Question

In Fig. 5-59,block Aand 6.0 kgblock Bare connected by a string of negligible mass ${\overrightarrow{\mathbf{F}}}_{\mathbf{A}}\mathbf{=}\mathbf{\left(}\mathbf{12}\mathbf{N}\mathbf{\right)}\hat{\mathbf{i}}$. Force acts on block A; force ${\overrightarrow{\mathbf{F}}}_{\mathbf{B}}\mathbf{=}\mathbf{\left(}\mathbf{24}\mathbf{N}\mathbf{\right)}\hat{\mathbf{i}}$acts on block B. What is the tension in the string?

Short answer

The tension in the string is 2.4 N

Step-by-step solution

Given information

It is given that,

The masses of blocks A and B $M_A=4\mathrm{kg},M_B=6\mathrm{kg}$,

Forces acting on blocks,$F_A=12\mathrm{N},F_B=24\mathrm{N}$

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 diagram the tension in the string can be found.

Formula:

Newton’s second law is,

$F_{net}=\underset{}{\sum Ma\left(1\right)}$

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

Determining the tension in the string

FBD for block A:

FBD for block B:

From FBD,

$$\begin{gathered} T+F_A=M_{A}a\left(1\right) \\ {F}_B-T=M_{B}a\left(2\right) \end{gathered}$$

Adding equation (ii) and (iii),

$$\begin{gathered} {\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}=({\mathrm{M}}_{\mathrm{A}}+{\mathrm{M}}_{\mathrm{B}})\mathrm{a} \\ \mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}+{\mathrm{M}}_{\mathrm{B}}} \\ \mathrm{a}=\frac{36\mathrm{N}}{10\mathrm{kg}}\times \left(\frac{\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right) \\ =3.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From equation (ii),

$$\begin{gathered} \mathrm{T}={\mathrm{M}}_{\mathrm{A}}\mathrm{a}-{\mathrm{F}}_{\mathrm{A}} \\ \mathrm{T}=(4\mathrm{ckg})(3.6\mathrm{m}/{\mathrm{s}}^{2)-(12\mathrm{N})} \\ \mathrm{T}=2.4\mathrm{N} \end{gathered}$$

Hence, the tension in the string is 2.4 N