# Block and cart friction problem

• xr250h
In summary, if you want to win the bet, you need to apply a minimum of 16.3 m/s^2 of force to the cart in order to keep the box from falling.
xr250h
Block and cart friction problem - SOLVED

## Homework Statement

You and your best pal make a friendly bet that you can place a 2.0-kg box against the side of a cart, and that the box will not fall to the ground, even though you guarantee to use no hooks, ropes, fasteners, magnets, glues, or adhesives of any kind. When your friend accepts the bet, you begin pushing the cart. The coefficient of static friction between the box and the cart is .60.

a.) Find the minimum acceleration for which you will win the bet.
b.)What is the magnitude of the frictional force in this case?
c.) Find the force of friction on the box if the acceleration is twice the minimum needed for the box to not fall
d.) Show that, for a box of any mass, the box will not fall if the magnitude of the forward acceleration is $$a \geq g/\mu$$ where $$\mu$$ is the coefficient of static friction.

## Homework Equations

$$F = ma$$

## The Attempt at a Solution

I'm not really sure where to start with this one. I set up a Fnetx and Fnety equation but I don't know how to relate the two.

Fnetx = Fa = m*ax where Fa = force applied

Fnety = Ff - Fg = m*ay = 0 (if the block isn't sliding down)

I don't really know where to go from here.

Last edited:
Ff - Fg = m*ay = 0
Great start! So Ff = Fg.
If you fill in the detailed formulas for Ff and Fg, you'll soon have the answers for (a) and (d). The masses can be confusing; be careful to know what mass you mean whenever you write an m.

Ok So I've got Ff - Fg = m*ay and Ff = $$\mu$$*Fn. But I don't know where my normal vector would go? Does it go in the same direction as my acceleration or opposite of Fg?

The normal force is the force that presses the two surfaces together, so it is horizontal. You must ask yourself how hard the cart must push on the block to make it accelerate at "a".

Ok I found out that the normal force is equal to the applied force, so I got Ff = $$\mu$$ma. Then, I used that to plug in for Ff. So I got $$\mu$$ma - mg = 0.

I then solved for a and got 16.3 m/s^2.

For part b, I know I could either use Ff=mg or Ff= u*ma and use my new found acceleration (both give the same answer). But for part c, does the acceleration even change the Ff? If I use Ff= mg for part c, I get the same answer as part b.

If I use Ff = u*ma , I get twice the force as part b.

(I had to use u as mu because latex isn't working right for some reason)

Your 16.3 looks good to me!

But for part c, does the acceleration even change the Ff?
I would say YES. With more acceleration, the cart pushes harder on the block, increasing Fn and Ff = μ*m*a so your second try is perfectly correct. The Ff = mg approach is wrong because the friction force is not equal to mg in the case where the acceleration has been doubled.

Alright so I changed my way of calculating Ff to Ff = u*ma = 19.6 N for part b.

And for part c, doubling the acceleration, I get 39.12 N.

Thank you so much for helping me! Time for sleep now, it's 1 AM here .

Thanks again!

## 1. What is a block and cart friction problem?

A block and cart friction problem is a physics scenario where a block or object is placed on top of a cart or surface, and the friction between the two surfaces is taken into consideration in order to determine the motion or behavior of the block.

## 2. How does friction affect a block and cart system?

Friction between the block and cart will cause resistance and make it harder for the block to move along the surface of the cart. It also plays a role in the acceleration and velocity of the block, as well as the force needed to move the block.

## 3. What factors affect the friction between a block and cart?

The factors that affect friction in a block and cart system include the type of surfaces in contact, the weight of the block, the force applied to the block, and the roughness or smoothness of the surfaces.

## 4. How is friction calculated in a block and cart problem?

The friction force can be calculated using the coefficient of friction, which is a constant value representing the interaction between two surfaces. The formula for friction force is F = μN, where F is the friction force, μ is the coefficient of friction, and N is the normal force between the surfaces.

## 5. Are there any real-life applications of block and cart friction problems?

Block and cart friction problems have practical applications in various industries, such as transportation and engineering. For example, understanding friction in train or car wheels can help improve efficiency and safety in transportation systems. It is also crucial in designing machinery and equipment to ensure smooth and efficient movement.

Replies
3
Views
812
Replies
6
Views
382
Replies
42
Views
1K
Replies
11
Views
2K
Replies
2
Views
320
Replies
19
Views
3K
Replies
8
Views
6K
Replies
21
Views
7K
Replies
15
Views
2K
Replies
2
Views
1K