the object and the surface beneath it... contact area increase, the friction force magnitude will also increase.
It turns out that the friction force does not depend on contact area. It seems counter-intuitive, but that is how it is.
I was stressed stuying this but watching the video somehow reminded me not to overthink and enjoy the ride thanks for the vid ❤
That is the spirit! Take your time and go through the examples, it will click. 🙂
Great video...quick question though, why use the angle 30 and not 330 since your resolved vertical force should be negative. ( The force itself kind of is in the 4th quadrant.)
In the real world negative signs indicated direction. Better to use angles less than 90 degrees and then determine direction if needed.
Good Day Prof Biezen, if a field with less gravity does it also mean less friction? Or other word expressed bearing on the moon with less gravity than earth will last longer in lifetime?
Have a nice weekend... 👍👍
That is an interesting way of looking at it and yes, purely on the friction alone, there would be less wear due to friction. On the other hand, the large temperature differences between day and night on the Moon would add additional material stress.
I want to ask if the force applied to the block is horizontal to right, that means no inclined angle , how to calculate the force needed to move the block ?
You would then have to overcome the friction force caused by the weight of the block alone. Force friction = m g mu where mu is the coefficient of friction
What should I do when I was asked "What is the smallest force that
could make the box slide along the table?" given only a 5kg box on a horizontal table, horizontal force of 15 N, and a 0.4 coefficient of friction?
You need a force just big enough to overcome the static coefficient of friction. F > Mg (mu)
@@MichelvanBiezen thankk youuu so much!!
How about condition inside U Boot Prof Biezen? Do we experience also some reduction of gravity force when u Boot are floating in water?
No, it is not the same as being in a spacecraft orbiting the Earth.
I'm from Iraq and I'm a student in electrical engineering and I'm watching your videos now because I have an exam in physics, thank you very much
Glad that you found our videos. Welcome to the channel and good luck on your exam!
@@MichelvanBiezen
Thank you very much, but I hope if you have a group on WhatsApp or Telegram for students, or I can communicate with you or with one of your students and be friends with him, and I repeat my thanks to you very much
The best way to get a hold of me is via the comments on the videos.
Sir , can you explain how we do the same problem but , when applied force and mass is given to find the angle of friction ?
Not sure what you mean by the "angle" of friction. But in general, regardless of what is being asked, the problem will be set up in the exact same way to derive the exact same equation, and instead of finding the acceleration, you'll be finding something else (like the coefficient of friction, etc.)
Dude, thank you so much I really appreciate it.
Glad it was helpful
@@MichelvanBiezen I’m indebted, I was struggling with my physics take home final. Still am but we move.
Thumbs up sir!
What happens when both the coefficient of static and dynamic frictions are equal?
The problem would be worked out the same. You would still find the net force and if the net force is greater than zero, the block would accelerate.
I think the calculation in the final answer equals to 22.4 right?
no
how can we lessen the force applied to an object being pushed horizontally?
Sorry, but I don't quite understand the question. What would be the purpose of lowering the force? The force applied is given in the problem and we can set it to any value we want.
Hello sir I have a question.
A 4.00 kg block is at rest on a horizontal surface, acted on by a 6.97 N horizontal force. The coefficients of static and kinetic friction are 0.795 and 0.158.
Moments after the horizontal force is applied, the magnitude of the force of static friction acting on the block, in N and to two decimal places, is
(If there is no static friction acting on the block, enter an answer of "0" with the appropriate number of decimal places.)
Since the applied force is not sufficient to move the block (while it is at rest), the static friction force will match the applied force and therefore the static friction force will be 6.97N
what if they are asking for the kinetic friction instead( Moments after the horizontal force is applied, the magnitude of the force of kinetic friction acting on the block, in N and to two decimal places, is)@@MichelvanBiezen
Thank you, I have been struggling with this for a bit now and finally understand what I'm doing. 😁
Smart teacher amazing!!!!!
an object of mass 6 kg sliding horizontal surface with a uniform velocity of 8 /s assuming force of friction offered by the surface to be zero the force required to maintain the motion of object with the same uniform velocity is
zero If there are no frictional forces to slow the object down, then it requires no force to keep it going (Newton's first law)
Tq for clr my concept 😢 really got some help
i love this video
Good sence of humer
the beginning is epic..breaking the monotony
The start isn't always as polished as we like it to be, but we are having fun doing it......🙂
Isn't it suppose to be N = mg - Fsin (theta)? The other videos taught it like that
Since the force is pushing down, we need to add it to the weight of the object resulting in a greater normal force.
@@MichelvanBiezen So you would add on a downward force and subtract on an upward force. Ok Thank you😊😊
That is correct.
