- #1
ADMINISTRAT0R
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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
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