Laws of Motion 3 Question 19
20. In the figure masses $m _1, m _2$ and $M$ are $20 kg, 5 kg$ and $50 kg$ respectively. The coefficient of friction between $M$ and ground is zero. The coefficient of friction between $m _1$ and $M$ and that between $m _2$ and ground is 0.3 . The pulleys and the strings are massless. The string is perfectly horizontal between $P _1$ and $m _1$ and also between $P _2$ and $m _2$.
The string is perfectly vertical between $P _1$ and $P _2$. An external horizontal force $F$ is applied to the mass $M$. (Take $g=10 m / s^{2}$.)
$(2000,10 M)$
(a) Draw a free body diagram of mass $M$, clearly showing all the forces.
(b) Let the magnitude of the force of friction between $m _1$ and $M$ be $f _1$ and that between $m _2$ and ground be $f _2$. For a particular force $F$ it is found that $f _1=2 f _2$. Find $f _1$ and $f _2$. Write equations of motion of all the masses. Find $F$, tension in the string and accelerations of the masses.
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Answer:
Correct Answer: 20. (b) $f _1=30 N, f _2=15 N, F=60 N, T=18 N, a=\frac{3}{5} m / s^{2}$
Solution:
- Given, $m _1=20 kg, m _2=5 kg, M=50 kg$,
$$ \mu=0.3 \text { and } g=10 m / s^{2} \text {. } $$
(a) Free body diagram of mass $M$ is given as
(b) The maximum value of $f _1$ is
$$ \left(f _1\right) _{\max }=(0.3)(20)(10)=60 N $$
The maximum value of $f _2$ is
$$ \left(f _2\right) _{\max }=(0.3)(5)(10)=15 N $$
Forces on $m _1$ and $m _2$ in horizontal direction are as follows
Now, there are only two possibilities :
(1) Either both $m _1$ and $m _2$ will remain stationary
(w.r.t. ground) or
(2) Both $m _1$ and $m _2$ will move (w.r.t. ground)
First case is possible when
| $T \leq\left(f _1\right) _{\max }$ | or | $T \leq 60 N$ | |
|---|---|---|---|
| and | $T \leq\left(f _2\right) _{\max }$ | or | $T \leq 15 N$ |
These conditions will be satisfied when $T \leq 15 N$
say $T=14 N$, then $f _1=f _2=14 N$
Therefore, the condition $f _1=2 f _2$ will not be satisfied.
Thus, $m _1$ and $m _2$ both cannot remain stationary.
In the second case, when $m _1$ and $m _2$ both move
$$ \begin{aligned} & f _2=\left(f _2\right) _{\max }=15 N \\ \therefore \quad & f _1=2 f _2=30 N \end{aligned} $$
Now, since $f _1<\left(f _1\right) _{\max }$, there is no relative motion between $m _1$ and $M$ i.e. all the masses move with same acceleration, say $a$
$$ \therefore \quad f _2=15 N \text { and } f _1=30 N $$
Free body diagrams and equations of motion are as follows
For $m _1, \quad 30-T=20 a$
For $m _2, \quad T-15=5 a$
For $M, \quad F-30=50 a$
Solving these three equations, we get
$$ F=60 N $$
$$ T=18 N \text { and } a=\frac{3}{5} m / s^{2} $$
NOTE
- Friction always opposes the relative motion between two surfaces in contact.
- Whenever there is relative motion between two surfaces in contact, always maximum friction (kinetic) acts, but if there is no relative motion then friction force ( $f$ ) may be less than its limiting value also. So, don’t apply maximum force.
