91Ó°ÊÓ

It is given that,

$$\begin{gathered} M=12kg \\ {F}_1=50N \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.Acceleration can be calculated by using Newton’s second law.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\underset{}{∑Ma}$

where, $F$ is the net force, $M$is mass and $a$ is an acceleration.

When$F_2=30Nand{F}_3=20N$

$a=\frac{Netforce}{mass}$

$a=\frac{F_1-F_2-F_3}{m}$

$a=\frac{50-30-20}{12}$

$a=0m/s^2$

Hence, the least acceleration for is$F_2=30Nand{F}_3=20N$is a$a=0m/s^2$

When $F_2=30Nand{F}_3=10N$,

$a=\frac{Netforce}{mass}$

$a=\frac{F_1-F_2-F3}{m}$

$a=\frac{50-30-10}{12}$

$a=0.83m/s^2$

Hence, the least acceleration for is$F_2=30Nand{F}_3=10N$is a$a=0.83m/s^2$

When, $F_2=F_3=30N$

Now, apply $F_{2}and{F}_3$in the opposite direction making an angle$\mathrm{θ}$with$x-axis$

For least acceleration:

$$\begin{gathered} \underset{}{∑}{F}_x=0 \\ 50-30×\cos \left(\mathrm{θ}\right)-30×\cos \left(\mathrm{θ}\right)=0 \\ 50=60\cos \left(\mathrm{θ}\right) \\ \mathrm{θ}=34° \end{gathered}$$

Forces $\stackrel{→}{F_2}$and $\stackrel{→}{F_3}$should be arranged with an angle $34°$with the negative horizontal axis to produce least acceleration. For $\mathrm{θ}=34°$least acceleration will be$a=0m/s^2$.