Second law

Question

Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed $\upsilon $. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)

Moderate

Solution

Let ${T}_{1}$ be tension in string connecting m and 2m and ${T}_{2}$ be tension in string connecting 2m and 3m.

Let a be common acceleration of the system.

$\therefore \text{\hspace{0.17em}\hspace{0.17em}}a=\frac{F-\left(m+2m+3m\right)g}{m+2m+3m}=\frac{F-6m}{6m}$

As the system moves with constant speed, therefore, a = 0

$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}F-6mg=0\text{\hspace{0.17em}\hspace{0.17em}}or\text{\hspace{0.17em}\hspace{0.17em}}F=6mg$

The free body diagram of block m is as shown in the figure.

The equation of motion of block of mass m is

$F-{T}_{1}-mg=0\phantom{\rule{0ex}{0ex}}6mg-{T}_{1}-mg=0\phantom{\rule{0ex}{0ex}}{T}_{1}=5mg\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}---\left(i\right)$

The free body diagram of block of mass 2m is as shown in the figure.

The equation of motion of block of mass 2m is

${T}_{1}-{T}_{2}-2mg=0\phantom{\rule{0ex}{0ex}}5mg-{T}_{2}-2mg=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(U\mathrm{sin}g\text{\hspace{0.17em}\hspace{0.17em}}\left(i\right)\right)\phantom{\rule{0ex}{0ex}}{T}_{2}=3mg$

The free body diagram of block of mass

3m is as shown in the figure.

The equation of motion of block of mass 3m is

${T}_{2}-3mg=0\phantom{\rule{0ex}{0ex}}{T}_{2}=3mg$

Net force on the block of mass 2m is

${F}_{net}={T}_{1}-{T}_{2}-2mg\phantom{\rule{0ex}{0ex}}=5mg-3mg-2mg=0$

Alternate solution

As all blocks are moving with constant speed, therefore, acceleration is zero. So net force on each block is zero.

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