Physics Olympiad

Block on a Conveyor Belt

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.


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 $$ s=l+vt=l+v\sqrt{\frac{2l}{\mu g}}$$

Physics Olympiad

Block on a Cart Problem

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: Block on a Cart

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
$$ F=ma$$


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


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

Physics Olympiad

The Leaning Ladder Problem

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?


The Leaning Ladder Problem
Figure showing the forces acting

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}

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Physics Olympiad

Frictional force between pen and paper

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?


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
$$ F=f_1+f_2+Ma
Now, (a=\mu_1 g)
Hence, $$ F=(m+M)(\mu_1+\mu_2)g$$