question about friction.. someone pls help!?

Question

All crates are in equilibrium ( no acceleration). From greatest to least, rank the amount of friction between the crate and the floor.
A= 100N (v=0)
B= 120N (v=0)
C= 130N (v=1 m/s^2)

A is one man pushing a crate that weighs 100N with velocity of 0.. B is one man pushing a crate that weighs 120N with a velocity of 0.. and C is one man pushing a crate that weighs 130N with a velocity of 1 m/s^2

PLEASE HELP! I cant seem to get this

Steve4Physics · Accepted Answer

There are a number of problems with the question, even with the extra information you supplied.
1) You can’t have a velocity of 1 m/s^2 because 1 m/s^2 is an acceleration, not a velocity.  And you have said the crates are not accelerating.  So I’ll assume you mean v = 1 m/s.

2) When you say v=0 do you simply mean the crate is not moving? Or do you mean it is about to start moving (on the point of slipping)?
If you have a 100N crate on a surface, the frictional force can be ANY value from 0 up to the limiting frictional force (when it starts to slip).  For example if the limiting frictional force is 80N, the frictional force can be any value up to 80N; if you push on the side with a 10N force, friction = 10N, v=0; if you push on the side with a 20N force, friction = 20N, v=0.  You can push with any force up to 80N before slipping starts.

3) Are the coefficients of friction the same for all 3 boxes?  There is no indication of what parameters are the same for the different boxes.
_________________________

Normal force, N = weight.
Limiting frictional force = μs.N where μs is coefficient of static friction.

Let’s assume that the coefficient of static friction, μs, in all cases is 0.5 for example.

For the 100N weight crate, the frictional force could be any force between 0N and (0.5x100 =) 50N.

For the 120N weight crate, the frictional force could be any force between 0N and (0.5x120) = 60N.

For the 130N weight crate, the velocity is constant so the frictional force on it equals the applied force.  The kinetic frictional force will be μk.N (where μk is coefficient of kinetic friction).  μk is usually less than μs, so the frictional force will be  less than (0.5x130=) 65N – but we don’t know how much less.

So there is no way to answer the question.

I suggest posting a new question, with the exact original wording.

Matt H · Answer

I don't think there is enough information to tell.  
You say all crates are in equilibrium (No Acceleration) but you then say one crate is accelerating (C 1m/Sec^2).

But if all crates are on the same surface and made of the same material then the one that weighs the most would be affected the most by friction etc..

The greater the weight, the greater the friction.  Assuming no use of wheels or anything.

question about friction.. someone pls help!?

2 Answers