MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details
Get extra R 2,800 off
USE CODE: askschfd09

				   

cause of friction is said to be electromagnetic forces between the surfaces of solids in contact. it is defined to be the horizontal component of the contact force. this implies that friction acts on a body at rest also; which is strange?


if such interparticle electromagnetic forces are responsible for friction why is it independent of surface area of contact?


for surfaces of given materials, friction is more in case the surfaces are smooth or rough?


can the value of coefficent of friction (static or dynamic) be greater than 1?

7 years ago

Share

Answers : (1)

										

Dear Raghav,


               As u explained the friction is due to electromagnetic forces forces but these are only interactions and they will be in equilibrium. Now when we try to move the object this equilibrium is lost. Thus an oppositie foce known to be frictional force is developed to regain the equilibrium.


But these forces depends only on the nature of the materials in contact but not the area(the force we experience is not the net effect of the electromagnetic interactions). The statement given below is Amontons' 2nd Law


"The force of friction is independent of the apparent area of contact". (Amontons' 2nd Law) (Amontons' 2nd Law does not work for elastic, deformable materials. For example, wider tires on cars provide more traction than narrow tires for a given vehicle mass because of surface deformation of the tire)


Friction is more in the case of rough materials because of both electromagnetic interactions and the interlocking of the surfaces in contact (but not due to the increase of surface area).


Yes it can be greater than 1 in case of nail pinned into a wall, while walking or driving on sand, etc.


 Dear STUDENT NAME,


----------------Solution Here ----------------


Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.



All the best !!!


 



Regards,


Askiitians Experts


Adapa Bharath

7 years ago

Post Your Answer

Other Related Questions on Mechanics

a packet is released from a balloon accelerating upward with acceleration a. the acceleration of packet just after the release is
 
 
Since the ballon was moving upward with acceleration a, there is a force of gravity on the packet too. The balloon is moving upward bexause the reaction force R>W, weight of the packet. As...
 
Shaswata Biswas 2 months ago
 
Just after the release the packet posses only the component of the velocity aquired. Here the accleration of the balloon has no effect on the acceration of the body. When the body is...
 
Shaswata Biswas 2 months ago
 
The accleration of the packet will be `g` acting downward as no external force other then gravity working on it.it will achieve the case of free fall. If I am wrong please ..explain the...
 
Chandan kumar mandal one month ago
The potential energy of a particle of mass 1kg moving along x-axis is given by U(x)=[x^2/2-x]J. If total mechanical energy of the particle is 2J find its maximum speed.
 
 
total mechanical energy = U+ K.E to attain maximum speed the obect must have maximum K.E as K.E will be maximum , U has to be minimum (conservation of energy )given - U(x) = x²/2-xfor this...
 
fizaparveen one month ago
 
Ans- 2Speed will be max when kinetic energy is maximum so potential energy will be minimum so differentiating the function put it equal to 0 so we obtain minimum for x=0 now kin potential...
 
rishabh doshi one month ago
 
We need to differentiate the equation and we will get value of x. x=1 then substituting x=1 in the given equation will give value of U. Substitute U and total energy=2 in TE = KE + PE. Then...
 
Dhanyashree one month ago
if a body move with u velocity then after collide with wall and remove with v velocity what is the impulse
 
 
Hi aman, I think this might help as we know that Impulse and also F=m(dv/dt) so (here m is the mass of body and dv/dt =acc) here p is the momentum ans delta p is change in momentum hence in ...
 
Ankit Jaiswal 2 months ago
what is work energy thereom...................................?
 
 
For any net force acting on a particle moving along any curvilinear path, it can be demonstrated that its work equals the change in the kinetic energy of the particle by a simple derivation...
 
rahul kushwaha 2 months ago
 
To change the kinetic energy of a particle we have to apply a force on it and the force must do work on it. This relationship between KE and WORK is called "work -energy theorem...
 
Ankur Saha 2 months ago
 
Relation bewteen KE and W: The work done on an object by a net force equals the change in kinetic energy of the object: W = KEf - KEi. This relationship is called the work-energy theorem.
 
Shreyas Patil 3 months ago
 
The amount of work done by the net force acting on a particle is equal change in it’s kinetic energy.
 
6 days ago
 
work energy therom is nothing but the total work done by all forces is equal to the change in its kinetic energy …. W=1/2 {kx 2 2 -kx 1 2 }
 
6 days ago
 
Total workdone by the Net Force on the body is equal to change in it’s kinetic energy . W total =1/2{kx 2 2 -kx 1 2 }
 
6 days ago
The cieling of long hall is 25 m high What is the maximum horizontal distance that a ball thrown with a speed of 40 ms can go without hitting the ceiling of wall? Plz explain with a DIAGRAM
 
 
Let the angle of throwing be Q with respect to the ground So, the range is R = 40*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is...
 
Tapas Khanda 3 months ago
 
Let the angle of throwing be Q with respect to the ground So, the range is R = 30*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is...
 
Tapas Khanda 3 months ago
 
Diagram is not required to solve this question. You can directly use the formula to find range : R = (v^2/g)*sin2Q Maximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8) So, ...
 
Tapas Khanda 3 months ago
a particle of mass m starts moving from origin along x axis and its velocity varies with position (x) as v=k√x. the work done by force acting on it during first "t" seconds is ???ans) (m k^4...
 
 
I want you to recheck your answer mr strange...you had written the value of a as k^2 ÷ 4 and you put that value at last as k^2 ÷ 2....Overall if u cann xplain it in brief i wud be...
 
Vaibhav 20 days ago
 
sry it was a mistake , this is right.......... :) …...........................................................................................
 
DR STRANGE 20 days ago
 
I dont understand how did you got (k dx)/(2√x dt).....and also this the integration of dx/√x should be x/2 but u wrote 2x how???
 
Vaibhav 20 days ago
View all Questions »

  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
Get extra R 2,600 off
USE CODE: askschfd09

  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Get extra R 2,800 off
USE CODE: askschfd09

More Questions On Mechanics

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details