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
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?
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
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.
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.
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.
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.
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.
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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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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?
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it help me to understand all concept of contact point system
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I get 2 block problems with a single force on one of them. Seems like some of these give you so much extra information.
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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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Made it easy thanks
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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?
best explanation ...keep it up...
Excellent
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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This is a really great video. Thank you!
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Thank you
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
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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.
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.
Bruh it took me 2 hours and I still didn’t understand and you taught me in 5 minutes
Really helpful ... Thanks a lot
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Nice one bro
This was a tremendous help!!!
dy/dx
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??
What is the net force of the 1st object when there are objects are attached
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?
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.
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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.
at 6:47 how did he get the contact force of block 2 on block 3? Why is it not 30 N, (45-15)
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Its awesome! But why is that so? Like we got HOW to calculate but why exactly we do that?
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Why is there "
Acceleration due to force" when you said there is no acceleration if velocity is constant?
What if there’s friction on the table
late but use example 3
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