Thanks
Hi Professor,
I wonder what the maximum angle of the force F is so that it is not possible to move the object at all. My guess is that this can be solved when the denominator of the expression you have derived is 0 (cos(theta)-0.4sin(theta) = 0). However, my intuition tells me that the only angle possible is 90 degrees because in that case there is no x-composant. Is it possible to calculate the angle in this case? Thank you!
Yes, set the component of the force in the x direction equal to the magnitude of the maximum friction force.
@@MichelvanBiezen Does this mean that the solution of cos(theta) - 0.4sin(theta) = 0 gives the maximum angle of the force F to not be able to move this object?
Correct
is there an example like this without friction
You would work it out exactly the same way and just set the coefficient of friction equal to zero. Only the horizontal component of the force will be pushing the block and F = ma becomes F(push) x cos (theta) = m a
What if the force applied is below the horizontal?
What is the question ?
What if the you applied force 30° below the horizontal?
thanks for this video but i have a question about it
sir, how can i find the critical angle which makes the box becomes impossible to push ?
Set the horizontal component of the force equal to the friction force caused by the weight of the block plus the vertical component of the push
@@MichelvanBiezen but sir, for the unknown force which is applied from the upper side with some angle i have to neglect mg if not, i reach this answer (Fcosx=(mg+fsinx) μ. Why i have to supposed to neglect the mg to find the critical angle?
Sometimes it helps to set limits to certain inputs, in this case the angle. Let the angle go to zero and what do you get?
Watching this in class in silent mode and wondering why the lecturer id laughing his ass off🤔😂
You'll have to turn on the sound after class! 🙂
I loughed loudly in the beginning :D:D:D
Thanks a lot! Pulchritudinous!
That is quite a word.
Watching this rn 2023
The principles of physics don't age.
A 40kg Wooden create rest on a horizontal wooden floor if the coefficient of static friction is 0.5 how much force is needed to set the create up
Please solve it
As much as it wants
If you mean start to move then greater then or equal to 200N
I made gravity 10 to can do it in my head
Thank you Prof.....
You are welcome
this guy is god
There is no God without Allah
@@RubelHossain-math r u mad😂😂
@@RubelHossain-math f** ur allah
May Allah guide of all you
Michel, if the angle of θ is increased to 70° and µ and mass remained unchanged (µ = 0.4, m = 5kg), for the example in this video, then the resulting answer is a negative number.
What does that mean?
Surely a LARGE enough force applied at θ = 70° would be able to move a 5 kg object?
If the angle is large enough then the friction force will increase to the point where the horizontal component of the applied force cannot move the block.
LMAO THANK GOD FOR THIS VIDEO
Excuse me, I have a question. Once you find Fx before it moves, how do you now find the Normal Force acting upon the object?
The normal force will be the sum of the weight and the y-component of F
thank you so much, your videos are way better than what my physics professor can teach in 2 hours.
THE GOAT
Not by any measure. But glad you like our videos.
May I ask Sir what if it is Pulling Force?
This force represented by the arrow is a pushing force
@@MichelvanBiezen Thanks Sir I just Make it a Minus Sine instead of + Sine degree , because my Problem Here is A pulling Force 😊, thanks For Teaching us Sir
Im confused I thought Fx would have to be greater or equal Ffriction
Only to get it moving (momentarily). Once an object is moving, it will continue to move if the net force = zero. (Newton's first law)
@@MichelvanBiezen If looking for the friction coefficient would we set it up the same and solve for coefficient instead of F?
Yes, you would use the exact same technique
Thank you sir
So nice of you
I still don't get it, why Fx < Fr ? As we know that if the force we apply less than the friction force that created by its object , so the object will not move.
If Fx is less than the friction force, the block will NOT move. (That is what that statement implies).
Where does 9.8 comes from???
That is the acceleration due to gravity. g = 9.8 m/sec^2
@@MichelvanBiezen thank you so much
made this look so easy opposed to my professor
Glad the video helped make it clear.
U really super
i thought the normal force should be mg - the vertical component force. this is how my teacher taught me?
The normal force is the force of the surface pushing back against all the forces pushing down on the surface. (In this case that is both the weight of the object AND the vertical component of the force pushing down).
ok thank you
Thumbs up
Thank you
I'm lost
Start with an earlier playlist (physics 4) PHYSICS 4.1 NEWTON'S LAWS EXAMPLES czcams.com/video/bLWpKeByGsQ/video.html PHYSICS 4 NEWTON'S LAWS OF MOTION czcams.com/video/ruyHfepjgcE/video.html
Oppenheimer
?
angrezi achhe se bolte ho sir
🙂
..............
You are good teacher but improve your video quality
Thank you. Yes we have improved our videos over the years. Some of the earlier ones could be better.
super
Yo farooki
Pogus here
Sir ji kabi ro hindi bol diya karo
what was the intro, so random and weird