Motion Of Two Bodies One Resting On The Other - Practice Questions & MCQ

Block on Block problems

Let us assume two blocks of masses m_1, and m₂ are kept on a smooth horizontal surface as shown in the figure. The coefficient of friction between the blocks is $\mu$. As the block of mass m₁ is being fulled by applying force F the system starts moving.

To solve problems based on block on block situations, we must begin by calculating the limiting friction at all the rough surfaces. These values will help us determine the possibilities of relative motion between bodies in contact. As the surface is smooth block m₁ will begin to move for any value of F. but the friction between the block will try to stop any relative motion between the blocks. Therefore, friction force between the blocks will take the any suitable direction and magnitude such that both blocks move with same acceleration (no relative motion). In this case the friction force between the blocks will be of static type. However, if we increase the value F, the requirement of friction force for no relative motion will also increase. The value of friction force can not be greater than the limiting value of friction.

To solve such problems we will initially assume that friction has successfully ensured that there is no relative motion between the blocks an determine the value of frictional force for such condition. If this value of frictional force is less than the limiting friction, we can say that our assumtion were right and all the results thus determined are the correct results.

But, if the value of frictional force comes to be more than limiting friction assuming no relative motion, then we know our assumption was incorrect as value of friction force can not be greater than limiting friction. In such case, since we have established that there is a relative motion between the blocks we must apply kinetic friction in suitable direction and determine different accelerations of the blocks.

Let us first assume that there is no relative motion between the blocks. Both the blocks are moving with common acceleration a_c and the friction force between the blocks is f.

F.B.D of both blocks m_1, and m₂ together-

$\begin{aligned} & \mathrm{F}=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{a}_{\mathrm{c}} \\ & \mathrm{a}_{\mathrm{c}}=\frac{\mathrm{F}}{\mathrm{m}_1+\mathrm{m}_2}\end{aligned}$

F.B.D of the block of mass m_2-

$\mathrm{N}=\mathrm{m}_2 \mathrm{~g}$
Limiting friction-

$
\mathrm{f}_{\mathrm{l}}=\mu \mathrm{N}=\mu \mathrm{m}_2 \mathrm{~g}
$

From newton's law-

$
\mathrm{f}=\mathrm{m}_2 \mathrm{a}_{\mathrm{c}}=\frac{\mathrm{m}_2 \mathrm{~F}}{\mathrm{~m}_1+\mathrm{m}_2}
$

Case-1

$
\begin{gathered}
\mathrm{f} \leq \mathrm{f}_1 \\
\text { or } \frac{\mathrm{m}_2 \mathrm{~F}}{\mathrm{~m}_1+\mathrm{m}_2} \leq \mu \mathrm{m}_{2 \mathrm{~g}} \mathrm{~g}
\end{gathered}
$

If this is the case then our assumption is correct and the blocks will move together with acceleration $\mathrm{a}_c$ and the friction force btween them is static and equal to $f$. Where-

$
\begin{aligned}
& a_c=\frac{F}{m_1+m_2} \\
& \mathrm{f}=\frac{\mathrm{m}_2 \mathrm{~F}}{\mathrm{~m}_1+\mathrm{m}_2}
\end{aligned}
$

Case-2

$
\begin{gathered}
\mathrm{f}>\mathrm{f}_1 \\
\text { or } \frac{\mathrm{m}_2 \mathrm{~F}}{\mathrm{~m}_1+\mathrm{m}_2}>\mu \mathrm{m}_2 \mathrm{~g}
\end{gathered}
$
 

If this is the case, then our assumption is wrong as static friction can not be greater than limiting friction.In this case the the friction force acting between the block will be kinetic fricton and both blocks will move with different accelerations. Let us assume the accelerations of blocks m₁ , and m₂ be a_1, and a₂ respectiveley.

Kinetic friction-

$
\mathrm{f}_{\mathrm{k}}=\mu \mathrm{N}=\mu \mathrm{m}_2 \mathrm{~g}
$

Along horizontal direction-

$
\begin{aligned}
& \mathrm{F}-\mathrm{f}_{\mathrm{k}}=\mathrm{m}_1 \mathrm{a}_1 \\
& \mathrm{a}_1=\frac{\mathrm{F}-\mu \mathrm{m}_{2 \mathrm{~g}}}{\mathrm{~m}_1} \\
& \mathrm{f}_{\mathrm{k}}=\mathrm{m}_2 \mathrm{a}_2 \\
& \mathrm{a}_2=\mu \mathrm{g}
\end{aligned}
$
 

• Case 2:- A force F is applied to the lower body,

Let's discuss possible 4 situations under this case:-

1. When there is no friction:-

B will move with acceleration (F/M) while A will remain at rest (relative to ground) as there is no pulling force on A.

$
a_B=\frac{F}{M} \text { and } a_A=0
$

As relative to $B$, A will move backward with acceleration (F/M) and so will fall from it in time $t$,

$
t=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 M L}{F}}
$

2. If friction is present between $A$ and $B$ only and
$\mathrm{F}^{\prime}($ Pseudo force on body A$)<\mathrm{F}_{\mathbf{1}}$ (limiting friction between body A and B ):-
Both the body will move together with common acceleration,
which is given by $a=\frac{F}{m+M}$
And the value of $\mathrm{F}^{\prime}$ will be

$
\begin{aligned}
& \quad F^{\prime}=m a=\frac{F}{m+M}=\frac{m F}{m+M} \\
& \text { and } F_l=\mu_s m g \\
& \text { as } F^{\prime}<F_l \Rightarrow \frac{m F}{m+M}<\mu_s m g \\
& \left.\Rightarrow F<\mu_s(m+M)\right) g
\end{aligned}
$

So both bodies will move together with acceleration,

$
\left.a_A=a_B=\frac{F}{m+M} \text { if } F<\mu_s(m+M)\right) g
$
 

3.     If friction is present between A and B only and F > F'_l(limiting friction between body A and B):-

               $\left.F_l^{\prime}=\mu_s(m+M)\right) g$

 Both the body will move with different acceleration. Here force of kinetic friction μkmg will oppose the motion of B while will cause the motion of A.

$\begin{aligned} & \quad m a_A=\mu_k m g \\ & \text { i.e } \quad a_A=u_k g\end{aligned}$

similarly

$
\begin{aligned}
F-F_k & =M a_B \\
\text { i.e. } \quad a_B & =\frac{\left[F-\mu_k m g\right]}{M}
\end{aligned}
$

As both, the bodies are moving in the same direction,
Acceleration of body A relative to B will be given by:-

$
a=a_A-a_B=-\left[\frac{F-\mu_k g(m+M)}{M}\right]
$

Negative sign implies that relative to B, A will move backward and will fall it after time,

$
t=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 M L}{F-\mu_k g(m+M)}}
$

4. If there is friction between $B$ and floor and $F>F_1$ " (limiting friction between body $B$ and surface):-

Here $F_l^{\prime \prime}=\mu_s(m+M) g$
The system will move only if $\mathrm{F}>\mathrm{F}_1$ then replacing F by $F-F_l^{\prime \prime}$
The entire situation (3) will be valid.
However if

$
F<F_l^{\prime \prime},
$
 

the system will not move.

Means friction between B and floor will be F while between A and B is zero.