A conveyor belt having a length (l) and carrying a block of mass (m)
moves at a velocity (v). Determine the distance covered by the block before it stops.

Discussion:

The initial velocity of the block relative to the ground is determined from the conditions $$ v_0=at$$ $$ l=v_0t-at^2/2$$
Now, acceleration of the block due to friction $$ a=\mu g$$
Hence, $$ v_0=\sqrt{2\mu gl}$$
The time of the motion of the block along the conveyor belt $$ t=\sqrt{\frac{2l}{\mu g}}$$
The distance covered by the block before it stops $$ s=l+vt=l+v\sqrt{\frac{2l}{\mu g}}$$

Try this beautiful Problem, useful for Physics Olympiad based on Block on a Cart.

The Problem: Block on a Cart

A block is placed against the vertical front of a cart. What acceleration must the cart have so that block A does not fall? The coefficient of static friction between the block and the cart is (\mu_s).

Discussion: 

A block is placed against the vertical front of a cart. Let the mass of the block be (m) and the common acceleration of the system be (a).

The force equation can be written as
$$ F=ma$$

Now,

$$f-mg=0$$ Since the coefficient of static friction is (\mu_s),

$$f=\mu_s$$

Therefore,$$ \mu_s ma=mg$$  $$\Rightarrow a=\frac{g}{\mu_s}$$

Let's discuss a beautiful problem useful for Physics Olympiad based on the leaning ladder.

The Problem: The Leaning Ladder

A ladder leans against a frictionless wall. If the coefficient of friction with the ground is \(\mu\), what is the smallest angle that the ladder can make with the ground and not slip?

Solution:

Let the ladder have mass m and length \(l\). We have three unknown forces: (i) the frictional force F
(ii) the normal forces \(N_1\) and \(N_2\). And we fortunately have three equations that will allow us to solve for these three forces:
\(\Sigma F_{vert}=0\), \(\Sigma F_{horiz}\) and \(\tau=0\).
Now, from the figure we can see that \(N_1=mg\).And then when we look at the horizontal forces, we see that \(N_2=F\).

We will now use \(\tau = 0\) to find \(N_2\) (or F).

So the number of unknowns are reduced frook three to one.
There are two forces acting at the bottom end of the ladder.
Balancing the torques due to gravity and \(N_2\), we have $$N_2lsin\theta=mg\Bigr(\frac{l}{2}\Bigr)cos\theta$$

( The factor 1/2 comes into play because the ladder behaves like a point mass located halfway up)

$$\Rightarrow N_2=\frac{mg}{2tan\theta}$$
This is also the value of the frictional force F. The condition
$$ F\mu N_1=\mu mg$$

therefore becomes

$$\frac{mg}{2tan\theta}\leq \mu mg$$ $$\Rightarrow tan\theta\geq\frac{1}{2\mu}
$$

Some Useful Links:

• Angular Momentum – Problem
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A pen of mass 'm' is lying on a piece of paper of mass M placed on a rough table. If the cofficient of friction between the pen and paper and, the paper and the table are (\mu_1) and (\mu_2) respectively, what is the minimum horizontal force with which the paper has to be pulled for the pen to start slipping?

Solution:

For pen to start slipping, maximum horizontal acceleration for pen and paper to start slipping is (f=\mu_1)g
Therefore, (a=\mu_1 g) is the common acceleration for pen and paper.
If (f_1) and (f_2) be the frictional forces for pen and paper respectively, the net force for the system
$$ F=f_1+f_2+Ma
F=\mu_1mg+\mu_2mg+Ma$$
Now, (a=\mu_1 g)
Hence, $$ F=(m+M)(\mu_1+\mu_2)g$$