Question

Figure shows a box of mass ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$on a frictionless plane inclined at angle $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}\mathbf{°}$. It is connected by a cord of negligible mass to a box of mass${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$on a horizontal frictionless surface. The pulley is frictionless and mass less.(a) If the magnitude of horizontal force$\overrightarrow{\mathbf{F}}$is 2.3 N, what is the tension in the connecting cord? (b) What is the largest value the magnitude of$\overrightarrow{\mathbf{F}}$may have without the cord becoming slack?

Short answer

(a) Tension in the connecting chord is 3.1N

(b) The largest value of the force F acting on$m_1$ without the cord becoming slack is 15N

Step-by-step solution

Given information

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$30°$

Determining the concept

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. 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’s second law is,

$F_{net}=\underset{}{\overset{}{\sum Ma}}$

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

(a) Determining the tension in the connecting chord

FBD for the system is,

Applying Newton’s second law,

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

$M_{2}g\sin \theta -T=M_{2}a$ (ii)

Add equations (i) and (ii),

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

Putting this value of a in equation (i),

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

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

(b) Determining the largest value of the force F acting on M1 without the cord becoming slack

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

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

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