Minimum acceleration needed for box to not fall from accelerating cart

[hkor]

Homework Statement

You place a 2kg box against the side of a cart. Nothing is used to stick the box to the cart which accelerates to the right. The coefficient of static friction is 0.6.

Find the minimum acceleration of the cart needed so that the box does not fall

Homework Equations

F=m.us
F=ma

The Attempt at a Solution

Im having trouble understanding the concepts behind this question. Maybe because we are using the static force of friction when the object is moving and not stationary.
Does the box remain on the cart when the friction force is equal to the acceleration?
what if the force is greater than fs?
or what if it is less?. Obviously I know that it will fall. but why? because shouldn't the box be stationary if fs>F

Any help or explanation would be great
Cheers :)

 

[hkor]

attempt at answering the question

Actually I do understand, now that i think about it.
So when the acceleration is greater than or equal to the static force we know that the box will stick to the cart.?

So this is what I've come up with..
F(y)=n-mg=0
therefore n-mg
mg=(2.5x9.8) =24.5

F to the right= f(s)
ma=us.n(=mg)
a=(0.06x24.5)/(2)
=5.88m/s/s?

Please correct me if I am wrong

 

[PhanthomJay]

Actually I do understand, now that i think about it.
So when the acceleration is greater than or equal to the static force we know that the box will stick to the cart.?

So this is what I've come up with..
F(y)=n-mg=0
therefore n-mg
mg=(2.5x9.8) =24.5

F to the right= f(s)
ma=us.n(=mg)
a=(0.06x24.5)/(2)
=5.88m/s/s?

Please correct me if I am wrong

Is it 2 kg or 2.5 kg for the mass?

The problem is not clearly worded: the box is placed on the front of the cart.

So when the acceleration is greater than or equal to the static force we know that the box will stick to the cart.?

acceleration and force are not the same

So this is what I've come up with..
F(y)=n-mg=0

the normal force does not act in the y direction.
I think you mean m = 2 kg

F to the right= f(s)
ma=us.n(=mg)
a=(0.06x24.5)/(2)
=5.88m/s/s?

the force on the block to the right is not the friction force. The friction force acts down. What force on the box acts right?

 

[hkor]

Sorry for the confusion the mass of the box is 2 kg and the box is in front of the cart :)

the force on the block to the right is not the friction force. The friction force acts down. What force on the box acts right?

I guess I still don't really understand the concept.
If the static friction of the box is acting downwards then why is it dependent on the force to the right whether the box stays stuck to the cart?

So:
F(x)=n-mg=Fs?

F(y) to the right
F=ma

 

[PhanthomJay]

Oh no you must look in the vertical y direction and horizontal x direction separately. Look at the box using a free body diagram. In the vertical direction, the weight of the box acts down, but it does not move down as noted in the problem. So another force must acting up . What is that force and how much is it?

 

[hkor]

So the friction force is acting upwards and the weight force is acting down?

therefore
(2kg)(9.8N) = Fs

Is the normal going in the horizontal direction?

 

[PhanthomJay]

So the friction force is acting upwards and the weight force is acting down?

therefore
(2kg)(9.8N) = Fs

yes correct

Is the normal going in the horizontal direction?

yes. Normal forces act perpendicular to the surfaces in contact.

 

[hkor]

ok so if F(s)= mg
F(s) =(2)(9.8) = 19.6N
Fs= 0.6xn
n= 19.6/0.6
n= 32.67

In order for the box to stick to the cart the force going in the positive x direction must be equal to or greater than the normal working in the -x direction?

F=n=ma

32.67=(2)(a)
a=16.335m/s/s?

 

[PhanthomJay]

Looks good!
Edit: Oh wait you said -x direction? Normal force is in +x direction to right. There are no other forces in the horizontal direction.