Does Wood on Wood Increase Frictional Force Compared to Other Surfaces?

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The discussion centers on the frictional force experienced by wood on wood compared to other surfaces like plastic and glass. An experiment indicated that a wooden block moves with less friction on plastic or glass than on wood. The presence of crests on wood raises questions about whether wood on wood would increase friction due to these interlocking features. Concerns were raised about trusting experimental results versus theoretical expectations. A homework question regarding the coefficient of kinetic friction for a wooden crate sliding across a floor was also presented, highlighting the urgency for assistance.
oliviecheng
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i have done an experiment to find the force of friction on several surfaces.but i found that the wooden block used can move in a smaller frictional force than on plastic or glass
Will the frictional force be smaller when the same material is used(wood on wood)?but there are many crests on wood, when the wooden block moves, crests will hit crests and make the block more difficult to move.
i suppose the frictiional force(wood on glass or plastic)is smaller then that of wood on wood
 
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It is not clear to me, what your question is.
 
Your only "problem" seems to be whether to trust your results. I would consider it very bad science to refuse to accept the results of an experiment because of some notion of what "should" happen.
 
need help with homework pleasez help

al right the problem is (If u use a horizontalforce of 30.0N to slide a 12.0-kg wooden crate across a flor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?) PLz help answr fast its due tmrw ahhhhh!
 
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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