Contact Force Between Blocks With Kinetic Friction - Physics Problems & Examples
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- Опубликовано: 25 авг 2024
- This physics video tutorial explains how to solve contact force problems between blocks with kinetic friction and without friction when the system is under constant acceleration by an applied force and when the masses are moving with constant velocity / speed. In some cases, the coefficient of kinetic friction may be provided. This video contains plenty of examples, practice problems and the equations / formulas needed to solve the contact force of block 1 on block 2 and the force exerted on block 3 by block 2 and vice versa. This video focuses a lot on newton's third law of motion which states that for every action force, there is an equal an opposite reaction force. This video also uses newton's second law of motion which equates to the formula net force equals mass times net acceleration.
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Hey man, just wanted to let you know I have my final tomorrow and you made my life 10x easier. Thanks bud.
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John Amadi he failed
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Watched these videos to help study for my physics exam and I got a 95! Thanks for your help!
In which grade are u??
@@khushisomerhalder9193 11th I presume
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Iwas trying to understand this concept for literally 2 hours and then you came ohhhhhh too good yr
1) Find Total Force
2) Find Total Acceleration
3) Find component Net Forces
4) Find the force of exertion by calculating Net minus Component.
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Very well explanation ! Love from kerala india
it help me to understand all concept of contact point system
Just trying to cram all this in before my AP physics final
Excellent dude. Gracias keep up the great work
Thank you soooo much Bro. Tomorrow I got an important test and I was total goner until I found this video. Thanks man.
Hi.
For the second example, you said that the amount of force M2 exerts on M1 is 35N.
Should not it M2 and M3 exert a force of 35N on M1?
This is exactly what i wanted
Saaammmee
Op! Tomorrow is my exam ! And thanks for helping
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it's really helpful for students
I get 2 block problems with a single force on one of them. Seems like some of these give you so much extra information.
Thanks very much!
its clear all my concepts that u so much
okay but why contact force has not changed.....i mean it will be 27N even if there is no friction and is 27N when friction is there in the last example.....please explain
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You don't make sense at 6:41, you said at the numerator,the force that has is being pushed should be on top,but the force that block 1 applies on 2 is 2*45/9=10N not the block being pushed=3*45/9=15N.
This was a tremendous help!!!
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in the last problem the 24 N for is the resultant force on the blocks. Then why is frictional force taken into consideration when we are calculating the final force. i think it is wrong
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Excellent
thank you sir. please also explain for 3 block friction problem kept one over another.
mohit hota i think the first explanation of three blocks was wrong it should wrote F2,1 not F1,2
If someone would make these videos but they get straight to the point and are no longer than two minutes
So, in the situation where all the blocks have the same mu with the ground, you can calculate the forces they exert on each other as if there were no friction forces, right?
Life saver ♥️
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dy/dx
நன்றி
You don't make sense at 6:41, you said at the numerator,the force that has is being pushed should be on top,but the force that block 1 applies on 2 is 2*45/9=10N not the block being pushed=3*45/9=15N.
In third problem, if you use formula for contact force then
Force exerted by m1 on m2 and m3=24*9/12
Shouldn't it be this??
thank u
My savour for life
What is the net force of the 1st object when there are objects are attached
Killed it
I subscribed to ur channel
Super dued
There's a question I am not able to solve, can you please help me with it?
Q. 3 blocks of mass 1kg, 4kg and 2kg are placed on a smooth horizontal plane. A force of 120 N is applied from the left side and a force of 50 N is applied from the right side. Calculate the contact force between 1 kg block and 4 kg block.
Net force on the system will be 120N-50N (if you take left applied force positive).
is it 40N?
makasih bang
gw juga mau bilang makasih bang
You don't make sense at 6:41, you said at the numerator,the force that has is being pushed should be on top,but the force that block 1 applies on 2 is 2*45/9=10N not the block being pushed=3*45/9=15N.
at 6:47 how did he get the contact force of block 2 on block 3? Why is it not 30 N, (45-15)
what if the constant force and the accelaration is unknown? and only the total mass is given.
what I mean is, what if the external force is unknown. we are also asked to find the acceleration. only the contact forces between block 1 and 2 are given.
චතුරාර්ය සත්ය අවබෝධ වෙත්වා
cool
Why is there "
Acceleration due to force" when you said there is no acceleration if velocity is constant?
(Better than a HIGH RANKED UNIVERSITY PROFESSOR
Its awesome! But why is that so? Like we got HOW to calculate but why exactly we do that?
Abubakar M that comes by experience...
What if there’s friction on the table
late but use example 3
If only I could just pay you instead of those credit hours man cause that prof is hopping around most of the steps...
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bakwas
Please use a proper mic. These lectures are needed for students like us but voice is ruining them.
Thanks a lot!!
Thanks !