Newtonian physics q

[chiddler]

---

Members don't see this ad.

A small box is on top of a lighter box is on a frictionless surface. There is friction between the two boxes.

When there is enough acceleration on the bottom box to overcome this friction force and cause the small box to begin sliding off, what is true of their accelerations at this moment?

A. The acceleration of the bottom box decreases and the acceleration of the top box decreases.
B. The acceleration of the bottom box decreases and the acceleration of the top box increases.
C. The acceleration of the bottom box increases and the acceleration of the top box decreases.
D. The acceleration of the bottom box increases and the acceleration of the top box increases.

Answer: C

My understanding: The friction force of the boxes points forward and the force due to the acceleration is backwards. So when friction switches from static to kinetic, the friction force decreases. The forward force decreases!

Why isn't B the answer?

 

[milski]

The friction force from top box on bottom box is pointed backwards. Friction when sliding is less than static friction, which makes the net force on the bottom box larger.

Same deal for the top box but there the friction from bottom on top is pointed forward. When it decreases, the acceleration will decrease too.

 

[howlovely]

I hope I understood the wording of your question correctly:

1. Box 1, the box on the bottom, is placed on a frictionless surface. A force, Fa, is being applied to Box 1 and points to the right. Static friction, fs, from Box 2 points to the left. Thus for Box 1, Fa - fs = ma, so a = (Fa - fs)/m.

2. Box 2, the box on the top, is placed on Box 1. Static friction, fs, from Box 1 must point to the right and its magnitude must equal the static friction that Box 2 exerts on Box 1 (due to Newton's 3rd Law). Since fs is the only force affecting Box 2, fs = ma, so a = fs/m.

The maximum force that can be applied to Box 1 (in which Box 2 stays still on Box 1) is equal to fsmax, which equals the normal force * coefficient of static friction. Like you said, when Fa becomes greater than fsmax, the static friction between the two boxes changes into kinetic friction, and thus the force friction exerts decreases.

Going back to the equations for the acceleration of the two boxes, you can see that since static friction was being subtracted from the applied force in Box 1, decreasing the force of friction will increase Box 1's acceleration. Similarly, since static friction was the only net force acting on Box 2, when static friction changes to kinetic friction, Box 2's acceleration will decrease.

So eventually, Box 2 will slide off of Box 1. Yay physics!

 

[chiddler]

The friction force from top box on bottom box is pointed backwards. Friction when sliding is less than static friction, which makes the net force on the bottom box larger.

Same deal for the top box but there the friction from bottom on top is pointed forward. When it decreases, the acceleration will decrease too.

ok i see my mistake.

i associated friction force with the wrong box.

thanks very much for both responses.

 

[chiddler]

wait why does the top box acceleration decrease? shouldn't it increase because it starts moving from rest?

 

[milski]

The force on the top changes from static to kinetic friction. Smaller force - less acceleration. Yes, the speed will continue to increase but at lower rate.


---
I am here: http://tapatalk.com/map.php?mzgprd

 

[chiddler]

ok thanks again.

 

[Morsetlis]

You can think of the top box's accel decreasing because, if it didn't, it would move with the bottom box (at a positive accel and velocity).

Instead, it is now going to fall off, at a lower accel and velocity. If you move the bottom box fast enough (think of pulling the paper out from underneath a book, for example), the top box (book) is going to stay still.

Instead of moving forward with the bottom box, the top box is now decreasing its acceleration from some positive value forward, to approaching 0.