Friction problem would really appreciate the help

  • Thread starter strobeao3s
  • Start date
  • #1

Homework Statement


Losing Cargo. A box of mass 12.1 kg rests on the flat floor of a truck. The coefficients of friction between the box and floor are mu_s = 0.190 and mu_k = 0.160. The truck stops at a stop sign and then starts to move with an acceleration of 2.18 m/s^{2}.

If the box is a distance 1.83 m from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck?


Homework Equations


f_smax= mu_s * N
f_k = mu_k * N
f_s + f_k = ma

x-x_0 = v_0x + 1/2 (a_x)t^2



The Attempt at a Solution


f_smax = .190(12.1*9.8) = 22.5

f_k = .160(12.1*9.8) = 19.0

22.5 + 19.0 =12.1(a)
41.5/12.1 = a = 3.43

1.83 = 1/2(3.43)t^2

t=1.03 seconds
 
Last edited:

Answers and Replies

  • #2
i'm totally lost

i'm having a really difficult time with friction problems any guidance would be really appreciated
 
  • #3
learningphysics
Homework Helper
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Important thing to note... the frictional force is pushing the box forward here...

The first question that needs to be answered is, does the box slide?

Calculate the force required to keep the box from sliding... ie assume the box isn't sliding and get the frictional force...

frictional force = ma = 12.1*2.18 = 26.378N

But the max. static frictional force is 22.5N as you calculated...

26.378N is beyond 22.5N so the box must be sliding...

So now we can ignore static friction and deal with kinetic friction.

You've calculated the frictional force = 19.0N... this is the only force acting on the box horizontally... hence 19.0N = ma

get the acceleration of the box...

now we have a kinematics problem.

the box is ahead of the rear of the truck by 1.83m, at t = 0.

the box has a certain acceleration... the rear of the truck has a certain accleration... at what time does the rear of the truck catch up to the box...
 

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