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Static Friction and Frictional Force Ranking Task 
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#1
Oct1411, 03:53 PM

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Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [the first value is the static friction and the second is the kinetic friction] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move.
Box 1= 600kg (Static friction0.8)(Kinectic friction0.5) Box 2=750kg (Static friction0.6)(Kinetic Friction0.5 Box 3=1500kg(Static friction0.3)(Kinetic friction0.1) Box 4=500kg(Static Friction0,6)(Kinectic friction0,3) Box 5=750kg(Static Friction0.4)(Kinectic Friction0.3) Box 6=250kg(Static Friction0.2)(Kinetic Friction0.1) I need to rank the crates on the basis of the frictional force acting on them The equation I would use here is the static friction equation (Static Friction is less than or equal to the miu times the normal force) Since the boxes arent moving I would just multiply the mass of the box times the coefficient of the static friction (For example the first box would be 600x0.8 and then rank those numbers i get from greatest to least correct? However I tried this and it doesnt like my answer. Am i doing something incorrect? 


#2
Oct1411, 06:43 PM

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#3
Oct1411, 07:27 PM

P: 13

I calculated for example the first box to be 600kg x the static coeffiecient which is 0.8. I have done this for all the boxes and my order from least to greatest was the 600 kg box being the greatest frictional force acting on it, then stacking box 2 and 3 because they had the same frictional force, and then stacking boxes 4 and 5 becuase they also had the same number and finally the least box was box 6 . To get the frictional force on these boxes that do not move i should use the static coefficient number given rather than the kinetic friction. So i did this for all these boxes and it did not like my order of answers I guess i am just still confused



#4
Oct1411, 07:27 PM

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Static Friction and Frictional Force Ranking Task
that order is from greatest to least, my apologies



#5
Oct1411, 09:14 PM

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Although you clearly stated in your first post that



#6
Oct1511, 12:10 AM

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The force of static friction is given by the following relation:
[itex]F_\text{static friction }\le\mu_sN\,.[/itex]Why is there an inequality sign, ≤, rather than an equal sign, = ? Can the force of static friction be greater than the external force being applied. 


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