Block and cart friction problem

1. Sep 22, 2010

xr250h

Block and cart friction problem - SOLVED

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

$$F = ma$$

3. 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: Sep 23, 2010
2. Sep 22, 2010

Delphi51

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.

3. Sep 22, 2010

xr250h

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?

4. Sep 22, 2010

Delphi51

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".

5. Sep 22, 2010

xr250h

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)

6. Sep 23, 2010

Delphi51

Your 16.3 looks good to me!

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.

7. Sep 23, 2010

xr250h

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!