Does Surface Area Affect the Coefficient of Friction?

AI Thread Summary
Surface area does not affect the coefficient of friction (µk) as it is determined solely by the nature of the surfaces in contact. While a larger mass increases the frictional force due to a higher normal force, the distribution of that force over a larger area does not change the total frictional force. When the normal force is constant, increasing the surface area reduces the normal force per unit area, but the overall frictional force remains unchanged. Thus, the relationship between frictional force and normal force is maintained regardless of surface area. The conclusion is that surface area has no impact on the coefficient of kinetic or static friction.
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Homework Statement


does surface area to affect the values of coefficient µk?


Homework Equations



frictional force =µk * normal force


The Attempt at a Solution


I know that if mass is larger then the frictional force is higher( f= ma)
but how about the surface area? does it affect the coefficient of kinetic or static force??
 
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Surface doesn't affect the friction. If the mass is larger then the frictional force is larger due to the increase of normal force. Especially:
Frictional force = μk * normal force
*μk depends only by the nature of the surfaces in touch
 
If the normal force is held constant, but distributed over a larger area, then the normal force per unit area decreases, but so also does the frictional force per unit area. The net result is that the overall frictional force remains the same.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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