Static and Kinetic Friction Problem, getting wrong answer?

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SUMMARY

The discussion centers on calculating the force of friction acting on a crate of oranges weighing 176 N on a truck bed, with static and kinetic friction coefficients of μs = 0.30 and μk = 0.20. The user incorrectly calculated the static friction force as 52.8 N, which is derived from multiplying the coefficient of static friction by the weight of the crate. The correct approach involves understanding that the crate does not accelerate when the truck moves at a constant velocity, leading to a frictional force equal to the static friction value. The user also inquired about the frictional force during acceleration and the maximum acceleration before sliding occurs.

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A crate of oranges weighing 176 N rests on a flatbed truck 2.0 m from the back of the truck. The coefficients of friction between the crate and the bed are μs = 0.30 and μk = 0.20. The truck drives on a straight, level highway at a constant 6.3 m/s.

a. What is the force of friction acting on the crate?
I multiplied .30 X 176= 52.8. Online assignment says it's wrong?

frictional force= co of friction times Normal force

b. If the truck speeds up with an acceleration of 4.0 m/s2, what is the force of the friction on the crate?

magnitude:
Direction:

I simply used the kinetic friction value times the normal force and still got it wrong.


frictional force= co of friction times Normal force

(c) What is the maximum acceleration the truck can have without the crate starting to slide?

I got this right at 3 m/s
 
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Alright, I got B and C, but I still can't get a.
 
The truck and the crate move together with a constant velocity along a straight line. Does the crate accelerate?

ehild
 

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