91影视

It is given that,

The mass of the box on the frictionless horizontal surface is$M_1=3kg$

The mass of the box on the frictionless inclined surface is $M_2=1kg$

The force applied on the mass$M_1$ is 2.3 N

The angle made by $M_2$with the horizontal is$30掳$

The problem is based on Newton鈥檚 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. Thus, using the free body diagramthe tension in the connecting chord and the maximum force acting without the cord becoming slack can be found.

Formulae are as follow:

Newton鈥檚 second law is,

$F_{net}=\underset{}{\overset{}{鈭�Ma}}$

where,$F_{net}$ is the net force, Mis mass and a is an acceleration.

FBD for the system is,

Applying Newton鈥檚 second law,

$F+T=M_{1}a$ (i)

$M_{2}g\sin 胃-T=M_{2}a$ (ii)

Add equations (i) and (ii),

$$\begin{gathered} F+M_{2}g\sin 胃=\left(M_1+M_2\right)a \\ a=\frac{F+M_{2}g\sin 胃}{M_1+M_2} \\ a=\frac{2.3+\left(9.8\right)\sin 30掳}{4} \\ a=1.8m/s^2 \end{gathered}$$

$$\begin{gathered} T=3\left(1.8\right)-2.3 \\ T=3.1N \end{gathered}$$

Hence, the tension in the connecting chord is 3.1N.

If maximum force acts on the box, the tension in the string becomes zero, so equations 1 and 2 become,

$$\begin{gathered} a=g\sin 胃 \\ F=M_{1}a \\ F=M_{1}g\sin 胃 \\ F=\left(3\right)\left(9.8\right)\sin 30掳 \\ F=14.7~15N \end{gathered}$$

Hence, the largest value of the force F acting on$M_1$ without the cord becoming slack is 15 N