Question

a block of mass M is being pulled with the help of a string of mass m and length L. The horizontal force applied by the man on the string is F.
Determine
$a$. Find the force exerted by the string on the block and acceleration of system.
$b$. Find the tension at the mid point of the string.
$c$. Find the tension at a distance x from the end at which force is applied.
Assume that the block is kept on a frictionless horizontal surface and the mass is uniformly distributed in the string.

Answer 1

$a$. Let the force applied by string to the block be T. For part (a), consider one system is block and other string. Let the acceleration of the system (block + string) be a.

Now we apply Newton's law to each of them.

For system II : $F-T=ma$......(i)

For system I : $T=Ma$......(ii)

AFter solving (i) and (ii), $a=\frac{F}{M+m}$, $T=\frac{MF}{M+m}$

$b$. Now we have to redefine our system. Choose system $1$ as block and half string and system $2$ as the other half string. On applying Newton's second law to system $1$ and system $2$, we have

System II : $F+T=\frac{m}{2}\times a$.......(iii)

Mass per unit length of string $=m/L$

Hence, mass of $L/2$ length of string $=\frac{m}{L}\times \frac{L}{2}=\frac{m}{2}$

System I : $T_1=(M+\frac{m}{2})a$......(iv)

After solving, (iii) and (iv), we get

$a=\frac{F}{M+m}$, $T_1=\frac{(M+m/2)F}{(M+m)}$

$c$. Now we can redefine our system in the block and string of length L-x (system I) and string of length x (system II)

System II : $F-T_2=(\frac{m}{L}\times x)a$........(v)

System I : $T_2=\{\frac{m}{L}(L-x)+M\}a$........(vi)

Solving, (v) and (vi), we get

$a=\frac{F}{M+m}$ and $T_2=\frac{\{m(L-x)/L+M\}F}{M+m}$

Here we see that acceleration in each part is same, but tension changes along the string.