IIT JEE , AIEEE Forum - Physics , Mechanics - how's friction directly propotional to N(contact force)?it should be inversely propotional

shikhin

in case of static friction...when some force is applied...how is the horizontal componet of the contact force i.e. static friction (f^{_{s) directly propotional to its vertical component i.e.}}

(normal force)? this is only possible when the contact force vector is inclined at an angle of 45⁰to both the axes.....in some books like H.C. Verma....its given that since f

..(friction proptional to normal force), f=

_s

??

sankydreams

Firstly let me tell u the coz of friction : it is due to interlockings between ruff surfaces, isnt it?
Now heavier the body, stronger the interlockings...its obvious
thats why static friction depends on normal reaction
rate if this helps u....

iitkgp_bipin

Actually the rough surfaces have grooves. When two surfaces comes in contact these grooves interlock with each other and if a heavier body is placed on a rough surface the interlocking is stronger because of the weight. As interlocking is increased friction is increased, hence static friction depends on normal reaction.

sudeep.kumar

Actually friction and normal .. both are components of contact force..

now if the contact force F is acting at an angle t with the vertical,
normal is N = Fcost and friction is f = Fsint...
Hence, friction, f = Ntant....
thus f and N are proportional

But as its difficult to mechanically measure F and t, and we are generally concerned with the components f and N, we use N and f directly instead of starting with F and t every time.

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