truck moves at 22,5 m/s
On board the truck, behind, is a box.
the static friction coefficient for the box, is 0,250 with regard to the friction between truck floor and the box floor.
calculate the shortest possible stopping distance, such that the box does not slide at all, on the floor of the truck.
The Attempt at a Solution
The problem looks very difficult to do with kinematics equation because we lack the necessary known values.
My reasoning process was a little bit clouded because of this. It seemed like a tough problem to crack without many known values beforehand.
When car decelerates in the first place, then all the objects tend to keep going forward. Like... if you are driving around in the night. Then a moose comes to the road and you need to avoid collision. You have a dangerous situation on the road, and you emergency brake with full power at the brakes. Then, your body goes forward, and the seat belt activates, and stops your body from hitting the windshield and/or the steering wheel.
The same thing would happen to the box possibly, indeed...
t0= 0 seconds
In the beginning scenario 0
the box is at constant speed, as well as the car is at the constant speed. The sum of forces for the box seems to be 0. The box is in balanced state, and box's relative speed with regard to the car floor is 0.
Box feels gravity and normal force of the floor. Also it could be reasoned that because the box is staying still, in a moving car. Then the friction is enough to keep it there.
We know from experience that in the beginning scenario 0
Friction force= frictioncoefficient * normal force
Ffriction = mboxg *0,25
for our purposes, g= 10m/s^2
Ffriction = mbox* 2,5 m/s^2
In the deceleration scenario 1
negative acceleration must be applied to the car. Force of breaking, causes negative acceleration for the car.
I suppose now the box would possibly fly toward the front of the car (mass keeps going forward.)
I suppose the Ffriction force, must resist this box's motion forwards? This seems to be the explanation based upon experience in real life and Newton's ideas...
(I'm little bit confused about static friction here, clarification could be useful)
I suppose with second thought. I had in my mind that the speed of the box, with regard to earth, and the speed of the car with regard to earth, must be equal speeds.If the box moves at greater ground speed, than the car does. Then the box slides forward on the floor. Therefore the relative speed of box with regard to the car should be zero.
I suppose that the overall acceleration (or deceleration) ought to be equal for the car and the box. It should be same value.
But then again forces themselves cause accelerations of any objects... Force causes acceleration.
Probably the box would require less force, to displace it by any amount. Then again boxes also tend to have hefty friction between the floor in many cases...
Probably the car, being the heavier object, usually, would require greater force, to decelerate.
I could use some clarification for my reasoning and problem solving process... I guess.
Yet I cannot escape the idea, that of course. The mass of the box, ought to keep moving forward, at the event of the deceleration (similar to the seat belt example). So, the friction must be enough to resist the forward velocity.
If there had been no friction in the beginning 0 scenario. Then to my mind it would be as though the box stands still with respect to the ground. And the car moves forward at 22,5m/s, from underneath the box... So the box would have slipped off the car. And the car would have gone forward without the box. But these are only my silly reflections about the subject.
In a deceleration scenario without enough friction. The box would have slipped forward toward the driver's cabin of the truck. Which would be dangerous indeed.
I tried to draw a freebody diagram but I don't know whether to focus on the forces of the box, or the forces of the car+ box system. Or something like that. Drawing a picture simply makes it more confusing to me, it seems.
If somebody could clarify the situation I would be happy.
deceleration a= delta v/ delta t
2.5m/s^2 = (0- 22,5 m/s) / (t1- 0 seconds)
t1= 9 seconds duration for the deceleration to 0 m/s
distance = average speed * time
s= [(22,5-0) / 2] * 9